Question

Two infinitely long parallel conducting wires \(A\) and \(B\) carry currents \(I\) and \(2I\), respectively, in the same direction. Wire \(A\) lies on an insulated floor while wire \(B\) is fixed at a height \(h\) above the floor. The minimum value of \(h\) so that wire \(A\) does not rise from the floor is:

• A. \(\dfrac{4\mu_0 I^2}{\pi \lambda g}\)
• B. \(\dfrac{\mu_0 I^2}{2\pi \lambda g}\)
• C. \(\dfrac{\mu_0 I^2}{\pi \lambda g}\)
• D. \(\dfrac{2\mu_0 I^2}{\pi \lambda g}\)

Hint:

Parallel currents in the same direction attract each other. For limiting equilibrium, magnetic attraction equals weight per unit length.

Answer 1

The Correct Option is C

Concept:

• Force per unit length between two parallel currents: \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi h} \]

• At the limiting condition, \[ \text{Magnetic force}=\text{Weight per unit length} \]

Step 1: Calculate magnetic force per unit length.
\[ \frac{F}{L} = \frac{\mu_0(I)(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h} \]

Step 2: Apply equilibrium condition.
\[ \frac{\mu_0 I^2}{\pi h} = \lambda g \]

Step 3: Solve for \(h\).
\[ h = \frac{\mu_0 I^2}{\pi\lambda g} \] \[ \boxed{\text{Option (C)}} \]