Question

Two long conductors, separated by a distance $ d $ carry currents $ I_1 $ , and $ I_2 $ in the same direction. They exert a force $F$ on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to $ 3d $ . The new value of force between them is

• A. $ -2F $
• B. $ \frac{F}{3} $
• C. $ -\frac{2F}{3} $
• D. $ -\frac{F}{3} $

Answer 1

Answer: (C)

Magnitude of magnetic force per unit length between parallel current carrying wires,
$F = \frac{\mu_0}{2\pi} \frac{i_1 i_2}{r}$
Initially, $F =\frac{\mu_0}{2\pi} \frac{i_1 i_2}{d}$
As both the wires carry current in same direction, this force is attractive.
Finally $i'_1 = 2i$ and $r = 3d$
So, $F' = \frac{\mu_0}{2\pi} \frac{(2i_1)i_2}{3d}$
$ =\frac{2}{3} F$
Since, the direction of current is reversed in one of the wire, hence $F'$ would be repulsive in nature.
$\therefore F' = - \frac{2}{3}F$