Question

A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is

• A. 1.2 × 10^–5 N directed towards wire X
• B. 1.2 × 10^–4 N directed away from wire X
• C. 1.2 × 10^–4 N directed towards wire X
• D. 2.4 × 10^–5 N directed towards wire X

Answer 1

The Correct Option is A

To find the force acting on the wire Y, we can use the formula for the force per unit length between two parallel current-carrying wires:

F/L = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{d}

Where:

• \mu_0 is the permeability of free space, approximately 4\pi \times 10^{-7} \, \text{N/A}^2.
• I_1 and I_2 are the currents in wires X and Y, respectively.
• d is the distance between the wires.
• L is the length of the wire on which force is calculated.

Given:

• I_1 = 2 \, \text{A}
• I_2 = 3 \, \text{A}
• d = 5 \, \text{cm} = 0.05 \, \text{m}
• L = 0.5 \, \text{m} (for wire X, as the force is calculated per unit length and X is shorter)

Substituting these values into the formula, we have:

F = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{2 \times 3}{0.05} \cdot 0.5

Simplifying, we get:

F = 2 \times 10^{-7} \cdot \frac{6}{0.05} \cdot 0.5

= 2 \times 10^{-7} \cdot 120

= 2.4 \times 10^{-5} \, \text{N}

The force is directed towards wire X because the currents are flowing in the same direction, leading to an attractive force between the wires.

Therefore, the force acting on wire Y is 2.4 \times 10^{-5} \, \text{N} directed towards wire X.