Question

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current ‘I’ along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire ‘B’ is given by

• A. \(\frac {μ_0I^2}{2\pi d}\)
• B. \(\frac {2μ_0I^2}{\pi d}\)
• C. \(\frac {\sqrt 2μ_0I^2}{\pi d}\)
• D. \(\frac {μ_0I^2}{\sqrt 2\pi d}\)

Answer 1

The Correct Option is D

To solve this problem, we will use the concept of magnetic force between two parallel current-carrying wires. According to Ampère's force law, the force per unit length \( F/L \) between two parallel wires carrying currents \( I_1 \) and \( I_2 \) and separated by a distance \( d \) is given by the formula:

F/L = \frac{μ_0 I_1 I_2}{2\pi d}

In this problem, we have three parallel wires carrying the same current \( I \) in the same direction. Let us label the wires as \( A \), \( B \), and \( C \), with wire \( B \) being the middle one.

1. Each pair of wires exerts a magnetic force on each other. The forces on the middle wire \( B \) due to the other wires \( A \) and \( C \) will be considered.
2. Using the formula above, the force per unit length on wire \( B \) due to wire \( A \) is given by: F_{BA}/L = \frac{μ_0 I^2}{2\pi d}
3. Similarly, the force per unit length on wire \( B \) due to wire \( C \) is: F_{BC}/L = \frac{μ_0 I^2}{2\pi d}
4. Since the currents are in the same direction, the forces exerted by wires \( A \) and \( C \) on wire \( B \) will be attractive and directed towards each wire.

We must find the net force per unit length on wire \( B \). The forces from both wire \( A \) and wire \( C \) are in opposite directions, hence:

The net force is the vector sum of these forces. Since they are in opposite directions, the total force will be:

F_{\text{net}}/L = 2 \times \frac{μ_0 I^2}{2\pi d} \cos 45^\circ

Given that \(\cos 45^\circ = \frac{1}{\sqrt{2}}\):

F_{\text{net}}/L = 2 \times \frac{μ_0 I^2}{2\pi d} \times \frac{1}{\sqrt{2}} = \frac{μ_0 I^2}{\sqrt{2}\pi d}

Therefore, the magnitude of force per unit length on the middle wire \( B \) is given by:

\(\frac{μ_0 I^2}{\sqrt{2} \pi d}\)