Physics: Unit-wise Subjective Question Bank

1. (a) The moment of inertia of a body about an axis is the sum \sum mr^{2} of the products of the mass of each particle and the square of its perpendicular distance from the axis. It is the rotational analogue of mass: it measures the body's resistance to a change in its rotation. Unit: \text{kg m}^{2}.

(b) For a uniform rod of mass M and length l, take an element of length dx at distance x from the end; its mass is dm = (M/l)\,dx.

(c) Kinetic energy and frictional couple:

At a steady speed the motor's power is used entirely against friction, P = \tau\omega:

2. (a) The radius of gyration k is the distance from the axis at which the whole mass of the body could be concentrated without changing its moment of inertia: I = Mk^{2}, so k = \sqrt{I/M}.

(b) Integrate from the centre outwards, x from -l/2 to +l/2:

(c) There is no external torque on the platform, so the angular momentum is conserved, I_1\omega_1 = I_2\omega_2. Since K = \tfrac12I\omega^{2} = \tfrac12L\omega and L is constant,

The kinetic energy rises 3.3 times; the extra energy is the work her muscles do pulling her arms in against the outward tendency of the mass.

3. (a) With no external torque on her, her angular momentum L = I\omega stays constant. Folding her arms brings the mass close to the axis, so I decreases; to keep L constant her angular velocity \omega must increase. Extending them again increases I and slows her.

(b) A particle of mass m at distance r from the axis, acted on by a tangential force F, has tangential acceleration a = r\alpha, so F = mr\alpha. Its torque about the axis is \tau = Fr = mr^{2}\alpha. For the whole rigid body, all particles share the same \alpha:

(c) The torque is \tau = Fr; a longer handle increases the perpendicular distance r, so a smaller force F gives the same torque.

4. (a) The Earth is subject only to the Sun's gravitational pull, which acts along the line joining them and so has zero torque about the Sun. Its orbital angular momentum L = mr^{2}\omega is therefore conserved. When r decreases, \omega must increase — equivalently the radius vector sweeps equal areas in equal times (Kepler's second law).

(b) The Earth's spin angular momentum L = I\omega is conserved, since no external torque acts. Shrinking decreases I (for a sphere I\propto R^{2}), so \omega increases and the day becomes shorter. For example, if R fell to half, I would fall to a quarter and the day would be 24/4 = 6\ \text{h}.

(c) The law of conservation of angular momentum: if the resultant external torque on a system is zero, its total angular momentum remains constant.

5. (a) For constant angular acceleration \alpha, starting with angular velocity \omega_0:

A point at distance r from the axis moves along an arc s = r\theta, so v = r\omega and the tangential acceleration is a = r\alpha.

(b) \omega_0 = 120\times2\pi/60 = 12.57\ \text{rad s}^{-1}, \omega = 0, t = 20\ \text{s}.

Number of revolutions = 125.7/2\pi = 20.

6. (a) A rigid body is made of particles m_1, m_2,\dots at distances r_1, r_2,\dots from the axis, all turning at the same \omega. Particle i moves with speed v_i = r_i\omega, so its kinetic energy is \tfrac12m_ir_i^{2}\omega^{2}. Summing over all particles,

(b) K = L^{2}/2I, so

1. (a) Moment of inertia I = \sum mr^{2} is the rotational analogue of mass. Radius of gyration k = \sqrt{I/M} is the distance from the axis at which the whole mass would give the same I.

(b) Let the rod have mass M and length l; a length dx at distance x from the centre has mass (M/l)dx.

The parallel-axis theorem states I = I_{cm}+Md^{2}, where d is the distance between the parallel axes. For the end of the rod d = l/2:

which agrees with the direct integration from one end.

(c) At release the weight Mg acts at the centre, distance l/2 from the pivot, so

The mass cancels: a heavier rod of the same length falls at the same rate.

2. (a) Consider particle i of mass m_i at distance r_i with tangential force F_i. Its tangential acceleration is r_i\alpha, so F_i = m_ir_i\alpha and its torque is F_ir_i = m_ir_i^{2}\alpha. Summing over all particles, since \alpha is common to all,

(b) A torque \tau turning a body through a small angle d\theta does work dW = F\,ds = F\,r\,d\theta = \tau\,d\theta. For a constant torque through angle \theta:

Dividing by the time taken:

(c) \alpha = \tau/I = 10/2.5 = 4.0\ \text{rad s}^{-2}.

The two agree, as they must when no energy is lost. The instantaneous power is

(the average power is W/t = 100\ \text{W}; the power grows linearly because \omega does).

3. (a) If the resultant external torque on a system is zero, its total angular momentum is constant.

(b) The putty adds a moment of inertia mr^{2} = 0.10\times(0.40)^{2} = 0.016\ \text{kg m}^{2}, so I_2 = 0.516\ \text{kg m}^{2}. No external torque acts on the disc-and-putty system:

The lost energy has gone to heat and sound in the inelastic sticking of the putty, exactly as in a linear perfectly inelastic collision: momentum (here angular momentum) is conserved, kinetic energy is not.

(c) In the air there is no external torque about the centre of mass, so L = I\omega is fixed. Tucking reduces I, so the diver rotates faster and completes the somersault; opening out before entry raises I and slows the rotation for a clean entry.

1. (a) Simple harmonic motion is the periodic motion of a particle whose acceleration is directly proportional to its displacement from a fixed point (the mean position) and is always directed towards that point: a = -\omega^{2}x.

(b) At equilibrium the spring is stretched by e so that ke = mg. If the mass is pulled a further distance y down, the net restoring force is

Newton's second law gives ma = -ky, so

This is the defining equation of SHM, and

(c) The time period is unchanged. The expression for T contains only m and k, not g. In the satellite the mass and the spring are both "weightless", but the spring still exerts -ky on the mass and its inertia is unchanged, so it oscillates with the same T. (A simple pendulum would not oscillate at all, since its restoring force is the weight.)

2. (a) Amplitude r is the maximum displacement from the mean position. With y = r\sin\omega t,

The acceleration is proportional to the displacement, opposite in direction, with maximum value r\omega^{2}.

(b) A spring of half the length is twice as stiff, because the same force now stretches each of only half as many coils: the force constant doubles (k' = 2k). The frequency f = \dfrac{1}{2\pi}\sqrt{k/m} therefore increases by the factor \sqrt{2}: f' = \sqrt{2}\,f.

3. (a) At displacement x the speed is v = \omega\sqrt{A^{2}-x^{2}}, so

The restoring force F = -kx does work -\int F\,dx in displacing the particle, stored as

which does not depend on x or t: the energy is constant.

(b) U = \tfrac12kx^{2} is a parabola with its minimum (U = 0) at x = 0, rising to \tfrac12kA^{2} at x = \pm A (see the figure below).

(c) T = 2\pi\sqrt{l/g} and g_{Moon}\approx g/6, so

The pendulum swings 2.45 times more slowly on the Moon.

4. (a) SHM is motion in which the acceleration is proportional to the displacement from the mean position and directed towards it.

(b) \omega = \sqrt{k/m} = \sqrt{18/0.15} = \sqrt{120} = 10.95\ \text{rad s}^{-1}, amplitude A = 0.04\ \text{m}.

(c) The total energy is

The kinetic energy \tfrac12k(A^{2}-x^{2}) is an inverted parabola, maximum at x = 0; the potential energy \tfrac12kx^{2} is an upright parabola. They add to a horizontal line at E (the figure below).

5. (a) Free oscillation: the system oscillates at its natural frequency with constant amplitude, with no resistive force and no external driving (an ideal case). Damped oscillation: a resistive force (air drag, friction) removes energy each cycle, so the amplitude decays with time; the frequency is slightly lower than the natural value. Forced oscillation: a periodic external force of frequency f_d drives the system; after a transient it oscillates at f_d, not at its natural frequency, with an amplitude that depends on how close f_d is to the natural frequency.

(b) Resonance occurs when the frequency of the driving force equals the natural frequency of the system; the amplitude of the forced oscillation is then a maximum (limited only by damping). Examples: a child's swing pushed in time with its motion; the shattering of a glass by a singer of the right pitch; radio tuning; bridges swaying under marching soldiers.

6. (a) The bob of mass m on a string of length l is displaced by a small angle \theta. The component of weight along the arc is the restoring force F = -mg\sin\theta\approx-mg\theta for small \theta (in radians). The displacement along the arc is x = l\theta, so

This has the form F = -kx with k = mg/l, so the motion is SHM and

(b) Squaring, T^{2} = \dfrac{4\pi^{2}}{g}\,l. A graph of T^{2} (vertical) against l (horizontal) is a straight line through the origin of slope 4\pi^{2}/g; hence g = 4\pi^{2}/\text{slope}. Using the whole line averages out reading errors.

1. (a) In SHM the acceleration is proportional to the displacement and opposite in direction. Here, with mass m,

which is the defining equation of SHM, so the motion is SHM with T = 2\pi\sqrt{m/A}.

(b) Let the displacement be x = A_0\sin\omega t. Then v = A_0\omega\cos\omega t, and

(c) The maximum speed is at x = 0:

(d) A building has its own natural frequency, and it is lower for a taller structure (a longer, more flexible "cantilever"). A low-frequency wave is therefore more likely to match the natural frequency of a tall building, producing resonance: the amplitude of the building's oscillation grows large and it may collapse. Short, stiff buildings have high natural frequencies and are driven far from resonance by such waves.

2. (a) See the derivation for the vertical spring in short answer 1(b): F = -ky, so

The weight only changes the equilibrium position; the oscillation is about the new equilibrium and T does not contain g.

(b) K+U = \tfrac12kA^{2} as in short answer 3(a); the figure shows the energy graphs: U and K are parabolas that exchange energy twice per cycle, their sum constant.

(c) \omega = 2\pi/T = 2\pi/0.63 = 9.97\ \text{rad s}^{-1}.

3. (a) Angular SHM is oscillation in which the restoring torque is proportional to the angular displacement, \tau = -C\theta, where C is the torsional constant. By \tau = I\alpha:

(b) For a bob of mass m on a light string of length l displaced by a small angle \theta, the torque of the weight about the pivot is \tau = -mgl\sin\theta\approx-mgl\theta, and the moment of inertia is I = ml^{2}. Then

This is the torsional result with C = mgl.

(c) l = \dfrac{gT^{2}}{4\pi^{2}} = \dfrac{9.8\times4.0}{39.5} = 0.99\ \text{m}. On the Moon,

1. (a) Surface tension is the force acting per unit length on an imaginary line drawn in the surface of a liquid, perpendicular to the line and tangential to the surface: T = F/l, unit \text{N m}^{-1}.

(b) Let a wire frame with a sliding bar of length l hold a soap film of one surface. The surface pulls the bar with force F = Tl. If the bar is moved a distance dx, the work done against surface tension is

where \Delta A = l\,dx is the increase in surface area. This work is stored as surface energy, so the surface energy per unit area equals the surface tension: E/A = T, in \text{J m}^{-2} which is identical to \text{N m}^{-1}.

(c) The terminal velocity is v_t = \dfrac{2r^{2}(\rho-\sigma)g}{9\eta}\propto r^{2}. Volume is conserved when two drops of radius r coalesce: \tfrac43\pi R^{3} = 2\times\tfrac43\pi r^{3}, so R = 2^{1/3}r. Then

For N drops v_2 = N^{2/3}v_1; with N = 2 and v_1 = 0.4\ \text{m s}^{-1} this gives v_2 = 0.63\ \text{m s}^{-1}.

2. (a) Stokes' law: the viscous drag on a small sphere of radius r moving with speed v through a fluid of viscosity \eta is F = 6\pi\eta rv. The coefficient of viscosity is the tangential force per unit area needed to maintain unit velocity gradient between layers: F = \eta A\,dv/dx; unit \text{Pa s}.

(b) Drop a small steel ball of known radius and density into a tall jar of the liquid; time its fall between two marks once it reaches terminal velocity, where weight = buoyancy + drag. Then

(c) Continuity: A_1v_1 = A_2v_2, so v_2 = 25\times3/15 = 5\ \text{m s}^{-1}. For water, \rho = 1000\ \text{kg m}^{-3} and Bernoulli's equation for a horizontal pipe gives

The pressure falls by 8.0\ \text{kPa} in the narrow section.

3. (a) In laminar flow every particle follows the same smooth path as the ones ahead of it, layers slide over each other without mixing, and the velocity at any point is constant; it occurs at low speeds (Reynolds number below about 2000). In turbulent flow the motion is irregular, with eddies and mixing, and the velocity at a point changes randomly; it occurs at high speed or in narrow constrictions.

(b) v_2 = 200\times2/50 = 8\ \text{m s}^{-1}. With \rho = 1000\ \text{kg m}^{-3}:

4. (a) Bernoulli's principle: for the steady, non-viscous, incompressible flow of a fluid, the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume is constant along a streamline: where the speed is high, the pressure is low.

(b) A_A = 1.0\ \text{cm}^{2} = 100\ \text{mm}^{2} and A_B = 20\ \text{mm}^{2}, so

5. (a) Consider a steady flow through a pipe: at the lower end (height h_1, area A_1, speed v_1, pressure P_1) and the upper end (h_2, A_2, v_2, P_2). In a short time \Delta t a volume \Delta V = A_1v_1\Delta t = A_2v_2\Delta t enters and leaves. Work done on the fluid by the pressures: (P_1-P_2)\Delta V. Change in energy: kinetic \tfrac12\rho\Delta V(v_2^{2}-v_1^{2}) plus potential \rho\Delta V g(h_2-h_1). By the work–energy theorem,

(b) The water leaves through a smaller area. By the equation of continuity its speed rises, and the range of a horizontal jet is proportional to its exit speed, so it goes further. (The flow rate hardly changes; the same volume per second is squeezed through a smaller opening.)

(c) For the steady flow of an incompressible fluid, the volume passing per second is the same at every cross-section: A_1v_1 = A_2v_2.

6. (a) The angle of contact is the angle between the tangent to the liquid surface at the point of contact and the solid surface, measured inside the liquid. For water on clean glass the adhesive force between water and glass exceeds the cohesive force between water molecules, the surface curves upward and \theta<90^{\circ}, so water is pulled up the tube. For mercury the cohesion is stronger, \theta\approx140^{\circ}>90^{\circ}, the meniscus is convex and the mercury is depressed.

(b)

The water rises 2.9\ \text{cm}.

7. (a) The volume of liquid flowing per second through a horizontal capillary tube is

where P is the pressure difference across the ends, r the radius, l the length and \eta the coefficient of viscosity.

(b) V/t\propto r^{4}. With the radius reduced by 10\%, r' = 0.9r:

The flow falls by 34\% — which is why narrowed arteries force the heart to raise the pressure sharply.

8. (a) A liquid column of height h and cross-section A has weight \rho Ahg, and this is carried by the pressure difference acting on the area A:

(b) Archimedes' principle: a body wholly or partly immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced. A floating iceberg has weight = upthrust, so the submerged fraction is 917/1025 = 0.895; the fraction above the surface is 1-0.895 = 0.105, about 10\%.

1. (a) The viscous drag on a small sphere of radius r moving at speed v in a fluid of viscosity \eta is F = 6\pi\eta rv.

(b) A sphere of density \rho falls in a fluid of density \sigma. Three forces act: the weight downward, W = \tfrac43\pi r^{3}\rho g; the upthrust upward, U = \tfrac43\pi r^{3}\sigma g; and the viscous drag upward, F = 6\pi\eta rv, which grows as v grows. The sphere accelerates until the net force is zero; that speed is the terminal velocity v_t:

The speed is proportional to r^{2} and inversely proportional to \eta.

(c)

(d) As a drop speeds up the drag 6\pi\eta rv increases until it (plus upthrust) balances the weight; the net force then vanishes and the drop falls at constant terminal velocity. (For real raindrops, the drag law itself changes at larger speeds, but the conclusion is the same.)

