Magnetic field

Before 1820 electricity and magnetism were two separate sciences. Then Oersted noticed a compass needle twitch beside a current-carrying wire, and within ten years Biot, Savart and Ampère had turned that twitch into a complete theory: a moving charge makes a magnetic field, and a magnetic field pushes a moving charge. This unit is the longest in the electricity section because every motor, loudspeaker, galvanometer and Hall sensor you will ever meet is an application of those two sentences.

16.1 Magnetic field lines and magnetic flux; Oersted's experiment

A magnetic field is the region around a magnet or a current in which a magnetic force can be detected. It is a vector field \vec{B}, called the magnetic flux density, and its SI unit is the tesla (T), where 1\ \text{T} = 1\ \text{N A}^{-1}\text{m}^{-1} = 1\ \text{Wb m}^{-2}.

A magnetic field line is a line drawn so that its tangent at every point gives the direction of \vec{B} there. Their properties:

• They run from N to S outside a magnet and from S to N inside it, so every line is a closed loop — unlike electric field lines, which begin and end on charges.
• The number of lines per unit area crossing a surface at right angles measures the strength of \vec{B}: crowded lines mean a strong field.
• Two field lines never intersect. If they did, the field would have two directions at one point, which is impossible.
• Lines are in tension along their length and repel sideways — which is why unlike poles attract and like poles repel.

Because every field line is a closed loop, every line that enters a closed surface must leave it again. Hence Gauss's law for magnetism:

The net magnetic flux through any closed surface is zero: isolated magnetic poles (monopoles) do not exist.

Oersted's experiment

Hans Christian Oersted placed a straight wire above a pivoted compass needle and parallel to it, i.e. along the magnetic north–south line. He found:

1. with no current, the needle rested along N–S;
2. with a current, the needle turned through a large angle and settled almost perpendicular to the wire;
3. reversing the current reversed the deflection;
4. the deflection increased with the current and decreased as the wire was moved further away.

Conclusion: an electric current produces a magnetic field around it.

16.2 Force on moving charge; Force on a conductor

A charge q moving with velocity \vec{v} in a field \vec{B} experiences

where \theta is the angle between \vec{v} and \vec{B}.

Three consequences follow straight from F = qvB\sin\theta:

• If \vec{v} is parallel or antiparallel to \vec{B} (\theta = 0 or 180^{\circ}) the force is zero — the particle goes straight on.
• The force is greatest when \vec{v} \perp \vec{B}, and then F = qvB.
• \vec{F} is always perpendicular to \vec{v}, so the magnetic force does no work. It can change the direction of motion but never the speed or the kinetic energy.

Direction. Use Fleming's left-hand rule for the force on a positive charge (or a current): stretch the first three fingers of the left hand mutually perpendicular — Forefinger = Field, Centre finger = Current, Thumb = Thrust. For a negative charge, reverse the answer.

Force on a current-carrying conductor

A current is a stream of moving charges, so a wire in a magnetic field feels a force.

16.3 Force and Torque on rectangular coil; Moving coil galvanometer

Consider a rectangular coil of N turns, length l and breadth b, carrying a current I in a uniform field \vec{B} parallel to the plane of the coil.

• \tau is maximum = NBIA when the plane of the coil contains \vec{B} (\alpha = 0, \theta = 90^{\circ}).
• \tau is zero when the plane is perpendicular to \vec{B} (\theta = 0) — the coil is then in stable equilibrium.
• The result is independent of the shape of the coil: any plane coil of area A gives \tau = NBIA\sin\theta.

Moving coil galvanometer

Construction. A rectangular coil of many turns of fine insulated copper wire is wound on a light non-magnetic aluminium frame. It hangs by a thin phosphor-bronze suspension strip between the concave pole pieces of a strong permanent horseshoe magnet, with a cylindrical soft-iron core fixed inside it (not touching). The suspension strip carries the current in and provides the restoring couple; a fine spring below takes the current out. A small mirror on the strip and a lamp-and-scale arrangement read the deflection.

Why the field must be radial. The concave pole faces plus the soft-iron core make the field in the gap point everywhere along a radius of the cylinder. So however far the coil turns, its plane always stays parallel to \vec{B}, i.e. \theta = 90^{\circ} and \sin\theta = 1 always.

Quantity	Definition	Expression
Current sensitivity	deflection per unit current	\varphi/I = NBA/c
Voltage sensitivity	deflection per unit p.d.	\varphi/V = NBA/cG

To make a galvanometer more sensitive: increase N, A and B, and decrease c (hence phosphor bronze, which has a very small torsional constant). The soft-iron core does double duty — it makes the field radial and concentrates the flux, increasing B.

