Question

Two long thin parallel wires carrying current \( I \) separated by a distance \( d \) exert force \( F \) on one another. The distance between them is doubled and the current is decreased to \( \frac{1}{3} \). The force they exert on one another is

• A. \( \frac{F}{6} \)
• B. \( \frac{F}{3} \)
• C. \( \frac{F}{18} \)
• D. \( \frac{F}{9} \)

Hint:

The force between two current-carrying wires depends on the distance between them and the currents, with the force decreasing as the distance increases and the current decreases.

Answer 1

The Correct Option is C

Step 1: Use the formula for the force between two current-carrying wires.
The force between two parallel wires carrying currents \( I_1 \) and \( I_2 \) separated by a distance \( d \) is given by Ampère's law: \[ F = \frac{\mu_0 I_1 I_2}{2 \pi d} \] where \( \mu_0 \) is the permeability of free space. Step 2: Calculate the effect of changing distance and current.
The force is directly proportional to the product of the currents and inversely proportional to the distance between the wires. If the distance is doubled and the current is reduced to one-third, the new force \( F' \) is: \[ F' = \frac{\mu_0 \left(\frac{1}{3} I_1\right) \left(\frac{1}{3} I_2\right)}{2 \pi (2d)} = \frac{F}{18} \] Step 3: Conclusion.
Thus, the force exerted between the wires is \( \frac{F}{18} \), which corresponds to option (C).