Question

Two thin long parallel wires, \(W_1\) and \(W_2\) separated by distance '\(a\)', carry currents \(i\) and \(3i\) respectively in the same direction. The magnitude of the force per unit length exerted by wire \(W_1\) on wire \(W_2\) is

• A. \(\frac{μ_0\,i^2}{2π\,a}\)
• B. \(\frac{μ_0\,i^2}{π\,a}\)
• C. \(\frac{3μ_0\,i^2}{2π\,a}\)
• D. \(\frac{2μ_0\,i^2}{3π\,a}\)

Hint:

The force per length between parallel wires is mu0 i1 i2 / (2 pi a).

Answer 1

The Correct Option is C

Step 1: Formula:
Force per unit length between two long parallel wires carrying currents \(i_1\) and \(i_2\), at separation \(a\), is
\[ \frac Fl=\frac{\mu_0i_1i_2}{2\pi a} \]

Step 2: Substitute:
With \(i_1=i\) and \(i_2=3i\):
\[ \frac Fl=\frac{\mu_0(i)(3i)}{2\pi a}=\frac{3\mu_0i^2}{2\pi a} \]

Step 3: Direction:
The currents are in the same direction, so the force is attractive. Its magnitude is the same on both wires (Newton's third law).

Step 4: Check the Options:
Option (A) uses \(i\cdot i\), forgetting the 3. Option (B) has \(\pi a\) instead of \(2\pi a\) with the 1 factor. Option (D) has no basis in the formula.

Final Answer:
The force per unit length is \(\dfrac{3\mu_0i^2}{2\pi a}\), option (C). \[ \boxed{\text{(C) } \frac{3\mu_0i^2}{2\pi a}} \]