16.4 Hall effect

Uses of the Hall effect.

• The sign of V_H gives the sign of the charge carriers — this is how we know that conduction in an n-type semiconductor is by electrons and in a p-type by holes.
• From R_H = 1/nq we get the carrier concentration n, and hence the drift velocity.
• A calibrated slab is a Hall probe, the standard instrument for measuring magnetic flux density.
• Hall sensors count the revolutions of motors and wheels and detect whether a phone's flip cover is closed.

16.5 Magnetic field of a moving charge

A single charge q moving with velocity \vec{v} sets up a magnetic field at a point P whose position vector from the charge is \vec{r}:

where \theta is the angle between \vec{v} and \vec{r}, and

is the permeability of free space. Points to note:

• B = 0 along the line of motion (\theta = 0 or 180^{\circ}) and is greatest in the plane through the charge perpendicular to \vec{v}.
• \vec{B} is perpendicular to both \vec{v} and \vec{r}; the lines are circles centred on the line of motion — the right-hand grip rule again.
• A stationary charge (v = 0) produces no magnetic field, only an electric one.

For example, an electron moving at 2.0\times10^{7}\ \text{m s}^{-1} produces, at a point 1.0\ \text{nm} away at right angles to its motion,

— an enormous field, which is why atomic-scale magnetism is so strong.

Adding up the fields of all the carriers in a short length dl of wire, with I = nAqv_d, converts this formula into the Biot–Savart law of the next section: q\vec{v} \rightarrow I\,d\vec{l}.

16.6 Biot and Savart law and application to circular coil, long straight conductor, long solenoid

So dB \propto I, dB \propto dl, dB \propto \sin\theta and dB \propto 1/r^{2}. The total field is found by integrating over the whole conductor.

(a) Long straight conductor

Let P be at perpendicular distance a from a straight wire. Taking an element at distance l from the foot of the perpendicular and measuring the angle \phi from that perpendicular, integration gives the standard result

where \phi_1 and \phi_2 are the angles subtended at P by the two ends. For an infinitely long wire \phi_1 = \phi_2 = 90^{\circ}, so

and for a semi-infinite wire ending at the foot of the perpendicular, B = \mu_0 I/4\pi a.

(b) Circular coil — field at the centre

(c) Long solenoid

For a solenoid with n turns per unit length, integrating the on-axis coil formula over all the turns gives, at an axial point,

where \alpha_1 and \alpha_2 are the semi-angles subtended by the two ends. Two limits matter:

Position	Angles	Field
Middle of a very long solenoid	\alpha_1 = \alpha_2 = 90^{\circ}	B = \mu_0 n I
At one end of a long solenoid	\alpha_1 = 0, \alpha_2 = 90^{\circ}	B = \frac{1}{2}\mu_0 n I

16.7 Ampere's law and applications to long straight conductor, straight solenoid, toroidal solenoid

Ampère's law is to magnetism what Gauss's law is to electrostatics: always true, but only useful where the symmetry lets you take B outside the integral.

(a) Long straight conductor

Choose a circle of radius r centred on the wire, in the plane perpendicular to it. By symmetry B has the same magnitude everywhere on it and is everywhere tangential, so \vec{B}\cdot d\vec{l} = B\,dl:

in two lines, where Biot–Savart needed an integration. Inside a solid cylindrical conductor of radius R carrying a uniform current, the loop of radius r < R encloses only the fraction I r^{2}/R^{2}, giving

so B rises linearly from zero on the axis to \mu_0I/2\pi R at the surface, then falls as 1/r outside.

(b) Straight solenoid

(c) Toroidal solenoid

A toroid is a solenoid bent into a ring of N turns. Take a circular Amperian loop of radius r running along the middle of the core. By symmetry B is constant and tangential along it, and the loop is threaded by all N turns:

Writing n = N/2\pi r for the turns per unit length along the core recovers B = \mu_0 n I. For a loop drawn inside the hole or outside the toroid, the enclosed current is zero, so B = 0 there: a toroid confines its field completely inside the windings — which is why transformer and inductor cores are toroidal.

16.8 Force between two parallel conductors carrying current; definition of ampere

Two long parallel wires a distance d apart carry currents I_1 and I_2.

