Question

A wire AB is carrying steady current \(I_1\) and is kept on the table. Another wire CD carrying current \(I_2\) is held directly above as shown in figure. When the wire CD is left free and it remains suspended at its position, its mass per unit length is (g=acceleration due to gravity, \(μ_0\) = permeability of free space)

• A. \(\frac{μ_0I_1I_2}{2πrg}\)
• B. \(\frac{μ_0I_1I_2}{4πrg}\)
• C. \(\frac{μ_0I_1I_2}{πrg}\)
• D. \(\frac{μ_0I_1I_2}{πr^2g}\)

Hint:

Opposite currents repel, so the upward magnetic force per unit length equals the weight per unit length, lambda g.

Answer 1

The Correct Option is A

Step 1: Field of the lower wire:
Wire AB produces a magnetic field at CD of magnitude $B = \frac{\mu_0 I_1}{2\pi r}$.

Step 2: Force on a length $L$ of CD:
$F = B I_2 L = \frac{\mu_0 I_1 I_2 L}{2\pi r}$. Because the currents are opposite, this force points up, away from AB.

Step 3: Equilibrium:
The weight of that length is $(\lambda L) g$. Setting $\lambda L g = \frac{\mu_0 I_1 I_2 L}{2\pi r}$ and cancelling $L$ gives $\lambda = \frac{\mu_0 I_1 I_2}{2\pi r g}$.

Step 4: Result:
This matches option (A). The other options have a wrong numeric factor or a wrong power of $r$.

Final Answer:
Option (A). \[ \boxed{\frac{\mu_0 I_1 I_2}{2\pi r g}} \]