2. (a) See short answer 5(a) for the derivation from the work–energy theorem,

Assumptions: the fluid is incompressible (constant density), non-viscous (no internal friction), and its flow is steady and streamline (laminar), and the equation is applied along a single streamline.

(b) A Venturi meter is a horizontal tube with a narrow throat, with vertical manometer tubes at the wide part (area A_1, pressure P_1) and throat (A_2, P_2). By continuity v_2 = v_1A_1/A_2. By Bernoulli (same level),

The measured height difference h in the manometers gives P_1-P_2 = \rho_{l}gh.

(c) The wing's shape and tilt make air travel faster over the top than below. By Bernoulli's principle the pressure above is lower, so the net upward pressure force on the wing is the lift.

3. (a) Water of surface tension T rises in a tube of radius r with contact angle \theta. The surface tension acts along the line of contact of length 2\pi r, at angle \theta to the wall; the vertical component of the pull is

This lifts a column of height h, whose weight is \pi r^{2}h\rho g. At equilibrium,

h\propto1/r: narrower tubes give higher rises. For \theta>90^{\circ} (mercury) h is negative — a depression.

(b) Applications: (i) the rise of oil in the wick of a lamp, or of water in a blotting paper or a towel; (ii) the rise of sap and water from roots in fine vessels of plants; (iii) the supply of water to soil by fine pores.

(c) Volume is conserved: 8\times\tfrac43\pi r^{3} = \tfrac43\pi R^{3}, so R = 2r = 2.0\ \text{mm}.

The surface energy decreases by 3.6\ \mu\text{J}; the released energy appears as a small rise in temperature.

1. (a) The first law: the heat supplied to a system equals the increase in its internal energy plus the work done by the system: Q = \Delta U + W.

(b) At constant volume, W = 0 so all heat raises the internal energy: Q_V = nC_V\Delta T = \Delta U. Since U of an ideal gas depends on T alone, \Delta U = nC_V\Delta T in every process. At constant pressure,

From PV = nRT at constant P, P\Delta V = nR\Delta T. Therefore

(Mayer's relation). The extra heat at constant pressure pays for the expansion work.

(c) The efficiency of a heat engine is the fraction of the heat absorbed from the source that is converted to useful work:

2. (a) An indicator diagram is a graph of pressure against volume (p–V diagram) for a system or an engine cycle.

(b) For a small expansion dV at pressure P against a piston of area A, dW = F\,dx = PA\,dx = P\,dV. Between V_1 and V_2,

which is the area under the curve. For a closed cycle the net work is the area enclosed.

(c)

3. (a) Energy can be neither created nor destroyed. Heat Q is energy supplied to the system; it either raises the internal (molecular) energy, \Delta U, or leaves the system as work, W done on the surroundings. Nothing is lost, so the accounts balance: Q = \Delta U+W.

(b)

4. (a) For n moles at constant temperature T, P = nRT/V:

Since \Delta U = 0, this equals the heat absorbed.

(b)

5. (a) In an adiabatic change dQ = 0, so dU+P\,dV = 0, i.e. nC_V\,dT = -P\,dV. Differentiating PV = nRT: P\,dV+V\,dP = nR\,dT. Eliminate dT:

Integrating, \ln P+\gamma\ln V = \text{constant}, so

(b) Using TV^{\gamma-1} constant with \gamma = \tfrac53:

6. For a monatomic gas C_V = \tfrac32R and C_P = \tfrac52R.

7. (a) A thermodynamic system is the part of the universe (a quantity of gas in a cylinder, say) chosen for study; everything else is the surroundings. An open system exchanges both matter and energy with its surroundings (a boiling kettle); a closed system exchanges energy only (gas in a sealed cylinder with a piston); an isolated system exchanges neither (an ideal thermos).

(b) The internal energy depends only on the present state (P, V, T) of the system, not on how it arrived there, so \Delta U between two states is fixed. Heat and work depend on the path taken between the states (for example, the area under different p–V curves joining the same two points differs), so they are not properties of a state.

1. (a) The molar heat capacity at constant volume, C_V, is the heat needed to raise the temperature of one mole by one kelvin with the volume held fixed; C_P is the corresponding heat at constant pressure. Units: \text{J mol}^{-1}\text{K}^{-1}.

(b) See short answer 1(b): C_P-C_V = R. The extra heat at constant pressure does the expansion work P\Delta V = R\Delta T per mole.

(c) Dividing by C_V: \gamma = C_P/C_V = 1+R/C_V. For a diatomic gas with five degrees of freedom (translation and rotation), C_V = \tfrac52R = 20.8\ \text{J mol}^{-1}\text{K}^{-1}, C_P = \tfrac72R = 29.1\ \text{J mol}^{-1}\text{K}^{-1} and \gamma = 7/5 = 1.4.

2. (a) Isothermal: the temperature is fixed (slow change, good thermal contact with a reservoir); PV = \text{constant} and \Delta U = 0. Adiabatic: no heat flows (rapid change or perfect insulation); PV^{\gamma} = \text{constant} with \gamma>1, so the adiabatic curve is steeper than the isotherm through the same point (see the figure below).

(b) W = \int_{V_1}^{V_2}P\,dV with P = KV^{-\gamma}, where K = P_1V_1^{\gamma} = P_2V_2^{\gamma}:

(c) T_2 = 189\ \text{K} (short answer 5(b)), \gamma-1 = \tfrac23:

Check: W = nC_V(T_1-T_2) = 12.47\times111 = 1.38\times10^{3}\ \text{J}.

3. (a)

(b) C lies on the isotherm through A, so P_CV_C = P_AV_A with V_C = V_B = 6\ \text{L}:

In A\to B, W_{AB} = P\Delta V = 1.0\times10^{5}\times(6-2)\times10^{-3} = 400\ \text{J}. In B\to C the volume is constant, so W_{BC} = 0. Total W = 400\ \text{J}. Because A and C are at the same temperature, \Delta U = 0 over the whole path, so

absorbed. (If the gas is monatomic, A\to B absorbs Q_{AB} = \tfrac52P\Delta V = 1000\ \text{J} and B\to C rejects Q_{BC} = \Delta U_{BC} = -600\ \text{J}; the net heat absorbed is 1000-600 = 400\ \text{J}, as found.)

1. (a) Kelvin–Planck statement: it is impossible to construct an engine, working in a cycle, whose sole effect is to take heat from a single reservoir and convert it completely into work. Clausius statement: heat cannot flow from a colder body to a hotter body by itself, with no other change.

(b) No. A refrigerator does take heat from a cold body and give more heat to a hot one, but only because work is done on it by an external agent (the compressor motor). The Clausius statement forbids transfer by itself; here the transfer is not the sole effect, since electrical work is consumed and also appears as heat in the room.

(c)

(the last form only for a reversible engine, since Q_2/Q_1 = T_2/T_1). The efficiency is zero when T_2 = T_1 (equivalently Q_2 = Q_1): with no temperature difference no work can be extracted.

2. (a) T_2 = 9+273 = 282\ \text{K}.

For \eta = 0.50:

(b) The first law, Q = \Delta U+W, says that the heat supplied to a system is exactly accounted for by the gain in its internal energy and the work it does: energy changes form but the total is conserved. In an engine cycle \Delta U = 0, so W = Q_1-Q_2: the work output is the net heat input.

3. (a) The Otto cycle (figure below) models a petrol engine: 1\to2 adiabatic compression of the fuel–air mixture from V_1 to V_2; 2\to3 heating at constant volume when the spark ignites the mixture (heat Q_1 absorbed); 3\to4 adiabatic expansion (the power stroke, which delivers the work); 4\to1 cooling at constant volume when the exhaust valve opens (heat Q_2 rejected). The net work is the enclosed area.

(b) For the Otto cycle, with compression ratio r = V_1/V_2,

For the diesel cycle the compression ratio is much higher (15 to 22, against 6 to 10 for petrol) because only air is compressed, so there is no pre-ignition. A higher r gives a larger \eta: with \gamma = 1.4, r = 8 gives 56\% while r = 16 gives 67\% for the ideal Otto formula. The diesel cycle's efficiency also depends on the cut-off ratio \rho at which fuel injection stops: \eta_{diesel} = 1-\dfrac{1}{r^{\gamma-1}}\,\dfrac{\rho^{\gamma}-1}{\gamma(\rho-1)}. At the same compression ratio the Otto cycle is the more efficient; but a real diesel engine runs at a far higher r, so it is more efficient in practice.

4. (a) Working substance: an ideal gas in a cylinder, between a source at T_1 and a sink at T_2. (i) Isothermal expansion at T_1: heat Q_1 absorbed from the source, work done by the gas. (ii) Adiabatic expansion: the gas cools from T_1 to T_2 with no heat exchange. (iii) Isothermal compression at T_2: heat Q_2 rejected to the sink. (iv) Adiabatic compression: the gas warms from T_2 back to T_1, completing the cycle.

(b) \eta = 1-T_2/T_1, with temperatures in kelvin. Carnot's theorem: no engine working between two given temperatures can be more efficient than a reversible engine between the same temperatures — otherwise it could be coupled with a reversed Carnot engine to transfer heat from cold to hot with no net work, violating the Clausius statement. \eta = 1 would need T_2 = 0\ \text{K}, which cannot be reached.

5. (a) The coefficient of performance of a refrigerator is the ratio of the heat removed from the cold space to the work supplied:

For a reversible refrigerator Q_1/Q_2 = T_1/T_2, so

(b) T_2 = 273\ \text{K}, T_1 = 300\ \text{K}:

6. (a) Kelvin–Planck and Clausius statements are given in short answer 1(a). Suppose the Kelvin–Planck statement were false, so that an engine X converts heat Q from a cold reservoir entirely into work. Use that work to drive a refrigerator that takes Q_c from the same cold reservoir and gives Q_c+Q to a hot one: the combined result is a transfer of heat from cold to hot with no outside work, violating Clausius. So a violation of one is a violation of the other: they are two forms of the same law.

(b) Entropy S is a state function that measures the disorder of a system; for heat \Delta Q absorbed reversibly at temperature T, \Delta S = \Delta Q/T. The second law in terms of entropy: in any spontaneous (irreversible) process in an isolated system the entropy increases; it stays constant only in a reversible one. This gives time a direction — a hot cup cools, a gas expands into a vacuum, and neither reverses by itself.

7. T_1 = 500\ \text{K}, T_2 = 300\ \text{K}, Q_1 = 2000\ \text{J}.

Check: Q_2/Q_1 = 1200/2000 = 0.6 = T_2/T_1.

1. (a) The cycle (figure below) consists of four reversible steps on an ideal gas: A\to B isothermal expansion at T_1; B\to C adiabatic expansion to T_2; C\to D isothermal compression at T_2; D\to A adiabatic compression to T_1.

(b) For the isothermal expansion A\to B (volumes V_A\to V_B): heat absorbed

For the isothermal compression C\to D: heat rejected

For the adiabats B\to C and D\to A, using TV^{\gamma-1} constant:

Dividing these gives V_B/V_A = V_C/V_D, so the logarithms are equal and

(c) T_1 = 373\ \text{K}, T_2 = 273\ \text{K}: \eta = 1-273/373 = 0.268, about 27\%.

2. (a) Otto cycle (petrol): two adiabats and two isochores, as explained in section 3.5 short answer 3(a). Diesel cycle: 1\to2 adiabatic compression of air alone to a high ratio; 2\to3 fuel injected and burns at constant pressure (heat Q_1 in, gas expands to V_3); 3\to4 adiabatic expansion; 4\to1 cooling at constant volume (Q_2 out). The diesel engine ignites the fuel by the heat of compression, so there is no spark plug.

(b) In the Otto cycle, heat is absorbed at constant volume 2\to3 and rejected at constant volume 4\to1:

For the adiabats, T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1} and T_4V_1^{\gamma-1} = T_3V_2^{\gamma-1}, so T_2 = T_1r^{\gamma-1} and T_3 = T_4r^{\gamma-1} with r = V_1/V_2. Then T_3-T_2 = r^{\gamma-1}(T_4-T_1) and

(c) \eta = 1-8^{-0.4} = 1-0.435 = 0.565: 56\% (an ideal upper limit; real engines reach about 25–30\%).

3. (a) A refrigerator is a heat engine run backwards. A compressor (driven by a motor, work W) compresses a refrigerant vapour, which gives up heat Q_1 in the condenser outside; the refrigerant then expands through a valve, cools and absorbs heat Q_2 in the evaporator inside the compartment; the cycle repeats. By energy conservation Q_1 = Q_2+W.

(b) T_2 = 270\ \text{K}, T_1 = 300\ \text{K}:

(c) The Carnot (ideal) efficiency is 1-300/600 = 0.50, so the engine's efficiency is \eta = 0.90\times0.50 = 0.45.

1. (a) A progressive wave is a disturbance that travels through a medium, transferring energy from one place to another without transferring the medium itself. A stationary (standing) wave is formed by the superposition of two identical progressive waves travelling in opposite directions; the pattern stays in place and no energy is transferred along the medium.

(b) A wave travelling in the +x direction:

Here y is the displacement of the particle at position x at time t, A the amplitude, \omega = 2\pi f the angular frequency (rad \text{s}^{-1}), and k = 2\pi/\lambda the angular wave number (rad \text{m}^{-1}). The wave speed is v = f\lambda = \omega/k. For a wave travelling in the -x direction, use \sin(\omega t+kx).

2. (a) A wave travelling along a string is reflected at the fixed end with a phase change of \pi. The incident and reflected waves, of the same frequency and amplitude, travel in opposite directions and superpose. At some points they always cancel (nodes); at others they always reinforce (antinodes). The pattern does not move, hence "stationary".

(b) Let the incident wave be y_1 = A\sin(\omega t-kx) and the reflected wave y_2 = A\sin(\omega t+kx). The resultant is

The amplitude at x is 2A\cos kx. Nodes are where \cos kx = 0:

Consecutive nodes (n and n+1) are at

(Antinodes, where |\cos kx| = 1, are also \lambda/2 apart and lie midway between nodes, \lambda/4 from each.)

3. Compare with y = A\sin(\omega t-kx): A = 0.05\ \text{m}, \omega = 40\ \text{rad s}^{-1}, k = 2\ \text{rad m}^{-1}.

4.

5. (a) Superposition: when two or more waves overlap, the resultant displacement at any point is the algebraic sum of the displacements that each wave would produce alone.

(b) As the pulses overlap, the string at the centre has height 2+3 = 5\ \text{cm} at the instant they coincide. They then pass through each other and emerge with their original shapes and sizes (waves do not collide like particles).

(c) The displacements +3 and -3\ \text{cm} cancel: at the instant of complete overlap the string is flat everywhere there, and the energy is momentarily all kinetic; the pulses then continue unchanged.

1. (a) Let the incident wave be y_1 = A\sin(\omega t-kx). At the fixed end the reflected wave is y_2 = -A\sin(\omega t+kx) for a reflection with a phase change of \pi (taking the end at x = 0). Adding,

Using \sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}:

The amplitude 2A\sin kx depends on position, and every particle vibrates in SHM of frequency f with its own amplitude.

(b) Nodes have zero amplitude, \sin kx = 0:

Antinodes have maximum amplitude, |\sin kx| = 1:

(c) Consecutive nodes: x_{n+1}-x_n = \dfrac{(n+1)\lambda}{2}-\dfrac{n\lambda}{2} = \dfrac{\lambda}{2}. For a string fixed at both ends of length L, both ends are nodes, which quantizes L = n\lambda/2 and leads to the harmonics f_n = nv/2L (section 3.8).

2. (a) A source at x = 0 vibrates as y = A\sin\omega t. The disturbance travels at speed v, so a point at distance x copies the source motion a time x/v later:

(b) The phase at x is \omega t-kx. For two points \Delta x = x_2-x_1 apart at the same time,

One wavelength corresponds to 2\pi.

