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

Step 1: Set up the balance.
Wire $B$ above carries current $2I$ in the same direction as wire $A$ on the floor. Parallel currents in the same direction attract, so $B$ pulls $A$ upward. For $A$ to just stay down, this magnetic pull must balance $A$'s weight per unit length.
Step 2: Write the force per unit length.
Between parallel wires, $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi h}$.
Step 3: Substitute the currents.
With $I_1 = I$ and $I_2 = 2I$: $\dfrac{F}{L} = \dfrac{\mu_0 (I)(2I)}{2\pi h} = \dfrac{\mu_0 I^2}{\pi h}$.
Step 4: Apply the limiting condition.
At the minimum height, the upward magnetic force per length equals the weight per length $\lambda g$: $\dfrac{\mu_0 I^2}{\pi h} = \lambda g$.
Step 5: Solve for h.
Rearranging, $h = \dfrac{\mu_0 I^2}{\pi \lambda g}$.
Step 6: Choose the option.
This is the minimum height and matches option C.
\[ \boxed{ h = \frac{\mu_0 I^2}{\pi \lambda g} } \]