Question

Two parallel wires in free space are 10 cm apart and each carries a current of 10 A in the same direction. The force exerted by one wire on the other, per metre length is

• A. $ 2 \times {10} ^{-4} N$ repulsive
• B. $ 2 \times {10} ^{-7} N$ repulsive
• C. $ 2 \times {10} ^{-4} N$ attractive
• D. $ 2 \times {10} ^{-7} N$ attractive

Answer 1

The Correct Option is C

To determine the force exerted by one wire on the other, we can use the formula for the force between two parallel current-carrying wires. The force per unit length between two parallel wires carrying currents \(I_1\) and \(I_2\), separated by a distance \(d\) in free space, is given by:

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

Where:

• \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}, the permeability of free space.
• I_1 = I_2 = 10 \, \text{A}, the currents in each wire.
• d = 0.1 \, \text{m}, the distance between the wires (converted from 10 cm to meters).

Substituting the given values into the formula:

F/L = \frac{{4\pi \times 10^{-7} \cdot 10 \cdot 10}}{{2\pi \cdot 0.1}}

Simplifying the expression:

F/L = \frac{{4 \times 10^{-7} \cdot 100}}{{0.2}}

Calculating further:

F/L = \frac{{4 \times 10^{-7} \cdot 1000}}{{2}}

F/L = 2 \times 10^{-4} \, \text{N/m}

Since the currents are flowing in the same direction, the force between the wires will be attractive. This is because parallel currents in the same direction attract each other.

Thus, the correct answer is: $ 2 \times 10^{-4} N$ attractive.