(c) A = 4.0\ \text{cm}, \omega = 50\ \text{rad s}^{-1}, k = 0.25\ \text{cm}^{-1}.

1. (a) Newton: sound passes through air so slowly that the compressions and rarefactions are isothermal, and the speed is

At 0\ ^{\circ}\text{C}, P = 1.01\times10^{5}\ \text{Pa} and \rho = 1.29\ \text{kg m}^{-3} give 280\ \text{m s}^{-1}, whereas the measured value is about 332\ \text{m s}^{-1}: a discrepancy of about 16\%.

(b) Laplace: the compressions and rarefactions in a sound wave occur so rapidly that heat has no time to flow between layers, and the changes are adiabatic, so the bulk modulus is \gamma P, not P:

With \gamma = 1.4 the speed is \sqrt{1.4}\times280 = 331\ \text{m s}^{-1}, agreeing with experiment.

(c) Sound in a gas is a series of compressions (pressure above the mean) and rarefactions (pressure below the mean) travelling outwards; the pressure at each point varies about the atmospheric value, and it is this pressure variation that the ear detects.

2. (a) Laplace's correction: the changes of pressure and volume in a sound wave are adiabatic, not isothermal, so in Newton's formula P must be replaced by \gamma P (because the adiabatic bulk modulus of an ideal gas is \gamma P):

(b) The speed varies with temperature as v_t\approx v_0+0.61\,t (t in ^{\circ}\text{C}), so

(Using v\propto\sqrt{T}: 330\sqrt{298/273} = 345\ \text{m s}^{-1}, the same.)

3. (a) For an ideal gas P/\rho = RT/M, so v = \sqrt{\gamma RT/M}\propto\sqrt{T}: the speed rises with temperature, about 0.61\ \text{m s}^{-1} per ^{\circ}\text{C} near room temperature. At constant temperature a change of pressure has no effect, because \rho\propto P and the ratio P/\rho stays constant.

(b) Humidity: moist air is less dense than dry air at the same P and T (water vapour, molar mass 18, replaces \text{N}_2 and \text{O}_2), so sound travels slightly faster in damp air.

(c)

4. (a) Solid: v = \sqrt{Y/\rho}, where Y is Young's modulus. Liquid: v = \sqrt{K/\rho}, where K is the bulk modulus. (A solid also carries transverse waves at \sqrt{\eta/\rho}, with \eta the rigidity modulus.)

(b)

5. From Laplace's formula with P = \rho RT/M for an ideal gas (M the molar mass),

1. (a) Let a piston in a long tube of fluid (cross-section A, density \rho, bulk modulus B) suddenly move with speed u for a time \Delta t. A compression pulse travels ahead at speed v, so in \Delta t it covers v\Delta t of fluid, which has mass \rho Av\Delta t and is brought from rest to speed u. The piston moves a distance u\Delta t, so the volume of the compressed part shrinks by Au\Delta t from Av\Delta t: the fractional change is \Delta V/V = -u/v. The excess pressure needed is, by the definition of the bulk modulus,

Newton's second law (impulse = change of momentum) applied to the pulse:

Equating the two expressions for \Delta P:

(b) For a gas held at constant temperature (isothermal) B = P, giving Newton's formula v = \sqrt{P/\rho}. Laplace showed that the changes are adiabatic, for which B = \gamma P, giving v = \sqrt{\gamma P/\rho}.

(c)

2. (a) (i) Temperature: v\propto\sqrt{T}, so heating air makes sound faster, by about 0.61\ \text{m s}^{-1} per degree. (ii) Pressure: no effect if T is constant, because P/\rho is constant. (iii) Humidity: moist air is less dense, so the speed is higher.

(b) v_{20} = 332+0.61\times20 = 344\ \text{m s}^{-1}. The echo covers the distance to the wall and back in 0.50\ \text{s}:

(c) The pressure and displacement amplitudes are related by p_{max} = Bk\,s_{max}, with k = 2\pi f/v:

1. (a) A sonometer wire has f\propto1/L. Let the fork's frequency be f; at resonance the 75\ \text{cm} wire also vibrates at f. The shortened wire (74.5\ \text{cm}) vibrates faster:

(b) With 3 loops, L = 3\lambda/2, so \lambda = 2L/3 = 0.80\ \text{m} and

2. (a) The lowest frequency of a vibrating system is the fundamental (first harmonic). Higher frequencies that are integer multiples of it are harmonics (second, third, …). An overtone is any mode above the fundamental: the first overtone is the second harmonic (for an open pipe or a string), the second overtone is the third harmonic, and so on.

(b)

The first overtone is the second harmonic, f_2 = 2f_1 = 158\ \text{Hz}.

(c) f\propto\sqrt{T}, so quadrupling the tension doubles every frequency: the first overtone becomes 2\times158 = 316\ \text{Hz}.

3. (a) An open pipe has antinodes at both ends and L = \lambda/2 for the fundamental:

(b) In music a harmonic is a component of a note whose frequency is an integer multiple of the fundamental; the mix of harmonics determines the quality (timbre) of the note.

(c) By the Doppler effect: while the bus approaches, each successive sound crest is emitted closer to you and arrives more often, so the frequency is higher; after it passes, successive crests start farther away, so the frequency is lower. The horn's actual frequency does not change.

4. (a) At an open end the antinode is not exactly at the mouth of the pipe but a little outside it, at a distance e\approx0.6\,r (for a pipe of radius r); this extra length is the end correction. The effective length is L+e.

(b) For a closed pipe, the first resonance is at L_1+e = \lambda/4 and the second at L_2+e = 3\lambda/4. Subtracting,

(Using the first resonance, e = \lambda/4-L_1 = 17.25-16.0 = 1.25\ \text{cm}.)

5. (a) For a string of length L, tension T and mass per unit length \mu vibrating in a given mode: Law of length — f\propto1/L (at fixed T and \mu). Law of tension — f\propto\sqrt{T} (at fixed L and \mu). Law of mass — f\propto1/\sqrt{\mu} (at fixed L and T). These combine in f = \dfrac{1}{2L}\sqrt{T/\mu}.

(b)

6.

1. (a) Open pipe: both ends are antinodes of displacement (pressure nodes). The three simplest modes have L = \lambda/2,\ \lambda,\ 3\lambda/2 (figure below, left). Closed pipe: the closed end is a node, the open end an antinode; the modes have L = \lambda/4,\ 3\lambda/4,\ 5\lambda/4 (figure below, right).

(b) Open pipe: L = n\lambda_n/2, so \lambda_n = 2L/n and

Closed pipe: L = (2n-1)\lambda_n/4, so \lambda_n = 4L/(2n-1) and

(c) Open pipe: L = \dfrac{v}{2f_1} = \dfrac{340}{2\times264} = 0.64\ \text{m}; first overtone = 2f_1 = 528\ \text{Hz}. Closed pipe: L = \dfrac{v}{4f_1} = \dfrac{340}{4\times264} = 0.32\ \text{m}; first overtone = 3f_1 = 792\ \text{Hz}.

2. (a) A transverse wave on a string travels at v = \sqrt{T/\mu}. For a string fixed at both ends of length L, both ends are nodes, so the stationary wave must fit an integral number of half-wavelengths: L = n\lambda_n/2, \lambda_n = 2L/n. Then

(b) Stress = Young's modulus \times strain:

With T = \sigma A and \mu = \rho A, the area cancels in v = \sqrt{T/\mu} = \sqrt{\sigma/\rho}:

The second mode (n = 2) has f_2 = \dfrac{2v}{2L} = \dfrac{v}{L} = \dfrac{506}{1.5} = 338\ \text{Hz}.

(c) \mu = \rho\pi r^{2} increases four times when r doubles, so v = \sqrt{T/\mu} falls to half.

3. (a) The laws are stated in short answer 5(a). Sonometer experiment: a wire is stretched over two bridges on a wooden box and kept under a fixed tension by a hanging load. A tuning fork of known frequency is sounded and a light paper rider is placed on the wire; the movable bridge is adjusted until the rider is thrown off, which shows resonance, and the vibrating length L is measured. Repeating with forks of several frequencies, one finds f\times L is constant, i.e. f\propto1/L.

(b) \mu = \rho\pi r^{2}, so for equal T and L, f\propto1/\sqrt{\mu}\propto1/r. With r_1:r_2 = 1:2, f_1:f_2 = 2:1: the thinner wire vibrates twice as fast.

(c) \mu = \dfrac{4.8\times10^{-3}}{0.60} = 8.0\times10^{-3}\ \text{kg m}^{-1}.

1. The passenger (observer) moves at u_o = 18\ \text{m s}^{-1} and the source at u_s = 30\ \text{m s}^{-1}. (i) Approaching:

(ii) Receding:

(b) The intensity varies as 1/r^{2}, so at four times the distance it is 1/16 of its value.

2. For a source moving at speed u past a stationary listener, the frequency on approach is f_1 = \dfrac{fv}{v-u} and on recession f_2 = \dfrac{fv}{v+u}.

With f = 400, v = 340: 272000\,u = 60\,(115600-u^{2}), which gives u^{2}+4533\,u-115600 = 0, so

As a check, since u\ll v, \Delta f\approx2fu/v gives u = 60\times340/800 = 25.5\ \text{m s}^{-1}. The car moves at about 25\ \text{m s}^{-1}, i.e. 90\ \text{km h}^{-1}.

3. The cliff first acts as a stationary observer receiving f_1 from the moving source, and then as a stationary source emitting f_1 to the moving driver:

It is higher because the train approaches the cliff twice over, in effect: the crests are compressed at the source (shorter wavelength) and the driver moves towards the returning crests, so both effects raise the frequency.

4. The relation p_{max} = Bk\,s_{max} with k = 2\pi f/v:

The displacement is tiny because a pressure variation of only 3\times10^{-2}\ \text{Pa} (out of 10^{5}\ \text{Pa}) corresponds to a fractional volume change of about 2\times10^{-7}, so each particle moves only about 10^{-8}\ \text{m}; yet the ear detects it.

5.

6. (a) The intensity level is \beta = 10\log_{10}(I/I_0) decibels, where I_0 = 10^{-12}\ \text{W m}^{-2} is the threshold of hearing.

(b)

(c) \Delta\beta = 10\log_{10}2 = 3.0\ \text{dB}, independent of the starting level.

1. (a) The Doppler effect is the apparent change in the frequency of a wave heard by an observer when there is relative motion between the source and the observer.

(b) Let the speed of sound be v, the source frequency f and its wavelength \lambda = v/f. Source moving at u_s towards a stationary observer. In one period T = 1/f the source moves u_sT, so the wavelength in front is shortened:

Observer moving at u_o towards a stationary source. The wavelength is unchanged, but the speed of the waves relative to the observer is v+u_o:

Combined (positive direction from observer to source):

Change the sign of u_s or u_o for motion away.

(c) The source recedes (+u_s in the denominator) while the observer approaches (+u_o in the numerator):

2. (a) See short answer 5: intensity (objective), loudness (subjective, depends on intensity and frequency), pitch (frequency) and quality (harmonic content).

(b) For a wave s = s_{max}\sin(\omega t-kx) in a fluid of bulk modulus B, the fractional volume change is \Delta V/V = \partial s/\partial x = -ks_{max}\cos(\omega t-kx). The pressure change is

so the pressure amplitude is

The pressure variation is \pi/2 out of phase with the displacement: pressure is greatest where the displacement is zero.

(c) Intensity obeys the inverse square law, I\propto1/r^{2}:

At 2\ \text{m} the level was 60\ \text{dB}; quadrupling the distance lowers it by 12\ \text{dB}.

1. (a) Huygens' principle: every point on a wavefront acts as a source of secondary wavelets, which spread out in all directions with the speed of the wave; the new wavefront after a time t is the forward envelope (common tangent surface) of these wavelets.

(b) When a plane wavefront reaches a small aperture (of size comparable with the wavelength), only a few points on it can send wavelets through. With so few sources the envelope is part of a sphere: the wavefront beyond the aperture is spherical, centred on the aperture, and the light spreads out (diffraction).

(c) A wavefront is the locus of all points of a wave that are in the same phase at a given instant. It is a sphere for a point source, a plane for a very distant source, and a cylinder for a line source. Rays are normal to wavefronts.

2. Let a plane wavefront AB fall obliquely on a plane mirror, with A touching the mirror first; let i be the angle of incidence. While the wavelet from B travels the distance BC = vt to the mirror, the wavelet from A travels the same distance AD = vt in the same medium. The reflected wavefront is the tangent CD from C to the wavelet centred at A. In the right-angled triangles ABC and ADC the side AC is common and BC = AD = vt. They are therefore congruent, and

The incident ray, the reflected ray and the normal lie in one plane. This proves the law of reflection.

3. (a) A plane wavefront AB is incident on the boundary between medium 1 (speed v_1) and medium 2 (v_2), at angle of incidence i. While the wavelet from B reaches the boundary at C in time t (BC = v_1t), the wavelet from A travels AD = v_2t in medium 2. The refracted wavefront is CD, making angle r with the boundary. From the two triangles with the common hypotenuse AC:

This is Snell's law; with n = c/v it reads n_1\sin i = n_2\sin r.

(b)

The frequency is unchanged: f = c/\lambda = 3.0\times10^{8}/(600\times10^{-9}) = 5.0\times10^{14}\ \text{Hz}.

4. (a) Explained by the wave theory: reflection and refraction (Huygens' construction), interference and diffraction (superposition, bending) and polarization (transverse waves). Not explained: the photoelectric effect, which needs energy in quanta (photons).

(b) Foucault measured the speed of light in water and found it less than in air. The wave theory predicts v = c/n<c; Newton's corpuscular theory predicted a greater speed in a denser medium. The result ruled out the simple corpuscular theory.

5. (a) The refractive index of a medium is the ratio of the speed of light in vacuum to its speed in the medium: n = c/v.

(b) \dfrac{v_w}{v_g} = \dfrac{n_g}{n_w} = \dfrac{1.50}{1.33} = 1.13. Since f is the same, \dfrac{\lambda_w}{\lambda_g} = \dfrac{v_w}{v_g} = 1.13.

(c) Wavefronts arriving at the boundary must leave it at the same rate, or they would pile up or be created at the boundary. The number of wavefronts per second — the frequency — is therefore the same in both media.

1. (a) See short answer 1(a).

(b) The constructions and the congruent and similar triangles are given in short answers 2 and 3(a); the results are i = r for reflection and \sin i/\sin r = v_1/v_2 for refraction, with both rays and the normal in one plane.

(c)

2. (a) On entering a denser medium the end of the wavefront that arrives first slows down, while the other end is still moving at the higher speed. The wavefront therefore swings round, so its normal (the ray) turns towards the normal to the surface: r<i.

(b) For light going from a denser medium 1 (v_1) to a rarer medium 2 (v_2>v_1), \sin i/\sin r = v_1/v_2. The critical angle i_c is the angle of incidence for which r = 90^{\circ}:

For glass–air, \sin i_c = 1/1.5, so i_c = 41.8^{\circ}. For i>i_c the refracted wavefront cannot be constructed: total internal reflection occurs.

(c) Every point of the wavefront in the slit acts as a secondary source. When the slit width is comparable with the wavelength there are few such sources and the envelope of their wavelets spreads well beyond the geometrical shadow edges: the wave diffracts. A wide slit leaves a large flat central part of wavefront, and the light bends only at the edges.

1. (a) Interference is the redistribution of energy produced when two or more coherent light waves overlap: bright fringes where they reinforce and dark fringes where they cancel.

(b) Coherent sources emit waves of the same frequency with a constant phase difference. In practice they are obtained by dividing a single wavefront (Young's two slits) or a single beam by amplitude (thin films), so that any random change in the source phase appears equally in both.

(c) Two bulbs emit light in independent bursts of about 10^{-8}\ \text{s} with random phases; the phase difference between them changes randomly and faster than the eye can follow, so the pattern averages to uniform brightness.

