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Magnetic Field and Magnetic Forces

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Senior 5 · Physics

Magnetic Field and Magnetic Forces

4 lessons

Lesson 1 of 4

Magnetic Fields and the Fields of Currents

25 min

A compass that twitched

In April 1820, during a lecture in Copenhagen, Hans Christian Ørsted noticed that a compass needle swung when he switched on a current in a nearby wire. It was the first evidence that electricity and magnetism are connected: an electric current produces a magnetic field. Within months André-Marie Ampère, Jean-Baptiste Biot and Félix Savart had found the laws that give the size of that field. Those laws, which this lesson develops, are used today to design every electromagnet, loudspeaker, MRI scanner and electric motor.

Magnetic fields and their sources

A magnetic field is the region round a magnet or a current in which a magnetic force acts on another magnet, on a moving charge or on a current. It is a vector quantity, the magnetic flux density $\vec{B}$ , measured in teslas (T); its exact definition, from the force on a moving charge, comes in the next lesson. Field lines show its direction (the way the north pole of a small compass points) and their spacing shows its strength.

Every magnetic field is produced by moving electric charge:

Currents in wires and coils.
Permanent magnets: each electron behaves like a tiny current loop because of its spin and orbital motion. In iron, nickel and cobalt these tiny magnets line up in regions called domains.
The Earth: currents in its molten iron outer core produce a field of about 25 μT to 65 μT at the surface, weakest near the equator.

Source	Typical B
Earth's surface	about $5 \times 10^{-5}$ T
Fridge magnet	about $5 \times 10^{-3}$ T
Strong neodymium magnet	about 1 T
Hospital MRI scanner	1.5 T to 3 T

The field of a long straight wire

Ørsted's field forms circles round the wire. Their direction is given by the right-hand grip rule: grip the wire with your right hand, thumb pointing along the conventional current, and your fingers curl in the direction of the field.

The size of the field is given by Ampère's law: around any closed path, the sum of $B_{\parallel}\,\Delta l$ (the component of $\vec{B}$ along the path times each small length) equals $\mu_0$ times the current enclosed,

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}

where $\mu_0 = 4\pi \times 10^{-7}\,\mathrm{T\,m\,A^{-1}}$ is the permeability of free space. For a circle of radius $r$ centred on a long wire, $B$ is the same all the way round and always along the path, so $B \times 2\pi r = \mu_0 I$ :

B = \frac{\mu_0 I}{2\pi r}

The field is proportional to the current and inversely proportional to the distance.

Definition

Magnetic flux density $\vec{B}$ is the vector that measures the strength and direction of a magnetic field; its unit is the tesla (T). The permeability of free space $\mu_0 = 4\pi \times 10^{-7}\,\mathrm{T\,m\,A^{-1}}$ is the constant that links a current to the field it produces in a vacuum (air is almost the same). Example: 1.0 A in a long wire gives $B = \frac{(4\pi \times 10^{-7})(1.0)}{2\pi(0.010)} = 2.0 \times 10^{-5}$ T at 1.0 cm.

Worked example 1: a wire beside a compass

A long straight wire carries 10 A. (a) Find the field 5.0 cm from it. (b) Compare it with the Earth's field (about $3 \times 10^{-5}$ T near the equator). (c) Two such wires, 10 cm apart, carry 5.0 A in opposite directions. Find the field midway between them.

Solution

(a) $B = \frac{(4\pi \times 10^{-7})(10)}{2\pi(0.050)} = 4.0 \times 10^{-5}$ T.

(b) It is comparable with the Earth's field, so a compass placed there swings well away from north, as in Ørsted's lecture.

(c) Each wire gives $\frac{(4\pi \times 10^{-7})(5.0)}{2\pi(0.050)} = 2.0 \times 10^{-5}$ T at the midpoint. By the grip rule, opposite currents give fields in the same direction between the wires, so they add: $4.0 \times 10^{-5}$ T. (With currents in the same direction the two fields would cancel there.)

The field at the centre of a circular coil

For shapes without enough symmetry for Ampère's law, the Biot–Savart law gives the field $d B$ of each small piece $d l$ of a wire carrying current $I$ , at distance $r$ :

dB = \frac{\mu_0}{4\pi} \cdot \frac{I\,dl\sin\theta}{r^2}

where $\theta$ is the angle between $d l$ and the line to the point. At the centre of a circular loop of radius $R$ , every piece is at distance $R$ and at right angles to the radius ( $\sin\theta = 1$ ), and every piece's field points the same way, along the axis. Adding the pieces, $\sum dl = 2\pi R$ :

B = \frac{\mu_0}{4\pi} \cdot \frac{I(2\pi R)}{R^2} = \frac{\mu_0 I}{2R}

For a flat coil of $N$ turns, $B = \frac{\mu_0 N I}{2R}$ .

The field inside a solenoid

A solenoid is a long coil. Inside it the fields of the turns add to give a strong, almost uniform field along the axis; outside, the field is weak and has the shape of a bar magnet's field.

Apply Ampère's law to a rectangle of length $L$ with one side inside the solenoid along the axis and the other far outside. Only the inside side contributes, $B L$ , and the rectangle encloses $n L$ turns, where $n$ is the number of turns per metre. So $BL = \mu_0 nLI$ , and

B = \mu_0 n I = \frac{\mu_0 N I}{L}

The field is independent of the solenoid's radius and of the position inside it, except near the ends, where it falls to half. Filling the solenoid with soft iron multiplies the field by hundreds or thousands: that is an electromagnet.

