Question

A long straight wire carrying a current of 25 A rests on the table. Another wire PQ of length 1 m and mass 2.5 g carries the same current but in the opposite direction. To what height ($h$) will wire PQ rise? ($\mu_0 = 4\pi \times 10^{-7}$ SI unit)}

• A. 3 mm
• B. 4 mm
• C. 5 mm
• D. 8 mm

Hint:

Logic Tip: Parallel wires with currents in opposite directions repel each other.

Answer 1

The Correct Option is C

Concept: The force per unit length between two parallel current-carrying wires is: \[ \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi h} \] For equilibrium, the magnetic force balances the weight of the wire: \[ F = mg \]
Step 1: Set up the equilibrium condition \[ \frac{\mu_0 I_1 I_2}{2\pi h} \cdot l = mg \] \[ \Rightarrow h = \frac{\mu_0 I_1 I_2 \, l}{2\pi mg} \]
Step 2: Substitute values Given: \[ I_1 = I_2 = 25\,\text{A}, \quad l = 1\,\text{m}, \quad m = 2.5\,\text{g} = 2.5 \times 10^{-3}\,\text{kg} \] \[ h = \frac{4\pi \times 10^{-7} \times 25 \times 25 \times 1}{2\pi \times 2.5 \times 10^{-3} \times 9.8} \]
Step 3: Simplify Cancel $\pi$: \[ h = \frac{4 \times 10^{-7} \times 625}{2 \times 2.5 \times 10^{-3} \times 9.8} \] \[ h = \frac{2.5 \times 10^{-4{4.9 \times 10^{-2 \approx 5 \times 10^{-3}\,\text{m} \] Final Answer: \[ \boxed{h = 5\,\text{mm \]