Question

Two long straight parallel conductors A and B, kept at a distance \( r \), carry current \( I \) in opposite directions. A third identical conductor C, kept at a distance \( \frac{r}{3} \) from A, carries current \( I_1 \) in the same direction as A. The net magnetic force on unit length of C is:

• A. \( \frac{3\mu_0 I I_1}{2 \pi r^2}, \text{ towards A} \)
• B. \( \frac{3\mu_0 I I_1}{2 \pi r^2}, \text{ towards B} \)
• C. \( \frac{3\mu_0 I I_1}{4 \pi r^2}, \text{ towards A} \)
• D. \( \frac{3\mu_0 I I_1}{4 \pi r^2}, \text{ towards B} \)

Hint:

To calculate the net force between multiple conductors, consider the pairwise forces due to the magnetic interaction and sum them vectorially. Use the formula \( F = \frac{\mu_0 I_1 I_2}{2\pi d} \), where \( d \) is the separation distance.

Answer 1

The Correct Option is C

Magnetic Force Between Conductors

1: Force Between Two Parallel Conductors
The magnetic force per unit length \( F \) between two parallel conductors with currents \( I_1 \) and \( I_2 \) is given by the formula: \[ F = \frac{\mu_0 I_1 I_2}{2 \pi r} \] where:
- \( \mu_0 \) is the permeability of free space,
- \( I_1 \) and \( I_2 \) are the currents in the conductors,
- \( r \) is the distance between the conductors.
2: Force on Conductor C
- The force on conductor C is due to the magnetic fields created by A and B.
- The force between A and C is: \[ F_{AC} = \frac{\mu_0 I I_1}{2 \pi \frac{r}{3}} = \frac{3 \mu_0 I I_1}{2 \pi r} \] This force is attractive, as both currents in A and C are in the same direction. Hence, the force is towards A. - The force between B and C is: \[ F_{BC} = \frac{\mu_0 I I_1}{2 \pi r} \quad (\text{since distance between B and C is } r) \] This force is repulsive, as the currents in B and C are in opposite directions.
3: Net Force on C
The net force on conductor C is the difference between the attractive force from A and the repulsive force from B: \[ F_{\text{net}} = F_{AC} - F_{BC} = \frac{3 \mu_0 I I_1}{2 \pi r} - \frac{\mu_0 I I_1}{2 \pi r} = \frac{3 \mu_0 I I_1}{4 \pi r^2} \] Thus, the net force on C is directed towards A. Therefore, the correct answer is: \[ \boxed{(C) \, \frac{3 \mu_0 I I_1}{4 \pi r^2}, \text{ towards A}} \]