Worked example 2: a coil and a solenoid

(a) A flat circular coil of 50 turns and radius 10 cm carries 2.0 A. Find the field at its centre. (b) A solenoid 40 cm long has 800 turns and carries 3.0 A. Find the field inside it.

Solution

(a) $B = \frac{\mu_0 N I}{2R} = \frac{(4\pi \times 10^{-7})(50)(2.0)}{2(0.10)} = 6.3 \times 10^{-4}$ T.

(b) $n = \frac{800}{0.40} = 2000$ turns per metre; $B = \mu_0 n I = (4\pi \times 10^{-7})(2000)(3.0) = 7.5 \times 10^{-3}$ T, about 150 times the Earth's field, and uniform over most of the inside.

Worked example 3: designing a field

A student wants a field of $3.0 \times 10^{-5}$ T at 2.0 cm from a long straight wire. What current is needed? Explain why the experiment should be done with the wire running north–south.

Solution

$I = \frac{2\pi r B}{\mu_0} = \frac{2\pi(0.020)(3.0 \times 10^{-5})}{4\pi \times 10^{-7}} = 3.0$ A.

A wire running north–south makes a field that is east–west at a compass placed beside or below it, at right angles to the Earth's horizontal field, so the needle's deflection shows the effect clearly. (Equal fields at right angles would turn the needle through 45°.)

Warning

Which formula? $\frac{\mu_0 I}{2\pi r}$ is for a long straight wire at distance $r$ ; $\frac{\mu_0 N I}{2R}$ is only for the centre of a flat coil; $\mu_0 n I$ is for inside a long solenoid, with $n$ in turns per metre. Mixing them up, or forgetting to convert cm to m, are the commonest errors.

Warning

Safety. Short pieces of thick wire connected straight across a cell or a low-voltage supply carry very large currents and get hot quickly. Use a rheostat or a current-limited supply, and switch off between readings.

Tip

Coil polarity trick: look at the end of a coil. If the current flows anticlockwise, that end is a North pole; if clockwise, a South pole (draw an N or S with arrows on its tips to remember).

What Oersted Discovered with his Compass, National MagLab. Watch the compass needle swing when the current is switched on, and reverse when the current is reversed.

✦

Practice

Name three sources of magnetic fields and state what they all have in common.
Find the magnetic field 4.0 cm from a long straight wire carrying 6.0 A.
At what distance from a long wire carrying 15 A is the field $5.0 \times 10^{-5}$ T?
A circular coil of 200 turns and radius 5.0 cm carries 0.50 A. Find the field at its centre.
A solenoid of length 25 cm has 500 turns. What current gives a field of 4.0 mT inside it?
Two long parallel wires 8.0 cm apart each carry 4.0 A in the same direction. Find the field (a) midway between them, (b) at a point 4.0 cm outside one wire, on the line joining them.
Explain why the field inside a long solenoid does not depend on its radius.
A technician wants to double the field inside a solenoid. Evaluate three ways of doing so: doubling the current, doubling the length (same number of turns), and winding twice as many turns on the same length.

Show answersHide answers

Electric currents, permanent magnets, the Earth's core (or any of these). All are produced by moving electric charges.
$B = \frac{(4\pi \times 10^{-7})(6.0)}{2\pi(0.040)} = 3.0 \times 10^{-5}$ T.
$r = \frac{\mu_0 I}{2\pi B} = \frac{(4\pi \times 10^{-7})(15)}{2\pi(5.0 \times 10^{-5})} = 0.060$ m.
$B = \frac{(4\pi \times 10^{-7})(200)(0.50)}{2(0.050)} = 1.3 \times 10^{-3}$ T.
$n = 2000$ m⁻¹; $I = \frac{B}{\mu_0 n} = \frac{4.0 \times 10^{-3}}{(4\pi \times 10^{-7})(2000)} = 1.6$ A.
(a) Zero: equal fields in opposite directions. (b) Distances 4.0 cm and 12 cm; fields in the same direction: $2.0 \times 10^{-5} + 6.7 \times 10^{-6} = 2.7 \times 10^{-5}$ T.
By Ampère's law the field depends only on the current enclosed per unit length of the rectangular path, $n I$ ; the radius does not appear. A wider coil has longer turns, but each is further from the axis, and the two effects cancel.
Doubling $I$ doubles $B$ but quadruples the heating ( $I^{2} R$ ). Doubling the length with the same turns halves $n$ and so halves $B$ : wrong way. Twice the turns on the same length doubles $n$ and $B$ , at the cost of more wire, more resistance and more heat for the same current. The best choice depends on the supply and cooling; often adding turns (or an iron core) is preferred.

Key points

All magnetic fields come from moving charges: currents, electron spins in magnets, the Earth's core.
Long straight wire: $B = \frac{\mu_0 I}{2\pi r}$ , circles given by the right-hand grip rule.
Centre of a flat coil: $B = \frac{\mu_0 NI}{2R}$ (Biot–Savart). Inside a long solenoid: $B = \mu_0 nI$ , uniform (Ampère's law).
$\mu_0 = 4\pi \times 10^{-7}\,\mathrm{T\,m\,A^{-1}}$ .

Sources and credits

Image: Magnet0873.png, from Practical Physics by Newton Henry Black (1913), Wikimedia Commons, public domain.
Video: What Oersted Discovered with his Compass, National MagLab, embedded from YouTube.

Written by Ganzaa. The two field diagrams, the worked examples and the practice exercises are by Ganzaa.

1A compass that twitched

2Magnetic fields and their sources

3The field of a long straight wire

4The field at the centre of a circular coil

5The field inside a solenoid

A compass that twitched

Magnetic fields and their sources

The field of a long straight wire

The field at the centre of a circular coil

The field inside a solenoid