2. Slits S_1 and S_2 are a distance d apart, and the screen is a distance D\gg d away. For a point P at height y above the central point O, the path difference is

Bright: \Delta = n\lambda, so y_n = \dfrac{n\lambda D}{d}, n = 0, 1, 2,\dots Dark: \Delta = (n+\tfrac12)\lambda, so y_n' = \left(n+\tfrac12\right)\dfrac{\lambda D}{d}. The separation of consecutive bright fringes is

and that of consecutive dark fringes is the same, \dfrac{\lambda D}{d}, independent of n. The fringes are equally spaced, with a dark fringe midway between successive bright ones.

3. (a) Two waves of amplitude a superpose with phase difference \phi: the resultant amplitude is R = 2a\cos(\phi/2), so I\propto R^{2} = 4a^{2}\cos^{2}(\phi/2), i.e. I_R = 4I\cos^{2}(\phi/2). Constructive (\phi = 0, 2\pi,\dots): I_{max} = 4I. Destructive (\phi = \pi, 3\pi,\dots): I_{min} = 0.

(b)

For I_1 = I_2 = I this becomes 2I(1+\cos\phi) = 4I\cos^{2}(\phi/2). The average intensity is 2I: energy is redistributed, not created.

4.

Ten fringes: 10\beta = 8.7\ \text{mm}.

5. (a) In a medium of refractive index \mu the wavelength becomes \lambda' = \lambda/\mu, so

The fringes become narrower (closer together).

(b) With white light the central fringe (n = 0) is white, since all colours have zero path difference there. On either side are a few coloured fringes: in each order, blue (small \lambda, small \beta) is nearest the centre and red farthest. After a few orders the colours overlap and the pattern fades to uniform white.

1. (a) Interference is the superposition of two coherent waves to give a stable pattern of maxima and minima of intensity. Conditions: the sources must be coherent; the amplitudes should be nearly equal for good contrast; the slit separation must be small compared with the distance to the screen; and the light should be nearly monochromatic.

(b) See short answer 2: \beta = \lambda D/d.

(c)

2. (a) Let the two waves at P be y_1 = a_1\sin\omega t and y_2 = a_2\sin(\omega t+\phi). Their sum is R\sin(\omega t+\delta) with

(from the phasor triangle). Intensity is proportional to the amplitude squared, so

For I_1 = I_2 = I_0: I = 2I_0(1+\cos\phi) = 4I_0\cos^{2}(\phi/2).

(b) The average of \cos\phi over the pattern is zero, so \bar{I} = I_1+I_2 (= 2I_0): energy is redistributed from the dark to the bright regions, not created or destroyed.

(c) The sheet adds an extra optical path (\mu-1)t to that slit. The central bright fringe, where the total path difference is zero, moves towards the covered slit to the point where the geometrical path difference cancels the extra path. The shift, in fringes, is

1. (a) The grating spacing is d = \dfrac{1}{6000}\ \text{cm} = 1.667\times10^{-6}\ \text{m} and \lambda = 5890\ \text{Å} = 5.89\times10^{-7}\ \text{m}. For the nth maximum d\sin\theta = n\lambda:

(b) For n = 3, \sin\theta_3 = 3\lambda/d = 1.06>1, which is impossible: the third order cannot be observed. The highest order satisfies n<d/\lambda = 2.8, i.e. n = 2.

(c) Diffraction is the bending of light around the edges of an obstacle or through a narrow aperture, spreading it into the geometrical shadow and producing a pattern of bright and dark fringes; it is marked when the size of the obstacle or aperture is comparable with the wavelength.

2. (a) d = \dfrac{1}{500}\ \text{mm} = 2.0\times10^{-6}\ \text{m}; n = 2, \theta = 30^{\circ}:

(b) A grating has a very large number N of slits. For a direction slightly away from a principal maximum the waves from the N slits have phases spread right round the cycle and cancel almost exactly, so each principal maximum is narrow and bright; a double slit has only two sources, so its fringes are broad and of gradually varying intensity.

3. (a) d = \dfrac{1}{3000}\ \text{cm} = 3.33\times10^{-6}\ \text{m}:

(b) The grating. The maxima are far apart (angles measured to a fraction of a degree, not fringes of 0.9\ \text{mm} read with a ruler), sharp and bright, and \lambda = d\sin\theta/n involves directly measured angles. In the double slit, \beta = 0.87\ \text{mm} is small and has a fractional reading error several times larger.

(c) The highest order obeys n<d/\lambda because \sin\theta\leq 1.

4. (a) A single slit of width a is lit by a plane wavefront. Each point in the slit is a source of secondary wavelets. In a direction \theta the paths from the two edges differ by a\sin\theta. If a\sin\theta = \lambda, divide the slit into two halves: for every point in the top half there is a point in the bottom half a/2 away whose wavelet is half a wavelength out of phase, so they cancel in pairs. The same holds for a\sin\theta = n\lambda:

The central maximum lies between the first minima at \theta = \pm\lambda/a.

(b) Width of the central maximum = 2\lambda D/a = \dfrac{2\times600\times10^{-9}\times1.0}{0.20\times10^{-3}} = 6.0\times10^{-3}\ \text{m} = 6.0\ \text{mm}.

5. (a) Rayleigh criterion: two point sources are just resolved when the central maximum of the diffraction pattern of one falls on the first minimum of the pattern of the other; for a circular aperture of diameter D the minimum angular separation is \theta_{min} = 1.22\,\lambda/D.

(b)

(c) Increase the diameter D of the objective, or use light of shorter wavelength (as in the electron microscope, where \lambda is about 10^{-11}\ \text{m}).

1. (a) Light from a narrow slit, instead of forming a sharp image, spreads out into a broad bright central maximum flanked by much weaker, narrower secondary maxima, separated by dark minima. The central band is twice as wide as each secondary one, and most of the energy is in it. The narrower the slit, the wider the pattern.

(b) The path difference between the edge rays is a\sin\theta. For the first minimum, a\sin\theta = \lambda: the slit divides into two halves whose wavelets cancel in pairs. The general minima are a\sin\theta = n\lambda. Secondary maxima lie approximately midway between successive minima, at

and are of rapidly decreasing intensity (4.5\%, 1.6\%, … of the central one).

(c) First minimum: \sin\theta = \lambda/a = 5.0\times10^{-7}/1.0\times10^{-4} = 5.0\times10^{-3}, so \theta = 5.0\times10^{-3}\ \text{rad}. The angular width of the central maximum is

2. (a) A plane transmission grating is a glass plate ruled with a great many parallel, equally spaced opaque lines: the width of each transparent slit is a, of each opaque line b, so the grating spacing is d = a+b. A parallel beam falls normally on it. In a direction \theta the rays from corresponding points of adjacent slits differ in path by (a+b)\sin\theta. They reinforce in every slit of the grating when this equals a whole number of wavelengths:

These are the principal maxima; n = 0 is the undeviated central image.

(b) d = \dfrac{1}{5000}\ \text{cm} = 2.0\times10^{-6}\ \text{m}.

The maximum order: n<d/\lambda = 3.3, so n_{max} = 3 (at \sin\theta_3 = 0.90, \theta_3 = 64.2^{\circ}).

(c) White light gives a white central image (n = 0) and on either side a series of spectra in which violet is nearest the centre and red farthest (because \sin\theta\propto\lambda); higher orders overlap.

1. (a) Polarization is the restriction of the vibrations of the electric vector of a transverse wave to a single plane (plane of vibration), perpendicular to the direction of travel.

(b) Look at glare (reflection from water or a road) through one lens and rotate the lens, or hold one pair of glasses in front of another and rotate it: polaroid lenses darken to almost black at 90^{\circ}; ordinary tinted glass shows no change in transmitted light.

(c) Only transverse waves can be polarized, so the phenomenon shows that light is a transverse wave.

(d) Polaroid sunglasses (cut glare), camera filters, liquid-crystal displays, 3-D cinema glasses, stress analysis of transparent models.

2. (a) Brewster's law: the tangent of the polarizing angle i_p of a transparent medium is equal to its refractive index: \tan i_p = \mu.

(b)

(c) At i_p, reflected and refracted rays are perpendicular, so r = 90^{\circ}-i_p = 37.6^{\circ}. Check: \sin52.4^{\circ}/\sin37.6^{\circ} = 0.792/0.610 = 1.30.

3. Brewster's law: \tan i_p = \mu. Proof: at the polarizing angle the reflected ray is completely polarized and is perpendicular to the refracted ray, so

Snell's law: \mu = \dfrac{\sin i_p}{\sin r}. With r = 90^{\circ}-i_p, \sin r = \cos i_p, so

Why the rays are perpendicular: the reflected light is re-radiated by electrons in the glass that are set vibrating by the refracted wave. For the component with its electric vector in the plane of incidence, these electrons vibrate perpendicular to the refracted ray. If the reflected ray were along that direction of vibration (which happens exactly when the reflected and refracted rays are perpendicular), no light could be radiated that way, because an oscillating charge does not radiate along its own axis. Only the component vibrating perpendicular to the plane of incidence is reflected, so the reflected light is completely plane polarized.

4. (a) Malus' law: the intensity of plane-polarized light transmitted by an analyser is I = I_0\cos^{2}\theta, where \theta is the angle between the polarization direction of the light and the axis of the analyser.

(b) The first polaroid passes half: I_1 = I_0/2. The second, at 60^{\circ}:

(c) With crossed axes, \theta = 90^{\circ} and I = 0: no light gets through.

5. (a) Light from a source passes through a tourmaline crystal T_1 (the polarizer) and then a second crystal T_2 (the analyser), with their axes initially parallel: light emerges from both. When T_2 is rotated about the beam, the emergent light dims, becoming zero when the axes are at 90^{\circ}, and brightens again at 180^{\circ}. This behaviour is possible only if the vibrations are perpendicular to the direction of travel — if light were longitudinal, rotation about the beam's axis would not matter by symmetry.

(b) Sound in air is longitudinal: the particles vibrate along the direction of propagation, so every direction perpendicular to it is equivalent, and there is no plane of vibration that could be selected.

1. (a) Unpolarized light has vibrations in all directions perpendicular to the ray (ordinary sunlight). Plane polarized light has vibrations in one plane only. Partially polarized light has a preference for one plane but with some vibrations in other planes (light reflected at any angle other than i_p).

(b) See short answer 3: \tan i_p = \mu follows from i_p+r = 90^{\circ} and Snell's law.

(c)

(d) At i_p the reflected light is completely plane polarized: through the rotating polaroid it goes from bright to extinction twice per full turn.

2. (a) A polaroid is a sheet containing long molecular chains aligned in one direction (for example iodoquinine sulfate crystals or stretched polyvinyl alcohol doped with iodine); the chains absorb light whose electric vector is along the chain and transmit the component perpendicular to it. Unpolarized light is thus turned into plane polarized light with half the intensity. Malus' law: I = I_0\cos^{2}\theta (defined in short answer 4).

(b) After the polarizer, I_1 = I_0/2. After the analyser at 45^{\circ},

(c) After the first polaroid: I_0/2. After the middle polaroid at 45^{\circ} to the first: \frac{I_0}{2}\cos^{2}45^{\circ} = I_0/4. The third is at 45^{\circ} to the second: \frac{I_0}{4}\cos^{2}45^{\circ} = I_0/8. The transmitted fraction is 1/8 = 12.5\% — whereas with no middle polaroid the two crossed polaroids transmit nothing.

1. (a) Junction (current) law: the algebraic sum of the currents at a junction is zero, \sum I = 0 — charge does not accumulate (conservation of charge). Loop (voltage) law: the algebraic sum of the e.m.f.s in any closed loop equals the sum of the products IR round it, \sum\mathcal{E} = \sum IR, i.e. the net change of potential round a closed loop is zero (conservation of energy).

(b) In the Wheatstone bridge four resistors P, Q, R, S form a quadrilateral ABCD with the battery across A–C and the galvanometer across B–D. At balance the galvanometer current is zero, so the same current I_1 flows in P and Q, and I_2 in R and S. The loop law for ABDA (potential of B equals that of D) gives

and for BCDB

Dividing,

2. (a) Principle: when a steady current flows through a wire of uniform cross-section, the potential difference between two points is proportional to the length of wire between them. A potentiometer compares or measures voltages by balancing them against this known fall along the wire. Voltmeter of infinite resistance: at the balance point no current flows in the cell (or circuit) being measured, so its terminal voltage equals its e.m.f. — unlike a voltmeter, which draws some current and reads less than the e.m.f.

(b) At balance \dfrac{R}{X} = \dfrac{l}{100-l} = \dfrac{20}{80} = \dfrac14. Doubling both keeps the ratio \tfrac14, so the balance point does not move. Interchanging them inverts the ratio:

(c) The balance point is unaffected: the Wheatstone bridge condition P/Q = R/S is symmetric with respect to the battery and galvanometer positions. (It is more sensitive with the more convenient arrangement, but the null point is the same.)

3. Let the ratio of resistances be in proportion to the wire lengths: R/S = l/(100-l).

After adding 5\ \Omega to S:

Equating, 1.5S = S+5, so S = 10\ \Omega and R = 15\ \Omega.

4. (a) A cell of e.m.f. E and internal resistance r is balanced on the potentiometer wire: on open circuit at length l_1, so E = kl_1 (k is the potential gradient). Then a resistance R is connected across the cell. A current I = E/(R+r) flows and the terminal voltage V = IR balances at length l_2: V = kl_2. Dividing,

(b)

5. (a) A low-resistance shunt S is connected in parallel with the galvanometer. Most of the current bypasses the coil and only the fraction it can stand flows through it; the scale is recalibrated. The combined resistance is very small, so the ammeter disturbs the circuit very little.

(b) The galvanometer carries I_g = 5\ \text{mA} and the shunt I-I_g = 1-0.005 = 0.995\ \text{A} at the same potential difference:

6. (a) A high resistance R is connected in series with the galvanometer. The current stays within I_g up to the maximum voltage V, and the scale is calibrated in volts. The high resistance means the voltmeter draws little current from the circuit.

(b)

7. (a) Joule's law: the heat produced in a conductor in time t is H = I^{2}Rt — proportional to the square of the current, to the resistance and to the time.

(b) A superconductor is a material whose resistance drops to exactly zero below a critical temperature T_c (for example mercury at 4.2\ \text{K}). Unlike a perfect conductor (only zero resistance), it also expels magnetic field from its interior, the Meissner effect, and this persists if the field is applied before or after cooling. Uses: powerful electromagnets in MRI, maglev trains, lossless transmission.

(c) An ohmmeter has a galvanometer in series with a cell and a variable resistor. With the terminals shorted the galvanometer gives full-scale deflection (marked 0\ \Omega) after adjusting the resistor; with the terminals open there is no deflection (\infty). An unknown resistance joined between the terminals reduces the current in proportion, so the scale can be marked in ohms; it is non-linear and crowded at the high end.

8. (a) E_1/E_2 = l_1/l_2, so

(b) At the balance point no current is drawn from the cell, so the reading is the true e.m.f. and is independent of the cell's internal resistance; a voltmeter draws current and reads less. The reading is obtained as a ratio of lengths, which can be measured very precisely, and needs no calibrated meter scale.

(c) Manganin has a high resistivity (so a long, thin wire fits in a short length) and a very small temperature coefficient of resistance, so the potential gradient stays steady as the wire warms.

1. (a) The meter bridge is a practical Wheatstone bridge: a uniform wire AC of length 100\ \text{cm} is stretched along a metre scale, with a known resistance R in the left gap and the unknown X in the right gap of a copper strip; a battery is across AC and a galvanometer connected between the junction B of R and X and a sliding jockey J on the wire. The jockey is moved until the galvanometer shows no deflection, at distance l from A. The resistances of the wire lengths are proportional to lengths, so

(b) Near the middle the bridge is most sensitive and the percentage error in (100-l)/l from a fixed error in reading l is smallest.

