Question

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

• A. 2 × 10^-4 Nm^-1 [attractive]
• B. 2 × 10^-7 Nm^-1 [attractive]
• C. 2 × 10^-4 Nm^-1 [repulsive]
• D. 2 × 10^-7 Nm^-1 [repulsive]

Answer 1

The Correct Option is A

Two parallel wires in free space are 10 cm apart and each carries a current of 10A in the same direction. We need to find the force exerted by one wire on the other per unit length.

The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ and separated by a distance $d$ is given by:

$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d}$

where:

• $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7}\,\text{N/A}^2$),
• $I_1$ and $I_2$ are the currents in the wires, and
• $d$ is the distance between the wires.

In this case, $I_1 = I_2 = 10\,\text{A}$ and $d = 10\,\text{cm} = 0.1\,\text{m}$. Substituting these values:

$\frac{F}{l} = \frac{(4\pi \times 10^{-7}\,\text{N/A}^2)(10\,\text{A})(10\,\text{A})}{2\pi (0.1\,\text{m})} = 2 \times 10^{-4}\,\text{N/m}$

Since the currents are in the same direction, the force is attractive.

The correct answer is (A) $2 \times 10^{-4}$ Nm^-1 [attractive].