Before considering the mathematical nature of the forces on currents in magnetic fields it is worth just looking at the simple magnetic field diagrams that give rise to these effects. These are shown in Figure 1. (a) is the field between two magnets, (b) the field due to a current in a straight wire and (c) the resulting field if they are put together. This last field is known as the "catapult" field because it tends to catapult the wire out of the field in the direction shown by the arrow.

If a coil carrying a current is placed in a magnetic field it will experience a force on two of its sides in such a way as to make the coil rotate. This effect is the basis of all electric motors and moving coil meters. Think of all the places where electric motors are used from stereos, disc drives, CD players, starter motors in cars, washing machines etc. etc. and you will realise how important this effect is! The forces are shown in Figure 2(a).

You can see why the coil will rotate from the 'double catapult' field diagram in Figure 2(b). Since the current moves along the two opposite sides of the coil in opposite directions the two sides receive a force in opposite directions also, thus turning the coil.

Mathematical consideration – formulae for the force on a coil in a magnetic field

Consider a rectangular coil with sides of length a and b placed in a magnetic field of flux density B and free to rotate about an axis perpendicular to the paper, as shown in Figure 3. A current of I Amps flows in the coil.

The field exerts a force on the sides b given by

Force (F) on side length b = BNIb

where N is the number of turns on the coil.

If the perpendicular to the coil is at an angle q to the field direction, then the torque exerted on the coil is Fd where d = a sin q.

Therefore the torque C is given by:

Torque (C) on the coil = BANI sinq

where A = ab, the area of one face of the coil.

The maximum torque occurs when the plane of the coil is lying along the field lines (q = 90

The minimum value of the torque is zero, when q = 0.

Calculate the torque needed to hold a coil of 20 turns and area 8cm

Torque = BANIsinq= 30

This is a very useful application of the force on a current in a
magnetic field. A steady magnetic field is placed across a tube carrying a conducting liquid, as
shown in Figure 4. A current is passed through the liquid at right angles to the magnetic field,
and therefore a force is exerted on the liquid which pushes it down the pipe.

This type
of pump is particularly useful since there are no moving parts. It has found application in two
widely different fields: for pumping liquid sodium coolant round a nuclear reactor, and for
pumping blood round the body if the heart is damaged.