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 calculated using the formula:\[F = \frac{\mu_0 I_1 I_2}{2 \pi r}\]where:
- \( \mu_0 \) signifies the permeability of free space.
- \( I_1 \) and \( I_2 \) represent the currents within the conductors.
- \( r \) denotes the distance separating the conductors.
2: Force on Conductor C
- Conductor C experiences force from the magnetic fields generated by conductors A and B.
- The force between conductors 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 because the currents in A and C flow in the same direction. Consequently, the force acts towards A.- The force between conductors B and C is:\[F_{BC} = \frac{\mu_0 I I_1}{2 \pi r} \quad (\text{as the distance between B and C is } r)\]This force is repulsive because the currents in B and C flow in opposite directions.
3: Net Force on C
The net force on conductor C is the result of subtracting the repulsive force from B from the attractive force from A:\[F_{\text{net}} = F_{AC} - F_{BC} = \frac{3 \mu_0 I I_1}{2 \pi r} - \frac{\mu_0 I I_1}{2 \pi r} = \frac{2 \mu_0 I I_1}{2 \pi r} = \frac{\mu_0 I I_1}{\pi r}\]Therefore, the net force on C is directed towards A. The correct answer is:\[\boxed{(C) \, \frac{\mu_0 I I_1}{\pi r}, \text{ towards A}}\]