Force on a Closed Circuit in a Magnetic Field

Extending the result to a closed circuit

The expression for the force on a wire of arbitrary shape carries straight over to a closed wire circuit C. Summing $d\vec{F}_m = I\,d\vec{l}\times\vec{B}$ round the circuit,

$$\vec{F}_m = \oint_C I\,d\vec{l}\times\vec{B}$$

Here the integral is taken over the whole closed circuit C.

Uniform field

If $\vec{B}$ is uniform it can be taken outside the integral:

$$\vec{F}_m = I\left(\oint_C d\vec{l}\right)\times\vec{B}$$

The term in the bracket is the sum of the length vectors taken all the way round a closed circuit. Going round a closed path returns you to the starting point, so this vector sum is zero.

$$\therefore \ \vec{F}_m = 0 \qquad (\vec{B} \text{ uniform})$$

The board's answer

The value of the net force on a closed circuit placed in a uniform magnetic field is zero.

What this does not say

• It does not say that the force on each side of the circuit is zero. Individual sides do feel forces; it is their vector sum that vanishes.
• It does not say the net torque is zero. Two equal and opposite forces acting along different lines form a couple, and that is exactly what happens to a current loop — the subject of Section 10.7. A current loop in a uniform field therefore experiences no net translation, but it does rotate.
• The result requires the field to be uniform. In a non-uniform field the sum need not vanish.

Standard results

• $\vec{F}_m = \displaystyle\oint_C I\,d\vec{l}\times\vec{B}$ for a closed circuit
• $\displaystyle\oint_C d\vec{l} = 0$, hence $\vec{F}_m = 0$ for uniform $\vec{B}$
• Net force zero, net torque in general non-zero.