First slide

Force on a current carrying wire placed in a magnetic field

Question

Two long parallel wires separated by a distance d have equal current i flowing in each. The magnetic field of one exerts a force F on the other. The distance d is increased to 2 d and current in each wire is reduced from i to i / 2 What is the force between them now ?

Easy

A

F
B

F/2
C

F/4
D

F/8

Solution

Force per unit length of the wire is given by
$F = \frac{μ_{0}}{4 π} \times \frac{2 i_{1} i_{2}}{d} = \frac{μ_{0}}{4 π} \times \frac{2 i^{2}}{d} (∵ i_{1} = i_{2} = i)$
When d is increased to 2 d and i is reduced to i / 2, then
$\begin{array}{l} F^{'} = \frac{m_{0}}{4 π} \times \frac{2 (i / 2)^{2}}{2 d} \\ = \frac{μ_{0}}{4 π} \times \frac{2 i^{2}}{8 d} = \frac{1}{8} [\frac{μ_{0}}{4 π} \times \frac{2 i^{2}}{d}] = F / 8 \end{array}$

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