Question

a wire AB is carrying a current of 12A and is lying on the table. another wire CD, carrying a current of 5A is arranged just above AB at a height of 1mm. what should be the weight per unit length of this wire so that CD remains suspended at its position? indicate the direction of current in CD.

Answer 1

The magnetic force acting on the wire CD should balance the gravitational force acting on it. Let the weight of wire CD per unit length be W. 

Given that I₁=12A, I₂=5A, d=1mm = 0.001m,

Magnetic force on unit length of CD = weight of CD per unit length = W

$W= \frac{ \mu_0 I_1I_2}{2 \pi d} = \frac{4 \pi *10^{-7}*12*5}{2 \pi *0.001} = 0.012\ N/m$