(c) X = 4.0\times\dfrac{100-40}{40} = 6.0\ \Omega. After interchanging, the left gap holds 6.0\ \Omega and the right 4.0\ \Omega:

2. (a) A potentiometer wire AB of uniform cross-section is connected in series with a driver cell (steady current) and a key. The positive terminals of the two cells E_1, E_2 are joined to A; their negative terminals go through a galvanometer and a two-way switch to a sliding jockey J (circuit in the figure below). With E_1 connected, slide J until the galvanometer shows zero deflection; note the length l_1. Repeat with E_2 to get l_2.

(b) With potential gradient k (volts per unit length) along the wire,

The driver e.m.f. must exceed both cell e.m.f.s and the current must stay constant.

(c) The potential difference across the wire is V = IR = 0.20\times20 = 4.0\ \text{V}; the gradient is

3. (a) (i) Ammeter: a small shunt resistance S in parallel with the galvanometer of resistance G diverts the excess current, so that only I_g goes through the coil when the total current is I: S = I_gG/(I-I_g). The resulting resistance GS/(G+S) is smaller than S. (ii) Voltmeter: a large resistance R in series limits the current to I_g at the full-scale voltage V: R = V/I_g-G. The resulting resistance G+R is very large.

(b) Shunt: S = \dfrac{I_gG}{I-I_g} = \dfrac{0.010\times30}{3.0-0.010} = 0.10\ \Omega. Series resistor: R = \dfrac{V}{I_g}-G = \dfrac{30}{0.010}-30 = 2970\ \Omega.

1. (a) The thermoelectric (Seebeck) effect: when the two junctions of a closed circuit of two different metals are kept at different temperatures, an e.m.f. (and a current) appears in the circuit.

(b) The neutral temperature \theta_n is the temperature of the hot junction at which the thermo-e.m.f. is a maximum (the graph of e against \theta has zero slope there). The inversion temperature \theta_i is the hot-junction temperature at which the thermo-e.m.f. falls back to zero and then reverses.

(c) \theta_i = 2\theta_n-\theta_c, and \theta_n is fixed. Lowering \theta_c by 10\ ^{\circ}\text{C} raises \theta_i by 10\ ^{\circ}\text{C}.

(d) Measuring temperatures (a thermocouple thermometer in furnaces, engines and laboratories), or detecting radiation in a thermopile; also thermoelectric generators.

2. (a) With the cold junction at 0\ ^{\circ}\text{C}, the thermo-e.m.f. first rises as the hot-junction temperature rises, reaches a maximum at \theta_n, then falls, reaching zero at \theta_i and reversing beyond it. The curve is a parabola with its peak at \theta_n and intercept at \theta_i = 2\theta_n (figure below).

(b) e = a\theta+\tfrac12b\theta^{2} (a>0, b<0). At the maximum de/d\theta = 0: a+b\theta_n = 0, so

and e = 0 again at \theta_i = -2a/b = 2\theta_n (for \theta_c = 0).

3. With \theta_c = 0: \theta_i = 2\theta_n-0 = 1000\ ^{\circ}\text{C}. With \theta_c = 20\ ^{\circ}\text{C}, \theta_n is unchanged: \theta_i' = 2(500)-20 = 980\ ^{\circ}\text{C}.

(a decrease of 2\%).

4. (a) \theta_i = 2\theta_n-\theta_c so

(b) \theta_c = 2(320)-600 = 40\ ^{\circ}\text{C}.

(c) The e.m.f. for a pair of metals is a function e = a(\theta_h-\theta_c)+\tfrac12b(\theta_h^{2}-\theta_c^{2}). Its maximum with respect to \theta_h is where a+b\theta_h = 0 — which does not contain \theta_c. Changing the cold junction changes the size of the e.m.f. but not the hot-junction temperature at which it peaks.

5. (a) The Peltier effect: when a current flows across a junction of two different metals, heat is absorbed at the junction for one direction of the current and evolved for the other; this is because the electrons carry different energy in the two metals.

(b) Joule heating I^{2}R is irreversible (the heat is always evolved) and proportional to I^{2}; the Peltier heat is proportional to I and reversible: reversing the current turns the heating at a junction into cooling.

(c) Small thermoelectric coolers for portable refrigerators and for cooling electronic detectors.

6. (a) A thermopile is a number of thermocouples connected in series, with all the "hot" junctions on one face and all the "cold" junctions on another; it is used to detect and measure radiant heat.

(b) The e.m.f.s of the individual junction pairs add, so a given small rise of temperature gives several times the e.m.f. of a single couple, increasing sensitivity.

(c) de/d\theta = 16-0.08\theta = 0 gives \theta_n = 200\ ^{\circ}\text{C}, and

The inversion temperature is 400\ ^{\circ}\text{C}.

1. (a) Two different metals, say copper and iron, are joined at two junctions to form a closed loop; one junction (hot) is kept at a higher temperature \theta_h than the other (cold) \theta_c. A galvanometer in the circuit shows a current. The e.m.f. depends on (i) the pair of metals (their positions in the thermoelectric series — the farther apart, the larger the e.m.f.), (ii) the temperatures of the two junctions, and to first order on their difference.

(b) With \theta_c = 0\ ^{\circ}\text{C}, raising \theta_h increases e roughly linearly at first; the rise slows, e reaches a maximum at the neutral temperature \theta_n, then decreases, reaching zero at \theta_i, and reverses in sign above it. The parabola e = a\theta+\tfrac12b\theta^{2} is symmetric about \theta_n. \theta_n depends only on the metals; \theta_i = 2\theta_n-\theta_c depends on \theta_c too.

(c) \theta_i = 2(270)-15 = 525\ ^{\circ}\text{C}.

2. (a) See short answer 5. When current I passes through a junction, the heat \Pi I per second is absorbed or evolved, where \Pi is the Peltier coefficient of the pair. Reversing I changes the sign of \Pi I: heat absorbed becomes heat evolved. By contrast I^{2}R does not change sign.

(b) See short answer 6(a). When radiation falls on the blackened hot junctions, they warm relative to the cold junctions; the summed e.m.f. is measured with a galvanometer and is proportional to the radiation received. Used to measure the intensity of infrared and heat radiation, e.g. in radiometers and some infrared thermometers.

(c) \theta_n = -a/b = 12/0.06 = 200\ ^{\circ}\text{C} (from 12-0.06\theta = 0), \theta_i = 400\ ^{\circ}\text{C}, and

1. (a) The Biot–Savart law: the magnetic field d\vec B due to a current element I\,d\vec l at a point with position vector \vec r from the element is

where \theta is the angle between d\vec l and \vec r; its direction is perpendicular to both d\vec l and \vec r.

(b) Take a coil of N turns and radius R, carrying I. Every element dl of the circle is perpendicular to the line \vec r to the centre, so \theta = 90^{\circ}, \sin\theta = 1 and r = R, and all the d\vec B at the centre point the same way (along the axis). Adding,

(c) Right-hand rule: curl the fingers of the right hand in the direction of the current in the coil; the thumb points along \vec B at the centre.

2. (a) Draw a circle of radius r around the wire, with the wire at its centre and perpendicular to its plane. By symmetry B is tangential and constant in magnitude round the circle. Ampère's circuital law: \oint\vec B\cdot d\vec l = \mu_0I_{enclosed}.

(b) F = BIl\sin\theta. Maximum at \theta = 90^{\circ}: F = 0.4\times2\times0.5 = 0.40\ \text{N}. Minimum at \theta = 0^{\circ} (wire along the field): F = 0.

3. (a) Voltage sensitivity is the deflection produced per unit voltage across the galvanometer: \dfrac{\phi}{V} = \dfrac{NBA}{CR}; it depends on the number of turns N, the field B, the coil area A, the torsional constant C (smaller is more sensitive) and the coil resistance R. Current sensitivity is \phi/I = NBA/C.

(b) A radial field (concave pole pieces with a soft-iron core) keeps the plane of the coil always parallel to \vec B, so the deflecting torque NIAB is the same at every angle and the deflection is proportional to the current — a linear scale.

(c)

4. (a) When a current-carrying conductor is placed in a magnetic field perpendicular to the current, the moving charge carriers are pushed sideways by the force qvB. Charge accumulates on one face, and a transverse electric field (Hall field) builds up until its force balances the magnetic force. The resulting transverse potential difference is the Hall voltage V_H. Its sign tells whether the carriers are electrons or holes.

(b) At balance eE_H = evB and V_H = E_Hw; with I = nevwt (where w is the width and t the thickness in the direction of \vec B), V_H = \dfrac{BI}{net}:

5. (a) Two long parallel wires a distance r apart carry currents I_1 and I_2 in the same direction. Wire 1 produces at wire 2 a field

directed perpendicular to the plane of the wires. Wire 2 (length l) carries I_2 perpendicular to this field, so the force on it is F = B_1I_2l:

For currents in the same direction the force is attractive; for opposite currents, repulsive.

(b) The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length and negligible cross-section, placed one metre apart in vacuum, produces between them a force of 2\times10^{-7}\ \text{newton per metre} of length. (From \mu_0/2\pi = 2\times10^{-7}\ \text{N A}^{-2} with I_1 = I_2 = 1\ \text{A}, r = 1\ \text{m}.)

6. (a) A charge q with velocity \vec v in a field \vec B experiences \vec F = q\vec v\times\vec B, of magnitude qvB\sin\theta and perpendicular to both \vec v and \vec B. For \vec v\perp\vec B the force is always perpendicular to the velocity: it changes the direction of motion but not the speed, and provides the centripetal force:

a constant radius: the path is a circle.

(b)

7. (a) Solenoid: take a rectangular Amperian loop with one side of length l inside the solenoid and parallel to its axis, the other outside where B\approx0. The enclosed current is nlI (with n turns per metre), and \oint\vec B\cdot d\vec l = Bl, so

Toroid: a circle of radius r inside the ring encloses N turns, each carrying I; with B tangential and constant,

Outside the toroid the enclosed current is zero, so B = 0.

(b)

8. (a) A magnetic field line is a curve whose tangent at every point gives the direction of \vec B, and whose density shows its strength. Magnetic flux through a surface is \Phi = BA\cos\theta, the number of lines crossing it, in weber.

(b) Oersted's experiment (1820): a compass needle held parallel to a straight wire deflects when a current is passed and reverses when the current is reversed. It showed that an electric current produces a magnetic field circulating round the wire — the first link between electricity and magnetism.

(c) At every point \vec B has a single direction, so a field line has a single tangent there; two lines crossing would give two directions at one point.

1. (a) The Biot–Savart law and the derivation of B = \mu_0NI/2R are given in short answer 1.

(b) n = 2.1\times10^{22}\ \text{m}^{-3} as in short answer 4(b).

(c) As in short answer 5: F/l = \mu_0I_1I_2/2\pi r, attractive for parallel currents; the ampere is defined through F/l = 2\times10^{-7}\ \text{N m}^{-1} for 1\ \text{A} at 1\ \text{m}.

2. (a) Let a rectangular coil of sides a (vertical) and b (horizontal) carry I in a uniform field \vec B, with the normal making an angle \theta with \vec B. The forces on the two vertical sides are equal and opposite, F = BIa each, forming a couple with arm b\sin\theta. The forces on the horizontal sides are equal, opposite and collinear, so they do not produce torque. The torque is

and for N turns

where m is the magnetic moment of the coil.

(b) Moving-coil galvanometer: a rectangular coil of many turns on a soft-iron core is suspended between the concave poles of a permanent magnet by a phosphor-bronze strip, which supplies the restoring torque C\phi. The field is radial. A current I gives a torque NBAI and the coil turns through \phi until NBAI = C\phi:

The deflection, read by a lamp-and-scale or pointer, is proportional to the current.

(c)

Here \theta is the angle between the normal to the coil and the field, as in the derivation.

3. (a) One ampere is the constant current which, flowing in each of two infinitely long straight parallel conductors one metre apart in vacuum, produces a force of 2\times10^{-7}\ \text{N} per metre of length between them.

(b) Adjacent turns of the solenoid carry current in the same direction, so, like parallel currents, they attract one another. The mutual attraction of all the turns compresses the coil along its axis, and the solenoid contracts (it would return to its length when the current is stopped).

(c) With zero tension in the supporting wires the upward magnetic force on the conductor just balances its weight. For a length l of conductor with linear density \mu = 0.2\ \text{g m}^{-1} = 2.0\times10^{-4}\ \text{kg m}^{-1}:

For the coil, with its plane parallel to the field the normal is perpendicular to \vec B, so \sin\theta = 1 and the torque is maximum:

(with A = 100\ \text{cm}^{2} = 1.0\times10^{-2}\ \text{m}^{2}).

1. (a) Magnetic permeability \mu of a medium is the ratio of the magnetic flux density produced in it to the magnetising field: \mu = B/H, unit \text{T m A}^{-1} (or \text{H m}^{-1}). It measures how readily the medium allows magnetic flux to be set up in it. Relative permeability \mu_r is the ratio of the permeability of the medium to that of free space:

It is a pure number, equal to the factor by which the material multiplies the flux density. Magnetic susceptibility \chi_m is the ratio of the magnetisation produced to the magnetising field: \chi_m = I/H, also a pure number. It measures how easily the material becomes magnetised.

(b) The total flux density inside a magnetised material is the sum of the part due to the magnetising field and the part due to the magnetisation of the material itself:

Divide throughout by \mu_0 H:

Now B/H = \mu, so the left side is \mu/\mu_0 = \mu_r, and I/H = \chi_m:

2. (a)

(b) Diamagnetic materials. Their induced magnetisation opposes the applied field (a consequence of Lenz's law at the atomic level), so a rod of bismuth suspended in a field sets itself across the field and is pushed towards the weaker region.

3. (a) Hysteresis is the lagging of the magnetisation (or flux density B) behind the magnetising field H when a ferromagnetic specimen is taken through a complete cycle of magnetisation. Because of this lag, B does not return along the original curve when H is reduced, and the B–H graph traces a closed loop (Figure 17). Starting from an unmagnetised specimen, B rises along OA to saturation. As H is reduced to zero, B falls only to OB_r, not to zero. Reversing H to -OH_c brings B to zero; further reversal saturates the specimen the other way, and the return path closes the loop.

(b) Retentivity (remanence) B_r is the flux density that remains in the material when the magnetising field is reduced to zero — the intercept on the B axis. It measures how much magnetism the material retains. Coercivity H_c is the reverse magnetising field needed to reduce that residual flux density to zero — the intercept on the H axis. It measures how hard the material is to demagnetise.

4. (a) When a ferromagnetic specimen is carried round one complete cycle of magnetisation, work must be done by the magnetising source to turn the domains against the internal friction that opposes their rotation and wall motion. This work is not recovered when H returns to zero, because the magnetisation lags behind: the material comes back to its starting point having absorbed net energy, which appears as heat. Quantitatively, the work done per unit volume in changing the flux density by dB is H\,dB, so over the whole cycle

and this integral is exactly the area enclosed by the B–H loop. A wide loop therefore means a large energy loss per cycle; the power wasted is (area) × (volume) × (frequency).

(b) (i) Transformer core — soft iron. Its hysteresis loop is tall but very narrow, so the area, and hence the energy lost per cycle, is small; this matters because the core is taken round the loop 50 times a second. Soft iron also has a very high permeability and low retentivity and coercivity, so it magnetises and demagnetises easily. (ii) Permanent magnet — steel (or alnico). Its loop is wide, with high retentivity (so it keeps its magnetism) and high coercivity (so stray fields and rough handling cannot demagnetise it). The large loop area does not matter because the magnet is not cycled.

