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)
Show Hint
Opposite currents repel, so the upward magnetic force per unit length equals the weight per unit length, lambda g.
Step 1: Understanding the Figure:
Wire AB lies on the table and carries current to the left. Wire CD is held above it, parallel, at distance \(r\), and carries current to the right. The currents are opposite, so the wires repel each other.
Step 2: Key Formula or Approach:
The magnetic force per unit length between two long parallel wires is \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \] For opposite currents it is repulsive.
Step 3: Direction of the force:
Wire CD is above AB, so the repulsion pushes CD upward. Gravity pulls it downward. For CD to stay suspended these two forces must be equal.
Step 4: Balance the forces:
Let \(\lambda\) be the mass per unit length of CD. Its weight per unit length is \(\lambda g\). \[ \lambda g = \frac{\mu_0 I_1 I_2}{2\pi r} \] \[ \lambda = \frac{\mu_0 I_1 I_2}{2\pi r g} \]
Step 5: Why the other options are wrong:
Options (B) and (C) carry factors of \(4\pi\) and \(\pi\) instead of \(2\pi\), which do not match the force formula for parallel wires. Option (D) has \(r^2\), but the force between parallel wires falls as \(1/r\), not \(1/r^2\).
Final Answer:
The mass per unit length is \(\frac{\mu_0 I_1 I_2}{2\pi r g}\), option (A).
\[ \boxed{\frac{\mu_0 I_1 I_2}{2\pi r g}} \]
