Question

A straight wire carrying current \( I \) is lying along the axis of a circular loop carrying current \( I \). The force on this wire due to the circular loop is proportional to (Assume the axis of the circular loop is perpendicular to the plane of the loop).

• A. \( I^2 \)
• B. \( I \)
• C. \( \frac{1}{r^2} \)
• D. \( \frac{1}{r^3} \)

Hint:

The force on a current-carrying wire in a magnetic field depends on the current, the length of the wire, and the magnetic field. In this case, the field depends on both currents and the distance between the wire and the loop.

Answer 1

The Correct Option is A

Consider a straight wire carrying current \( I \) along the axis of a circular loop also carrying current \( I \). According to Ampère’s Law and the Biot–Savart law, the magnetic field at a point along the axis of a circular loop of radius \( r \) carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2r} \quad \text{(on the axis of the loop)} \] where: - \( \mu_0 \) is the permeability of free space, - \( r \) is the radius of the circular loop, - \( I \) is the current in the loop. Now, the force on a current-carrying wire in a magnetic field is given by: \[ F = I L B \sin \theta \] where: - \( I \) is the current in the wire, - \( L \) is the length of the wire, - \( B \) is the magnetic field, and - \( \theta \) is the angle between the direction of the magnetic field and the wire. Since the wire is along the axis of the circular loop, \( \theta = 90^\circ \), so \( \sin \theta = 1 \). The length of the wire \( L \) can be assumed to be infinitesimal for a small section of the wire. Thus, the force on the wire due to the loop's magnetic field is: \[ F = I L \times \frac{\mu_0 I}{2r} \] Therefore, the force is proportional to \( I^2 \), as the current in the wire and the loop are both involved in the interaction, and their effects are squared in the formula. Thus, the correct answer is (A) \( I^2 \).