Check it: with I_1 = I_2 = 1\ \text{A} and d = 1\ \text{m},

A modern footnote. Since 20 May 2019 the ampere has been defined instead by fixing the elementary charge at e = 1.602176634\times10^{-19}\ \text{C}, so that 1\ \text{A} is the current corresponding to 1/e elementary charges per second. As a result \mu_0 is now a measured quantity, though it still equals 4\pi\times10^{-7}\ \text{T m A}^{-1} to within a few parts in 10^{10}. NEB papers still ask for the force-based definition above.

Chapter summary

• Magnetic flux \Phi = BA\cos\theta (weber); field lines are closed loops, so \oint\vec{B}\cdot d\vec{A} = 0 — there are no magnetic monopoles.
• Oersted: a current produces a magnetic field. Direction by the right-hand grip rule (or SNOW).
• Force on a moving charge \vec{F} = q\vec{v}\times\vec{B}, F = qvB\sin\theta; it does no work. Perpendicular entry gives a circle with r = mv/qB, T = 2\pi m/qB, independent of speed.
• Force on a conductor F = BIL\sin\theta; direction by Fleming's left-hand rule.
• Torque on a coil \tau = NBIA\sin\theta = NBIA\cos\alpha, with dipole moment m = NIA. A radial field makes \sin\theta = 1, so a moving-coil galvanometer has \varphi = (NBA/c)I and a uniform scale.
• Hall effect: V_H = IB/nqt with R_H = 1/nq; it gives the sign and the concentration of the carriers.
• Biot–Savart: dB = (\mu_0/4\pi)(I\,dl\sin\theta/r^{2}). Results: straight wire \mu_0I/2\pi a; centre of a coil \mu_0NI/2a; on the axis \mu_0NIa^{2}/2(a^{2}+x^{2})^{3/2}; long solenoid \mu_0nI (half that at an end).
• Ampère: \oint\vec{B}\cdot d\vec{l} = \mu_0I_{enc}, which gives the straight wire, the solenoid (\mu_0nI) and the toroid (\mu_0NI/2\pi r, zero outside) in a few lines each.
• Parallel wires: F/L = \mu_0I_1I_2/2\pi d; like currents attract. This force defines the ampere as 2\times10^{-7}\ \text{N m}^{-1} for 1\ \text{A} at 1\ \text{m}.

Practice questions

Group A — Multiple choice (1 mark each)

1. The SI unit of magnetic flux is [1]ateslabweberchenrydampere-metre
2. A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is [1]aqvBbzerocqvB\sin 45^{\circ}dmaximum
3. A charged particle enters a uniform magnetic field at right angles to it. Its path is [1]aa straight lineba parabolaca circleda helix
4. The magnetic flux density at the centre of a circular coil of N turns and radius a carrying a current I is [1]a\mu_0NI/2\pi ab\mu_0NI/2ac\mu_0NI/4\pi ad\mu_0NIa
5. Two long parallel wires carry currents in the same direction. They [1]aattract each otherbrepel each othercexert no forcedrotate about each other
6. The magnetic flux density inside a very long solenoid of n turns per metre carrying current I is [1]a\mu_0nI/2b\mu_0nIc\mu_0nI/2\pidzero
7. The field of a moving-coil galvanometer is made radial so that [1]athe coil does not rotatebthe deflection is proportional to the currentcthe coil resistance is reduceddthe torque is zero
8. The Hall voltage of a slab of thickness t is proportional to [1]atb1/tct^{2}d\sqrt{t}

Answers to Group A

1. (b) — \Phi = BA\cos\theta, measured in weber = T m².

2. (b) — \sin 0^{\circ} = 0, so F = qvB\sin\theta = 0.

3. (c) — the force stays perpendicular to v and constant, so it is centripetal.

4. (b) — B = \mu_0NI/2a; the 2\pi form belongs to a straight wire.

5. (a) — like currents attract.

6. (b) — Ampère's law on a rectangular loop gives B = \mu_0nI.

7. (b) — with \sin\theta = 1 always, \varphi = (NBA/c)I, a uniform scale.

8. (b) — V_H = IB/nqt.