5. (a) Magnetising field (magnetic field intensity) H is the field produced by the free current alone, independent of the medium; for a solenoid H = nI, unit \text{A m}^{-1}. Magnetisation (intensity of magnetisation) I is the magnetic moment developed per unit volume of the material, I = M/V, also in \text{A m}^{-1}. Flux density B is the total field inside the material, unit tesla. They are related by

(b)

6. (a) In a ferromagnetic material the atomic magnetic moments are not independent: a strong quantum-mechanical exchange interaction locks neighbouring atomic moments parallel to one another over small regions called domains, typically 10^{-6} to 10^{-4}\ \text{m} across and containing about 10^{17} atoms. Each domain is therefore magnetised to saturation even with no external field. In an unmagnetised specimen the domains point in random directions, so their moments cancel and the specimen shows no net magnetism. When an external field is applied, two things happen: at low fields the domain walls move, so domains already pointing near the field direction grow at the expense of the others; at high fields the remaining domains rotate bodily into the field direction. When all domains are aligned, the specimen is saturated and no further increase is possible. Domain-wall motion is not perfectly reversible — walls get stuck at impurities and grain boundaries — and this is the microscopic origin of hysteresis.

(b) The Curie temperature T_c is the temperature above which a ferromagnetic material loses its ferromagnetism and becomes merely paramagnetic (770\ ^{\circ}\text{C} for iron, 358\ ^{\circ}\text{C} for nickel, 1131\ ^{\circ}\text{C} for cobalt). Above T_c the thermal energy of the atoms exceeds the exchange energy holding the moments parallel, so the domain structure is destroyed; the susceptibility falls sharply and thereafter obeys the Curie-Weiss law \chi_m = C/(T-T_c). On cooling below T_c the domains reform and ferromagnetism returns.

7. (a) Soft iron for electromagnets: it has very high permeability, so a modest current gives a very strong field; very low retentivity, so the magnetism disappears almost completely the moment the current is switched off — essential for a crane magnet or a relay; and low coercivity with a narrow loop, so little energy is wasted if the field is reversed repeatedly. Steel for permanent magnets: it has high retentivity, so it keeps almost all its magnetisation after the magnetising field is removed, and high coercivity, so it is not demagnetised by stray fields, mechanical shock or moderate heating. Its wide loop means large hysteresis loss, but a permanent magnet is never cycled, so that is irrelevant.

(b) A changing magnetic flux in a solid iron core induces circulating eddy currents in the core itself, and these dissipate energy as heat (I^{2}R) and heat the transformer. Laminating the core — building it from thin sheets insulated from one another by varnish or oxide, with their planes parallel to the flux — breaks the conducting paths available to those currents. The induced e.m.f. in each thin lamination is small and its resistance is high, so the eddy currents, and the power lost (which varies as the square of the lamination thickness), are greatly reduced.

8. (a) Magnetic flux through a surface is \Phi = \vec{B}\cdot\vec{A} = BA\cos\theta, the total number of field lines crossing it, measured in weber. Gauss's law for magnetism: the total magnetic flux through any closed surface is zero,

Every line that enters a closed surface must leave it again.

(b) Because magnetic monopoles do not exist. Electric field lines begin on positive charges and end on negative ones, so a closed surface round a single charge has a net outward flux — that is Gauss's law for electricity. But the smallest unit of magnetism is the dipole: cut a bar magnet in half and you get two complete magnets, each with its own north and south pole, never an isolated pole. With no sources or sinks for \vec{B} to start or stop on, every field line must close on itself, running from N to S outside the magnet and from S to N inside it. The mathematical statement of this is \oint\vec{B}\cdot d\vec{A} = 0.

1. (a) Permeability \mu = B/H measures how readily a medium carries magnetic flux. Relative permeability \mu_r = \mu/\mu_0 is the pure number by which the medium multiplies the flux density that the same magnetising field would produce in vacuum. Magnetisation I is the magnetic moment per unit volume of the material. Susceptibility \chi_m = I/H measures how readily the material becomes magnetised. Inside a magnetised material the total flux density is the sum of the contribution of the magnetising field and that of the aligned atomic dipoles:

Divide both sides by \mu_0H:

On the left, B/H = \mu, so B/\mu_0H = \mu/\mu_0 = \mu_r. Therefore

This single relation classifies all materials: \chi_m slightly negative gives \mu_r < 1 (diamagnetic); \chi_m small positive gives \mu_r just above 1 (paramagnetic); \chi_m very large gives \mu_r \gg 1 (ferromagnetic).

(b) See the table in short answer 2 above: the comparison runs over the sign and size of \chi_m, the value of \mu_r, the atomic origin, the behaviour in a non-uniform field, the temperature dependence, and examples.

(c)

2. (a) Hysteresis is the lag of B behind H through a complete cycle of magnetisation. Figure 17 shows the loop: OA is the initial magnetisation curve rising to saturation; AB_r is the fall to the retentivity B_r as H returns to zero; B_rH_c is the reverse field to the coercivity H_c at which B = 0; then saturation in the reverse sense and the symmetrical return path, closing the loop.

(b) Consider unit volume of the material. When the flux density changes by dB under a magnetising field H, the source must supply work dW = H\,dB per unit volume. Over one complete cycle,

which is, by definition of an integral round a closed curve in the H–B plane, the area enclosed by the loop. Since the loop is traversed once per cycle, the power dissipated in a core of volume V at frequency f is

Microscopic origin. The magnetisation changes by domain-wall motion and domain rotation. Walls do not move smoothly: they are pinned at impurities, dislocations and grain boundaries, and break free in sudden jumps (the Barkhausen effect). Each jump dissipates energy irreversibly as lattice vibrations, i.e. as heat. That is why the return path does not retrace the outward one and why the loop has area at all.

(c) Transformer core — soft iron (or silicon steel): narrow loop, so small area and small hysteresis loss at 50\ \text{Hz}; also high \mu_r so the flux linkage is large. Electromagnet — soft iron: high permeability for a strong field, but very low retentivity so the magnetism collapses when the current stops, which is the whole point of a lifting magnet or a relay. Permanent magnet — steel, alnico or a rare-earth alloy: high retentivity so it stays magnetised, and high coercivity so it resists demagnetisation; the large loop area is irrelevant because it is not cycled.

3. (a) A ferromagnetic material is divided into domains — small regions, about 10^{-6} to 10^{-4}\ \text{m} across, within which an exchange interaction holds all the atomic magnetic moments rigidly parallel, so each domain is spontaneously magnetised to saturation. In an unmagnetised specimen the domains point in random directions and their moments cancel, giving no net magnetism. Magnetisation and saturation. Applying a weak field makes the domain walls move, so those domains already lying near the field direction grow at the expense of their neighbours; this stage is largely reversible. In stronger fields the remaining domains rotate bodily into line with the field. Once every domain is aligned, no further alignment is possible and B increases only as fast as it would in vacuum: the specimen has reached saturation. Hysteresis. Domain-wall motion is impeded by impurities, grain boundaries and internal strains, which pin the walls. Freeing a wall requires energy and happens in abrupt jumps, so the process is irreversible. When H is reduced, the walls do not return to their original positions, and some alignment survives at H = 0 — the retentivity. A reverse field must be applied to unpin them and restore B = 0 — the coercivity. The energy lost in these irreversible jumps is the area of the loop.

(b) The Curie temperature is the temperature above which thermal agitation destroys the domain structure and the material ceases to be ferromagnetic, becoming ordinarily paramagnetic. For iron it is 770\ ^{\circ}\text{C}. Above T_c the susceptibility follows \chi_m = C/(T-T_c).

(c) Eddy currents are circulating currents induced in the body of a conductor when the magnetic flux through it changes; by Lenz's law they flow so as to oppose the change, and they dissipate energy as heat. Uses: (i) induction furnaces, where a high-frequency field melts a metal charge by the heat of its own eddy currents; (ii) electromagnetic damping in moving-coil galvanometers, balances and induction-type energy meters, making them dead-beat; (iii) eddy-current brakes in trains and in exercise machines, which are smooth and wear-free; (iv) induction cooking hobs and metal detectors. Disadvantages: (i) they waste energy as heat in transformer and motor cores, lowering efficiency; (ii) the heating can damage insulation and demands cooling. Lamination. The core is built of thin sheets insulated from one another, their planes parallel to the flux. Each lamination has a small cross-section, so the flux through it and hence the induced e.m.f. is small, while its resistance to the circulating path is large. Since eddy-current power varies as the square of the lamination thickness, using sheets 0.3\ \text{mm} thick instead of a solid core cuts the loss by orders of magnitude. (Hysteresis loss is unaffected by lamination; that is reduced by choosing silicon steel.)

1. (a) First law: whenever the magnetic flux linked with a circuit changes, an e.m.f. is induced in it, and it lasts only as long as the change lasts. Second law: the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage:

(b) Lenz's law: the induced current always flows in such a direction as to oppose the change producing it; this is the meaning of the minus sign in Faraday's law. Energy argument. Suppose the north pole of a magnet is pushed towards a coil. By Lenz's law the induced current makes the near face of the coil a north pole, which repels the magnet, so the experimenter must do work to continue pushing. That work is exactly the source of the electrical energy that appears in the coil. Now suppose the law were the other way round: the induced current would make the near face a south pole, which would attract the magnet, pulling it in faster, increasing the flux change, increasing the current, and so on without limit. We should get unlimited electrical energy and unlimited kinetic energy from nothing, which violates the conservation of energy. Lenz's law is therefore the statement of energy conservation for induced currents.

2. (a) Let a conducting rod PQ of length l slide with velocity v on a pair of rails in a uniform field B directed into the plane, the velocity being perpendicular to both the rod and the field. In a time dt the rod sweeps out an area l\,v\,dt, so the flux through the circuit changes by

By Faraday's law the magnitude of the induced e.m.f. is

(Equivalently: each free electron in the rod feels a force evB along the rod, which drives charge to one end until the electrostatic field E = V/l balances it, giving V = Blv.) If \vec{v} makes an angle \theta with \vec{B}, then e = Blv\sin\theta.

(b)

3. (a) Construction. An armature — a rectangular coil of N turns of insulated copper wire wound on a laminated soft-iron core — is mounted on an axle between the poles of a strong permanent magnet or electromagnet, so that it can be rotated in the field. The two ends of the coil are joined to two separate slip rings fixed on the axle, and two carbon brushes press against the rings and lead the current to the external circuit. Working. As the coil is rotated, the flux linked with it changes continuously, so by Faraday's law an e.m.f. is induced. The sides of the coil cut the field lines in opposite senses, and each side reverses its sense of cutting twice per revolution, so the induced e.m.f. and current alternate in direction with the same frequency as the rotation. The slip rings keep each end of the coil permanently connected to its own brush, so the alternation is passed on to the external circuit unchanged.

(b) Let the coil of N turns and area A rotate with uniform angular velocity \omega in a field B, and let \theta = \omega t be the angle between the normal to the coil and \vec{B}, measured from the position where the coil's plane is perpendicular to the field. The flux linkage is

By Faraday's law,

So e = e_0\sin\omega t with peak value e_0 = NBA\omega = 2\pi fNBA.

4. (a) Self-inductance L of a coil is the property by which it opposes any change of current through itself, by inducing a back e.m.f. Numerically it is the flux linkage per unit current, or the back e.m.f. per unit rate of change of current:

Its SI unit is the henry (H): a coil has an inductance of one henry if a current changing at 1\ \text{A s}^{-1} induces a back e.m.f. of 1\ \text{V}.

(b) Let a long solenoid of length l, cross-sectional area A and N turns (so n = N/l per metre) carry a current I. The field inside is

The flux through one turn is \Phi = BA, so the total flux linkage is

Hence

where V = Al is the volume. With a core of relative permeability \mu_r, multiply by \mu_r.

5. (a) See the derivation box below. Summary: while the current is being built up from 0 to I, the source must work against the back e.m.f. L\,dI/dt; the total work is

and this is stored in the magnetic field of the coil.

(b)

6. (a) Mutual inductance M of a pair of coils is the flux linkage produced in the second coil per unit current in the first:

It is measured in henry, and M_{12} = M_{21}. Factors: (i) the number of turns of each coil; (ii) the area of cross-section and the size of the coils; (iii) the distance and orientation between them — M is greatest when they are close and coaxial, and zero when their axes are perpendicular; (iv) the permeability of the medium or core between them — a common soft-iron core greatly increases M.

(b)

7. (a) A transformer changes an alternating voltage from one value to another. It works on the principle of mutual induction: an alternating current in one coil produces a changing flux, which links a second coil and induces an alternating e.m.f. in it. Two coils, the primary of N_p turns and the secondary of N_s turns, are wound on a common closed core of laminated soft iron, which carries virtually all the flux from one coil to the other. Since the same flux \Phi links every turn of both coils,

Dividing one by the other, d\Phi/dt cancels:

If k > 1 it is a step-up transformer; if k < 1, step-down. For an ideal (loss-free) transformer the power is conserved, V_pI_p = V_sI_s, so

— voltage is raised only at the cost of current.

(b)

(a step-down transformer).

8. (a) (i) Copper (I²R) loss — heat in the resistance of the windings; reduced by using thick copper wire of low resistance. (ii) Iron (eddy-current) loss — currents circulating in the core; reduced by laminating the core with insulated sheets. (iii) Hysteresis loss — energy spent taking the core round its B–H loop 50 times a second; reduced by making the core of soft magnetic material with a narrow loop, such as silicon steel. (iv) Flux leakage — some flux from the primary does not link the secondary; reduced by winding the two coils on the same limb, one over the other, and by using a closed core of high permeability. (Humming due to magnetostriction is a fifth, small loss, reduced by clamping the laminations tightly.)

(b) Because a transformer depends entirely on a changing flux. A steady direct current produces a steady flux, so d\Phi/dt = 0 and no e.m.f. at all is induced in the secondary. Worse, the primary of a transformer has very low resistance and, with d.c., no back e.m.f. or inductive reactance to limit the current, so a large current flows and the primary winding burns out.

9. (a) Eddy currents are the circulating currents induced in the body of a solid conductor when the magnetic flux through it changes; they flow in closed loops in planes perpendicular to the flux and, by Lenz's law, oppose the change.

(b) Uses: induction furnaces for melting metals; electromagnetic damping in galvanometers and balances (dead-beat instruments); eddy-current brakes in trains; induction motors, induction cookers and metal detectors. Disadvantages: wasted energy as heat in the cores of transformers, motors and dynamos, reducing efficiency; the heat produced can damage the insulation and requires cooling arrangements.

(c)

1. (a) Faraday's first law: an e.m.f. is induced in a circuit whenever the magnetic flux linked with it changes, and it lasts only while the change lasts. Second law: the induced e.m.f. equals the rate of change of flux linkage, e = -N\,d\Phi/dt. The flux can be changed by moving the magnet, moving the coil, changing the area, rotating the coil, or changing the current in a neighbouring coil. Lenz's law: the induced current always opposes the change that causes it. Pushing a north pole towards a coil makes the near face a north pole, which pushes back; pulling it away makes the near face a south pole, which pulls back. The law fixes the direction of the current and supplies the minus sign in Faraday's law. It is required by the conservation of energy: the mechanical work done against the opposing force is exactly the electrical energy generated. If the current helped the motion instead, the system would accelerate itself and produce energy from nothing.

(b) Construction and working: see short answer 3(a) above — armature coil, field magnet, slip rings and brushes; rotation changes the flux linkage continuously and produces an alternating e.m.f. Derivation. Let the coil have N turns of area A and rotate with angular velocity \omega about an axis perpendicular to a uniform field B. At time t let the normal to the coil make an angle \theta = \omega t with \vec{B}. The flux through one turn is

so the flux linkage is N\Phi = NBA\cos\omega t. By Faraday's law,

The e.m.f. is therefore sinusoidal, with peak value e_0 = NBA\omega, reached when \omega t = 90^{\circ} — that is, when the plane of the coil is parallel to the field and its sides are cutting the lines of force fastest. The e.m.f. is zero when the plane is perpendicular to the field, even though the flux is then greatest.

(c)

2. (a) Self-inductance L is the flux linkage per unit current, L = N\Phi/I, equivalently the back e.m.f. induced per unit rate of change of current, e = -L\,dI/dt; unit the henry. Solenoid. For a long solenoid of length l, area A and N turns carrying I, the internal field is uniform:

The flux through each turn is BA, and there are N turns, so the flux linkage is

Dividing by I,

So L depends only on the geometry of the coil and the core material — never on the current.

