Question

Prove that the force acting on a current-carrying wire, joining two fixed points a and b in a uniform magnetic field, is independent of the shape of the wire.

Answer 1

Given:
Uniform magnetic field existing in the region of the wire = B
Let the electric current flowing through the wire be i.
Length of the wire between two points a and b = l
Magnetic force is given by
$$\begin{gathered} \overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B} \\ \overrightarrow{F}\mathit{=}ilB\sin \theta \end{gathered}$$
Let us consider two wires of length l, one straight and the other circular.
The circular wire is of radius a such that $2\mathrm{\pi }a=l$
Suppose the magnetic field is pointing along the z direction and both the wires are lying in the xy plane, so that the angle between the area vector and the magnetic field is $90°$.
For the straight wire of length l lying in a uniform magnetic field of strength B:
Force, F = $ilB\sin (90°)$ = ilB
For the circular wire:
Length, l = $2\mathrm{\pi }a$
Angel between the area vector and magnetic field will again be $90°$.
Force acting on the circular wire,
F = $i\left(2\mathrm{\pi }a\right)B\sin (90°)$
= $i2\mathrm{\pi }aB\mathit{}\mathit{=}\mathit{}ilB$
Both the forces are equal in magnitude. This implies that the magnetic force is independent of the shape of the wire and depends on the length and orientation of the wire.

Therefore, the magnetic force is independent of the shape of the wire.