Group B — Short answer (5 marks each)

1. Define magnetic flux and state its SI unit. Describe Oersted's experiment and state the conclusion drawn from it. [5]
2. Starting from the force on a single moving charge, derive the expression F = BIL\sin\theta for the force on a current-carrying conductor in a magnetic field. [5]
3. An electron (m = 9.1\times10^{-31}\ \text{kg}, e = 1.6\times10^{-19}\ \text{C}) moves at 3.0\times10^{7}\ \text{m s}^{-1} at right angles to a magnetic field of 0.20\ \text{T}. Find the radius of its path and its period of revolution. [5]
4. State the Biot–Savart law and use it to find the magnetic flux density at the centre of a circular coil of N turns and radius a carrying current I. [5]
5. Explain the Hall effect and show that V_H = IB/nqt. A metal strip of thickness 1.0\ \text{mm} with n = 5.0\times10^{28}\ \text{m}^{-3} carries 20\ \text{A} in a field of 1.5\ \text{T}; find the Hall voltage. [5]
6. Two long parallel wires 20\ \text{cm} apart carry currents of 10\ \text{A} and 15\ \text{A} in opposite directions. Find the force per metre between them and state its nature. Hence define the ampere. [5]

Answers to Group B

3.

r = \frac{mv}{eB} = \frac{(9.1\times10^{-31})(3.0\times10^{7})} {(1.6\times10^{-19})(0.20)} = \frac{2.73\times10^{-23}}{3.2\times10^{-20}} = 8.53\times10^{-4}\ \text{m} = 0.853\ \text{mm}

T = \frac{2\pi m}{eB} = \frac{2\pi(9.1\times10^{-31})}{3.2\times10^{-20}} = 1.79\times10^{-10}\ \text{s}

5. Derivation as in §16.4. Numerically, nq = (5.0\times10^{28})(1.6\times10^{-19}) = 8.0\times10^{9}, so

V_H = \frac{IB}{nqt} = \frac{(20)(1.5)}{(8.0\times10^{9})(1.0\times10^{-3})} = \frac{30}{8.0\times10^{6}} = 3.75\times10^{-6}\ \text{V} = 3.75\ \mu\text{V}

6.

\frac{F}{L} = \frac{\mu_0I_1I_2}{2\pi d} = \frac{(2\times10^{-7})(10)(15)}{0.20} = \frac{3.0\times10^{-5}}{0.20} = 1.5\times10^{-4}\ \text{N m}^{-1}

The currents are opposite, so the force is repulsive. Definition of the ampere: see §16.8.

Group C — Long answer (8 marks each)

1. (a) State the Biot–Savart law and use it to derive the magnetic flux density at a perpendicular distance a from a long straight conductor carrying a current I. [5](b) A long straight wire carries 20\ \text{A}. Find the flux density 10\ \text{cm} from it. A circular coil of 100 turns and radius 10\ \text{cm} carries 2.0\ \text{A}; find the flux density at its centre. [3]
2. (a) Derive an expression for the torque on a rectangular coil carrying a current in a uniform magnetic field. Explain, with a diagram, why the field of a moving-coil galvanometer is made radial and show that its scale is uniform. [5](b) A galvanometer coil of 200 turns and area 2.0\times10^{-4}\ \text{m}^{2} hangs in a radial field of 0.20\ \text{T}. The torsional constant of the suspension is 1.0\times10^{-6}\ \text{N m} per degree. Find the deflection produced by 5.0\ \text{mA} and the current sensitivity. [3]
3. (a) State Ampère's circuital law and apply it to find the magnetic flux density inside a long straight solenoid and inside a toroidal solenoid. [5](b) A toroid of mean radius 15\ \text{cm} has 500 turns and carries 3.0\ \text{A}. Find the flux density along its mean circle and the flux density outside the toroid. [3]

Answers to Group C

1. (b) Straight wire:

B = \frac{\mu_0 I}{2\pi a} = \frac{(2\times10^{-7})(20)}{0.10} = 4.0\times10^{-5}\ \text{T}

Circular coil:

B = \frac{\mu_0 NI}{2a} = \frac{(4\pi\times10^{-7})(100)(2.0)}{2(0.10)} = \frac{2.513\times10^{-4}}{0.20} = 1.26\times10^{-3}\ \text{T}

2. (b)

\tau = NBIA = (200)(0.20)(5.0\times10^{-3})(2.0\times10^{-4}) = 4.0\times10^{-5}\ \text{N m}

\varphi = \frac{\tau}{c} = \frac{4.0\times10^{-5}}{1.0\times10^{-6}} = 40\ \text{degrees}

Current sensitivity = \dfrac{NBA}{c} = \dfrac{(200)(0.20)(2.0\times10^{-4})}{1.0\times10^{-6}} = 8.0\times10^{3} degrees per ampere, i.e. 8 degrees per milliampere.

3. (b)

B = \frac{\mu_0 NI}{2\pi r} = \frac{(2\times10^{-7})(500)(3.0)}{0.15} = \frac{3.0\times10^{-4}}{0.15} = 2.0\times10^{-3}\ \text{T}

Outside the toroid (and in the central hole) the Amperian loop encloses no net current, so B = 0.