(b) See the derivation box below. In building the current from 0 to I the source does work

Where it resides: in the magnetic field in the space inside the solenoid. Substituting L = \mu_0n^{2}Al and I = B/\mu_0n,

so the energy per unit volume is u = B^{2}/2\mu_0 — an expression containing only the field itself. The field is the store, exactly as an electric field stores \tfrac12\varepsilon_0E^{2} per unit volume.

(c)

3. (a) Principle: mutual induction. A changing current in one coil sets up a changing flux which, linking a second coil, induces an e.m.f. in it. Construction. Two coils — the primary of N_p turns, fed from the a.c. supply, and the secondary of N_s turns, feeding the load — are wound on a closed, laminated soft-iron core, usually one over the other on the same limb so that almost all the flux is shared. The laminations are insulated from each other and the core metal is silicon steel. Working. The alternating primary current produces an alternating flux in the core. The core guides essentially the whole of this flux through the secondary, so the same d\Phi/dt acts on every turn of both coils:

Dividing the second by the first cancels d\Phi/dt:

For an ideal transformer there are no losses, so the power in equals the power out:

A step-up transformer raises the voltage but lowers the current in exactly the same ratio; it cannot create power. This is why electrical energy is transmitted at very high voltage — the current, and so the I^{2}R loss in the line, is small.

(b) Copper loss (I^{2}R in the windings) — use thick, low-resistance copper conductors, and oil-cool large transformers. Eddy-current loss — laminate the core, insulating the sheets from one another. Hysteresis loss — use a magnetically soft core (silicon steel) with a narrow B–H loop. Flux leakage — use a closed core of high permeability and interleave the primary and secondary windings on the same limb. With these measures a large power transformer reaches an efficiency of 98–99\%.

(c)

For an ideal transformer, V_pI_p = V_sI_s:

Check: output power = 10\times2.0 = 20\ \text{W}; input power = 240\times0.083 = 20\ \text{W}. ✓

1. (a) The peak (maximum) value I_0 is the largest value the alternating current reaches in either direction during a cycle. The r.m.s. (root-mean-square, or virtual) value is that steady direct current which would produce the same heating effect in a given resistance in the same time — it is the square root of the mean of the squares of the instantaneous values over one complete cycle. The mean value over a complete cycle is zero, because the positive and negative halves cancel; the useful mean is therefore taken over half a cycle, and equals 2I_0/\pi = 0.637I_0.

(b) Let I = I_0\sin\omega t. The heat produced in a resistance R in time dt is I^{2}R\,dt, so the mean of I^{2} over one period T is

Using \sin^{2}\omega t = \tfrac12(1-\cos2\omega t), the mean of \cos2\omega t over a whole number of cycles is zero, so the mean of \sin^{2}\omega t is \tfrac12:

2. (a)

(b)

(c) Because the r.m.s. value is the one that determines the heating and the power: an a.c. of r.m.s. value 220\ \text{V} delivers exactly the same power to a heater as a steady 220\ \text{V} d.c. supply, so all the d.c. formulae P = V^{2}/R = I^{2}R can be used unchanged. Ordinary a.c. ammeters and voltmeters (hot-wire and moving-iron types) respond to the mean square of the current and so read r.m.s. directly. The mean value over a full cycle would be useless, since it is zero. (The peak value still matters for insulation design, which must withstand 311\ \text{V}, not 220\ \text{V}.)

3. (a) Let V = V_0\sin\omega t be applied to a pure inductance L of zero resistance. The back e.m.f. balances the applied voltage:

Integrating,

So the current has the same frequency but its phase is \pi/2 behind the voltage: the current lags the voltage by 90^{\circ}. Physically, the self-induced e.m.f. opposes the growth of the current, so the current cannot rise until after the voltage has. Comparing with I = I_0\sin(\omega t-\pi/2) gives I_0 = V_0/\omega L, so the quantity \omega L plays the part of resistance. Inductive reactance X_L = \omega L = 2\pi fL, measured in ohms, is the opposition offered by an inductor to alternating current. It is proportional to frequency: an inductor passes d.c. (f=0, X_L=0) freely but chokes high frequencies.

(b)

4. (a) Let V = V_0\sin\omega t be applied to a pure capacitance C. The charge is

and the current is the rate of flow of charge:

The current is \pi/2 ahead of the voltage: the current leads by 90^{\circ}. Physically, the current is largest when the capacitor is empty and the voltage across it is zero, and falls to zero when the capacitor is fully charged and the voltage is greatest. Comparing with I = I_0\sin(\omega t+\pi/2) gives I_0 = V_0/(1/\omega C), so capacitive reactance is

in ohms. It is inversely proportional to frequency: infinite at f = 0, so d.c. is blocked completely, and small at high frequency, so a capacitor passes a.c. easily.

(b)

5. (a) In a series circuit the same current flows through all three elements, so the current phasor I is drawn as the reference along the positive x-axis. Then V_R = IR is in phase with I (drawn along I); V_L = IX_L leads I by 90^{\circ} (drawn upwards); V_C = IX_C lags I by 90^{\circ} (drawn downwards). See Figure 18. V_L and V_C are opposite, so they combine to a single phasor of magnitude |V_L-V_C|. The applied voltage is the resultant of this and V_R, at right angles:

(b)

6. (a) Resonance is the condition in which the inductive and capacitive reactances of a series LCR circuit become equal and cancel, so that the impedance falls to its minimum value R, the current rises to its maximum V/R, and the current is exactly in phase with the applied voltage (power factor = 1). It occurs at one particular frequency, the resonant frequency. At resonance X_L = X_C:

Because the current is greatest there, the series resonant circuit is called an acceptor circuit; it is used to tune a radio receiver to one station.

(b)

7. (a) Power factor is the cosine of the phase angle between the applied voltage and the current, \cos\phi = R/Z. It is the fraction of the apparent power V_{rms}I_{rms} that is actually consumed. Wattless (idle) current is the component of the current at 90^{\circ} to the voltage, I\sin\phi; it flows in the circuit but consumes no average power.

(b) Let V = V_0\sin\omega t and I = I_0\sin(\omega t-\phi). The instantaneous power is

Using 2\sin A\sin B = \cos(A-B)-\cos(A+B),

Averaged over a complete cycle the term \cos(2\omega t-\phi) gives zero, so

A pure inductor has R = 0, so \phi = 90^{\circ} and \cos\phi = 0, giving P_{av} = 0. Energy is not destroyed but exchanged: during a quarter cycle the inductor draws energy from the source and stores it in its magnetic field, and during the next quarter cycle it returns all of it. The average over a full cycle is exactly zero. The same is true of a pure capacitor, which stores and returns energy in its electric field.

8. (a) Resistance R is the opposition of a resistor; it is independent of frequency, the current and voltage are in phase, and it dissipates energy as heat. Reactance (X_L = \omega L or X_C = 1/\omega C) is the opposition of an inductor or capacitor; it depends on frequency, produces a 90^{\circ} phase difference, and dissipates no energy — it only stores and returns it. Impedance Z is the total opposition of a circuit containing both, Z = \sqrt{R^{2}+(X_L-X_C)^{2}}, with a phase angle between 0 and 90^{\circ}. All three are measured in ohms.

(b) The dielectric of a capacitor is an insulator, so no charge ever crosses the gap. With d.c., current flows only for the brief instant of charging; once the capacitor is charged to the supply voltage the current stops, so the capacitor blocks d.c. — consistent with X_C = 1/2\pi fC \to \infty as f \to 0. With a.c. the supply voltage reverses continually, so the capacitor charges, discharges, charges the other way and discharges again, 50 times a second. Charge surges back and forth in the leads, and this counts as a current in the external circuit even though no charge crosses the dielectric. The higher the frequency, the more often this happens and the larger the current, so X_C falls as f rises.

9. (a) The quality factor Q measures the sharpness of resonance. It is the ratio of the voltage developed across the inductor (or the capacitor) at resonance to the applied voltage:

Equivalently Q = f_0/\Delta f, where \Delta f is the width of the resonance curve at 1/\sqrt{2} of the maximum current. A high Q means a small R, a tall and very narrow resonance peak, and large voltage magnification across L and C — so the circuit is highly selective, accepting one frequency and rejecting its neighbours. This is exactly what a radio tuner needs.

(b)

1. (a) See the derivation box below for the full argument. In outline: the same current I = I_0\sin\omega t flows through R, L and C, so I is the reference phasor. V_R = IR is in phase with I; V_L = IX_L leads by 90^{\circ}; V_C = IX_C lags by 90^{\circ} (Figure 18). Since V_L and V_C are anti-parallel, their resultant is (V_L - V_C) along the V_L direction, perpendicular to V_R. By Pythagoras,

If X_L > X_C the circuit is inductive and the current lags; if X_C > X_L it is capacitive and the current leads; if they are equal the circuit is purely resistive and in resonance.

(b)

Since X_L > X_C, the current lags the voltage by 53.1^{\circ}.

Check by the other route: P = I^{2}R = (4.0)^{2}\times30 = 480\ \text{W}. ✓ All the power is dissipated in R; L and C dissipate none.

2. (a) Resonance is the state of a series LCR circuit in which X_L = X_C, so the two reactances cancel completely. Then Z has its least possible value R, the current its greatest value V/R, and \tan\phi = 0, so the current is exactly in phase with the applied voltage and the power factor is 1. Physically, the energy pumped into the magnetic field of L during one quarter cycle is exactly what the electric field of C gives back, so the source has only to supply the I^{2}R heat.

The resonance curve (Figure 18, right) is a plot of I against f: it rises to a sharp maximum at f_0 and falls away on both sides. A smaller R gives a taller, narrower peak.

(b) The quality factor Q = \omega_0L/R = (1/R)\sqrt{L/C} is the voltage magnification across L or C at resonance, and equals f_0/\Delta f, where \Delta f is the bandwidth at 1/\sqrt{2} of the peak current. Significance: a high Q means a sharp, selective response — the circuit accepts a narrow band of frequencies and rejects all others, which is how a radio set picks out one station from many. It also means large voltages appear across L and C individually, which may greatly exceed the supply voltage and must be allowed for in the insulation.

(c)

At resonance Z = R = 20\ \Omega, so

Check: Q = V_L/V = 3160/200 = 15.8. ✓ Note that V_L is nearly 16 times the supply voltage, and V_C is equal and opposite to it, so the two cancel in the series sum.

3. (a) Pure resistor. With V = V_0\sin\omega t, Ohm's law gives I = (V_0/R)\sin\omega t at every instant: the current is in phase with the voltage, \phi = 0. The opposition is the resistance R, independent of frequency. Pure inductor. V_0\sin\omega t = L\,dI/dt, so integrating,

The current lags the voltage by 90^{\circ}, and the opposition is the inductive reactance X_L = \omega L = 2\pi fL, which grows with frequency. Pure capacitor. q = CV_0\sin\omega t, so

The current leads the voltage by 90^{\circ}, and the opposition is the capacitive reactance X_C = 1/\omega C = 1/2\pi fC, which falls as frequency rises.

(b) With V = V_0\sin\omega t and I = I_0\sin(\omega t-\phi), the instantaneous power is

Over a complete cycle the mean of \cos(2\omega t-\phi) is zero, so

Wattless current. Resolve I into a component I\cos\phi along V and a component I\sin\phi perpendicular to it. Only the first contributes to P_{av}; the perpendicular component I\sin\phi flows in the wires, heats them, and loads the generator, but contributes zero average power. It is therefore called the wattless or idle current. In a pure inductor or capacitor \phi = 90^{\circ} and the whole current is wattless.

(c) The supply must carry the full current I = P/(V\cos\phi), so a low power factor means a larger current for the same useful power. That larger current causes greater I^{2}R heating in the cables and transformers, a larger voltage drop along the line, and requires thicker conductors and bigger switchgear — all wasted cost. Industrial loads are largely inductive (motors), so capacitor banks are connected in parallel to bring \cos\phi back close to 1; electricity boards fine consumers whose power factor falls too low.

1. (a) (i) Parallel entry. The force eE is along the line of motion, so the electron moves in a straight line with uniform acceleration a = eE/m. If it enters against the field it decelerates uniformly, stops and returns. The speed changes but the direction does not. (ii) Perpendicular entry. The force is always at right angles to the initial velocity, so the horizontal component of velocity stays constant while the vertical component grows uniformly. This is exactly the situation of a projectile under gravity, and the path is a parabola. Once the electron leaves the field region it continues in a straight line along the tangent.

(b) Let the electron enter along x with speed v, and let the field E be along -y, so the force on the electron (charge -e) is eE along +y. The acceleration is a = eE/m, constant. Horizontally, x = vt, so t = x/v. Vertically, starting from rest in that direction,

Everything in the bracket is constant, so y \propto x^{2}, which is the equation of a parabola with its vertex at the point of entry.

2. (a) An electron of charge e starting from rest and falling through a potential difference V has work eV done on it by the field. Since no other force acts, by the work-energy theorem all of this appears as kinetic energy:

(b)

(About 2.8\% of the speed of light, so the non-relativistic formula is safe.)

3. (a) A charge e moving with velocity v at right angles to a field B feels a force

By Fleming's left-hand rule (or \vec{F} = q\vec{v}\times\vec{B}) this force is perpendicular to \vec{v}. A force always perpendicular to the velocity can never change the speed — it does no work — it only changes the direction. A constant-magnitude force always at right angles to the motion is exactly a centripetal force, so the path is a circle. Equating the magnetic force to the required centripetal force:

The period T = 2\pi m/eB is independent of the speed — the principle of the cyclotron.

(b)

4. (a) Thomson's apparatus. Electrons from a hot cathode are accelerated through a potential difference V towards an anode with a fine hole, which selects a narrow beam. The beam then passes between a pair of parallel plates that provide a uniform electric field E (vertical), and through the same region a magnetic field B is applied at right angles both to E and to the beam. The beam finally strikes a fluorescent screen, where its deflection is measured. Procedure. With the magnetic field off, the electric field deflects the spot. The magnetic field is then switched on and adjusted until the spot returns exactly to its undeflected position. At that setting the two forces balance:

Combining this with the energy equation for the accelerating stage, eV = \tfrac12mv^{2},

Thomson found e/m \approx 1.76\times10^{11}\ \text{C kg}^{-1}, the same for every gas and every cathode material — which is how he concluded that the electron is a universal constituent of all matter.

(b) See the boxed expression above. Crossed fields are needed because a single deflection measurement contains two unknowns, the speed v and the ratio e/m, and one equation cannot give both. Balancing an electric deflection against a magnetic one removes the deflection from the problem entirely and yields v = E/B by itself; only then can e/m be extracted. It also makes the result independent of the length of the plates and of the exact geometry of the screen.

5. (a) One electronvolt is the kinetic energy gained by an electron when it is accelerated through a potential difference of one volt.

It is a convenient unit because atomic and nuclear energies are absurdly small in joules; ionisation energies are a few eV, X-ray photons a few keV, and nuclear binding energies a few MeV.

(b)

(The second answer needs no calculation: an electron falling through V volts gains exactly V electronvolts — that is the whole point of the unit.)

6. (a) Millikan's oil-drop experiment. A fine spray of oil is blown above a pair of horizontal parallel plates; a few droplets fall through a small hole into the space between them. The droplets acquire charge by friction during spraying, or by ionisation of the air with X-rays. A single droplet is watched through a travelling microscope. With the plates uncharged, the droplet falls and quickly reaches a terminal velocity v, at which the viscous drag balances the weight; Stokes' law then gives the radius, and hence the mass, of the droplet:

A potential difference is then applied so that the electric force lifts the droplet, and V is adjusted until the droplet hangs stationary. Conclusion. Repeating this for hundreds of droplets, Millikan found that the charge on every droplet was a whole-number multiple of one basic amount, 1.6\times10^{-19}\ \text{C}. Charge is therefore quantised, q = ne, and e is the smallest free charge in nature. Combined with Thomson's e/m, this also gives the mass of the electron.

