Question

Two long straight wires are arranged parallel to each other and are kept \(20\) cm apart in vacuum. They carry currents of \(2\) A and \(4\) A respectively in the same direction. What will be the magnetic force on a length of \(10\) cm of either wire?
(\(μ_0 = 4π\times 10^{-7}\) SI units)

• A. \(10^{-7}\) N
• B. \(2\times 10^{-7}\) N
• C. \(4\times 10^{-7}\) N
• D. \(8\times 10^{-7}\) N

Hint:

Force on length \(l\) is \(F=\frac{\mu_0I_1I_2l}{2\pi d}\).

Answer 1

The Correct Option is D

Step 1: Plan:
Find the magnetic field of one wire at the other and then the force on its length.

Step 2: Steps:
Field of the $4$ A wire at $20$ cm: $B = \frac{\mu_0I}{2\pi d} = 2\times10^{-7}\times\frac{4}{0.2} = 4\times10^{-6}$ T.
Force on $10$ cm of the $2$ A wire: $F = BIl = 4\times10^{-6}\times2\times0.1 = 8\times10^{-7}$ N.

Final Answer:
The force is $8\times10^{-7}$ N, option (D). \[ \boxed{8\times10^{-7}\ \text{N}} \]