(b) For a stationary drop the upward electric force balances the weight less the upthrust of the air:

where d is the plate separation, \rho the density of the oil and \sigma that of air. In the simplified school version the upthrust is neglected and the condition is qV/d = mg.

7. (a) A velocity selector is an arrangement of mutually perpendicular electric and magnetic fields that allows only particles of one particular speed to pass straight through, whatever their charge or mass. A charged particle moving with speed v perpendicular to both fields feels an electric force qE one way and a magnetic force qvB the other. It passes undeflected only if these are equal:

The charge q cancels, so the selected speed depends only on the ratio of the field strengths. Faster particles are pushed aside by the magnetic force, slower ones by the electric force, and both are stopped by a slit.

(b) In a mass spectrometer the ions produced by the source emerge with a wide spread of speeds. If they were sent straight into the analysing magnetic field, the radius r = mv/qB would depend on both m and v, and ions of the same mass would arrive spread out over the detector while ions of different mass overlapped. The crossed-field velocity selector removes the speed spread first, so that every ion entering the analyser has the same v; then r \propto m/q alone and each isotope gives a sharp, well-separated line.

8. (a) In a cathode-ray oscilloscope the electron beam is produced by an electron gun: a heated cathode emits electrons by thermionic emission, a cylindrical grid at a negative potential controls how many get through (this sets the brightness), and two anodes at high positive potential accelerate the beam and, acting as an electrostatic lens, focus it to a sharp spot (the focus control). The beam then passes between two pairs of deflecting plates at right angles to each other: the Y-plates deflect it vertically in proportion to the signal voltage applied to them, while the X-plates receive a sawtooth "time-base" voltage that sweeps the spot across the screen at a steady rate and flies it back. The combination traces the graph of the signal against time on a fluorescent screen, which glows where the electrons strike it. A conducting coating returns the charge and prevents the screen charging up.

(b) Uses: (i) to display and measure the waveform, amplitude, frequency and period of a varying voltage; (ii) to compare the phase or frequency of two signals (Lissajous figures); (iii) as a very high-resistance voltmeter that does not load the circuit; (iv) in medicine and radar, for example to display an electrocardiogram or an echo trace.

1. (a) Apparatus. Electrons emitted thermionically from a hot cathode are accelerated towards a perforated anode by a potential difference V; the hole selects a narrow beam. The beam passes between two horizontal parallel plates that provide a uniform electric field E = V_p/d, and a pair of coils provides a uniform magnetic field B over the same region, at right angles both to E and to the beam. The beam finally strikes a fluorescent screen, where the position of the spot is read off a scale. The whole tube is highly evacuated. Derivation. In the accelerating stage all the work done on the electron becomes kinetic energy:

In the deflecting region the electric force is eE and the magnetic force is evB, and they are arranged to act in opposite directions. The magnetic field is adjusted until the spot returns exactly to its undeflected position, at which setting

The charge e cancels from both sides, so

Returning to the energy equation and dividing both sides by mV,

Substituting v = E/B,

Thomson found the same value whatever gas filled the tube and whatever metal formed the cathode — proof that this particle is a constituent of all matter.

(b) Because one deflection measurement contains two unknowns, v and e/m, and a single equation cannot determine both. Balancing an electric deflection against a magnetic one removes the deflection from the problem altogether and gives v = E/B on its own; only then can e/m be found. The null method is also far more accurate than measuring a deflection, and it makes the answer independent of the exact length of the plates and of the distance to the screen.

(c)

Check: this agrees with the accepted value, and v/c = 0.028, so the non-relativistic treatment is justified. ✓

2. (a) Let the electron enter along the x-axis with speed v, and let a uniform field E act at right angles to it. The force eE is constant in size and direction and acts only along y, so the motion splits into two independent parts: uniform velocity along x and uniform acceleration a = eE/m along y. This is exactly the situation of a projectile. Along x: \quad x = vt \;\Rightarrow\; t = x/v. Along y, starting from rest in that direction:

The bracket is a constant, so y \propto x^{2} — a parabola, with its vertex at the point of entry. Once the electron leaves the plates no force acts, so it travels on in a straight line along the tangent to the parabola at the exit point. Note that the speed changes, because the electric force does work on the electron.

(b) In a magnetic field the force on a charge moving at right angles to \vec{B} is

and, by \vec{F} = q\vec{v}\times\vec{B}, it is always perpendicular to \vec{v}. A force perpendicular to the velocity does no work, so the speed never changes; it only turns the velocity. A force of constant magnitude always perpendicular to the motion is precisely a centripetal force, so the path is a circle. Equating:

The period is the circumference divided by the speed:

The v cancels, so T is independent of the speed (and of the radius): a faster electron simply travels round a proportionately larger circle in the same time. This is the principle on which the cyclotron works, since a fixed-frequency accelerating voltage stays in step with the particles as they gain energy.

(c)

3. (a) Millikan's oil-drop experiment. Oil is sprayed as a fine mist above two horizontal parallel metal plates; a few droplets drift through a small hole into the space between them. The spraying, or a beam of X-rays, leaves each droplet with a small charge. One droplet is watched through a travelling microscope with a calibrated eyepiece scale, and the space is illuminated from the side and kept at constant temperature. Stage 1 — find the mass. With the plates uncharged, the droplet falls and quickly reaches a terminal velocity v, at which the viscous drag given by Stokes' law balances the weight less the upthrust:

Measuring v gives the radius r and hence the mass of the droplet. Stage 2 — balance it. A potential difference V is now applied so that the electric force is upward, and V is adjusted until the droplet hangs stationary:

from which the charge q is calculated. Conclusion. Repeating this for hundreds of droplets, and changing the charge on a droplet by irradiating it, Millikan found that every value of q was a whole-number multiple of one basic quantity:

No droplet ever carried a fraction of it. Electric charge is therefore quantised, and e is its indivisible unit. Combined with Thomson's e/m, this gives the mass of the electron, 9.1\times10^{-31}\ \text{kg}.

(b) A velocity selector is a region of mutually perpendicular electric and magnetic fields through which only particles of one particular speed can pass undeflected. A particle of charge q and speed v moving perpendicular to both fields feels an electric force qE and a magnetic force qvB in opposite directions; it goes straight only if

The charge and the mass both cancel, so the selected speed depends only on the ratio of the field strengths. Use: it is placed in front of the analysing magnet of a mass spectrometer, so that all the ions entering the analyser have the same speed; then the radius r = mv/qB depends on m/q alone and each isotope gives a sharp, separate line.

(c) One electronvolt is the kinetic energy acquired by an electron in falling freely through a potential difference of one volt:

For an electron accelerated through 200\ \text{V}:

1. (a) A photon is the quantum, or smallest indivisible packet, of electromagnetic radiation: light of frequency \nu can be emitted or absorbed only in whole units of energy h\nu. Properties: (i) its energy is E = h\nu = hc/\lambda, fixed entirely by the frequency and independent of the intensity; (ii) it has zero rest mass and always travels at c in vacuum; (iii) it carries momentum p = E/c = h/\lambda even though it has no rest mass; (iv) it is electrically neutral, so it is undeflected by electric and magnetic fields; (v) the intensity of a beam is the number of photons crossing unit area per second, not the energy of each; (vi) in an interaction with an electron, the photon is absorbed whole — it cannot give up part of itself.

(b)

2. (a) Laws of photoelectric emission. (i) For a given metal and a given frequency, the number of photoelectrons emitted per second (the photoelectric current) is directly proportional to the intensity of the incident light. (ii) For every metal there is a minimum frequency, the threshold frequency \nu_0, below which no emission occurs however intense the light or however long it shines. (iii) The maximum kinetic energy of the emitted electrons depends only on the frequency of the light and on the metal, and is completely independent of the intensity. (iv) Emission is instantaneous — within about 10^{-9}\ \text{s} of the light arriving — even for very feeble light.

(b) Laws (ii), (iii) and (iv) defeat the wave theory. On the wave picture, light energy arrives spread continuously over the whole surface, and an electron absorbs it gradually. Then: (ii) any frequency should eventually work, provided the light is intense enough or is left on long enough, so there should be no threshold at all; (iii) a more intense wave carries a larger electric field, so it should shake the electrons harder and eject them with more energy — the maximum kinetic energy should depend on intensity, and should not depend on colour; (iv) with very weak light an electron would need minutes or hours to collect enough energy from the spread-out wave, so emission should be delayed, not instantaneous. Experiment contradicts all three. Einstein's resolution is that light arrives as individual photons of energy h\nu, and one electron absorbs one whole photon in a single event.

3. (a) Einstein's photoelectric equation:

h\nu is the energy of the single incident photon, which is absorbed completely by one electron. W is the work function of the metal, the least energy needed to free an electron from the surface. \tfrac12mv_{max}^{2} is the maximum kinetic energy with which the electron emerges — it is a maximum because electrons that come from below the surface lose extra energy in collisions on the way out. Using W = h\nu_0, the equation may be written

which shows at once why no emission is possible for \nu < \nu_0: the kinetic energy would have to be negative.

(b) Work function W is the minimum energy required to remove an electron from the surface of a metal, usually quoted in eV. Threshold frequency \nu_0 = W/h is the lowest frequency of light that can just cause emission, with the electron leaving with zero speed. Threshold wavelength \lambda_0 = hc/W = c/\nu_0 is the corresponding longest wavelength; light of wavelength greater than \lambda_0 can never eject electrons from that metal.

4. (a) The stopping potential V_s is the least negative potential that must be applied to the collecting anode to stop even the fastest photoelectron from reaching it, so that the photoelectric current falls to exactly zero. An electron of charge e moving against a retarding potential V_s loses kinetic energy eV_s. The fastest electron is just stopped when

Substituting this into Einstein's equation h\nu = W + \tfrac12mv_{max}^{2},

(b) Photon energy at 400\ \text{nm} is 3.11\ \text{eV} (from question 1), so

5. (a) From V_s = (h/e)\nu - W/e, a graph of stopping potential against frequency is a straight line. It does not pass through the origin: it meets the \nu-axis at \nu_0 and, if extended backwards, meets the V_s-axis at -W/e. - The slope is h/e. Measuring the slope and multiplying by the known e gives Planck's constant: h = e\times\text{slope}. This is how Millikan measured h in 1916, and it agreed with the value from black-body radiation. - The intercept on the V_s-axis is -W/e, so W = e\times|\text{intercept}|. A striking feature is that the line has the same slope for every metal; changing the metal only shifts the line sideways. That universality is strong evidence that h is a constant of nature, not a property of the material.

(b) The intercept on the frequency axis is the threshold frequency \nu_0 of that metal — the frequency at which the stopping potential is zero, meaning the electrons just barely escape with no kinetic energy. It is related to the work function by W = h\nu_0.

6. (a)

(b) No. The incident wavelength 700\ \text{nm} is greater than the threshold wavelength 622\ \text{nm}, so its frequency is below \nu_0. Checking directly, the photon energy is

which is less than the work function 2.0\ \text{eV}. A single photon cannot free an electron, and since an electron absorbs only one photon at a time, no amount of extra intensity and no length of exposure will help — increasing the intensity merely sends more photons, each still too weak.

7. (a) A photon has zero rest mass, so the relativistic energy-momentum relation E^{2} = p^{2}c^{2}+m_0^{2}c^{4} reduces to E = pc. Hence

(Equivalently, using the equivalent mass m = E/c^{2} = h\nu/c^{2} and p = mc gives the same result.) The existence of this momentum is confirmed by radiation pressure, by the Compton effect and by the tails of comets, which are pushed away from the Sun.

(b) Energy of one photon:

Number emitted per second:

The number is enormous, which is exactly why the graininess of light is never noticed in everyday life.

8. (a) The photoelectric effect shows three things a wave cannot do. (i) The threshold. If the energy of a wave were spread out, faint red light shining for long enough would accumulate enough energy to free an electron; in fact red light never works on a metal whose threshold lies in the green, however bright. This is only sensible if the energy comes in indivisible lumps of size h\nu, and a single lump must be big enough by itself. (ii) Energy depends on colour, not brightness. Doubling the intensity doubles the number of electrons but leaves their maximum energy exactly unchanged; changing the colour changes the energy. So each electron deals with one photon, and the photon's energy is fixed by \nu. (iii) No time delay. Even at intensities so low that classical theory predicts an hour's wait, the first electrons appear within nanoseconds — because the energy is delivered in a single concentrated event, not gathered gradually. Together these force the picture of light as a stream of particles, each carrying energy h\nu and momentum h/\lambda. (Interference and diffraction equally force the wave picture, and both are true: this is wave-particle duality.)

(b) Applications: (i) in photocells used for automatic street lighting, burglar alarms, automatic doors and counting devices on production lines; (ii) in light meters for photography and in solar cells for electrical power; (iii) in the sound track of film projectors, where the varying light through the track becomes a varying current; (iv) in photomultiplier tubes for detecting very faint light in astronomy and in nuclear particle detectors; (v) in television camera tubes and image intensifiers for night vision.

1. (a) The four laws are as stated in short answer 2(a): current proportional to intensity; a threshold frequency below which nothing happens; maximum kinetic energy determined by frequency alone; and instantaneous emission. The wave theory explains only the first. It fails on the other three. A wave delivers energy continuously and over the whole illuminated area, so (i) there should be no threshold — low-frequency light, made intense enough or left on long enough, should always work; (ii) a more intense wave has a larger amplitude and therefore a larger electric field, so it should give the ejected electrons more kinetic energy, whereas experiment shows the maximum energy is completely independent of intensity and rises only with frequency; (iii) with a very feeble beam an electron would have to collect energy for minutes, so there should be a measurable delay, but none is found.

(b) Derivation. Einstein assumed that light of frequency \nu consists of photons of energy h\nu, and that a photoelectron absorbs one whole photon in a single act. The energy so received is spent in two ways: a minimum amount W must be spent in freeing the electron from the attraction of the metal surface, and whatever remains appears as kinetic energy. For an electron that happens to be right at the surface and loses nothing on the way out, the remainder is a maximum:

Work function W is the minimum energy needed to liberate an electron from the metal surface. Threshold frequency \nu_0 is the frequency for which h\nu_0 = W exactly, so the electron just escapes with zero kinetic energy; with W = h\nu_0,

For \nu < \nu_0 the right-hand side is negative, which is impossible — hence no emission at all, at any intensity.

(c)

Check: 622\ \text{nm} > 400\ \text{nm}, so the incident light is indeed energetic enough — consistent with a positive KE_{max}. ✓

2. (a) The stopping potential V_s is the smallest retarding potential difference applied to the anode that reduces the photoelectric current to zero, by turning back even the fastest electron. As an electron of charge e climbs through a retarding potential V_s it loses energy eV_s; the fastest electron is just brought to rest when

Einstein's equation is

Rearranging for V_s and dividing throughout by e:

(b) This is of the form y = mx+c, so the graph of V_s against \nu is a straight line of slope h/e and intercept -W/e on the V_s-axis, cutting the \nu-axis at \nu_0 = W/h. For two metals A and B with W_A < W_B, the lines are parallel (same slope h/e), with B's line displaced to the right and its threshold frequency larger. - To find h: measure the slope of either line and multiply by e; h = e\times\text{slope}. - To find W: read the intercept on the V_s axis and multiply by e, W = e\times|{-W/e}|; or read \nu_0 from the \nu-axis and use W = h\nu_0. The fact that both lines have the same slope, whatever the metal, is direct experimental proof that h is a universal constant.

(c)

3. (a) A photon is the quantum of electromagnetic radiation — the smallest indivisible packet in which light of a given frequency can be emitted or absorbed. Energy. Planck's hypothesis, which Einstein extended to free radiation, gives