Question

A circular loop of wire, carrying a current 'I' is lying in the xy-plane with its centre coinciding with the origin. It is subjected to a uniform magnetic field pointing along the +z-axis. The loop will:

• A. move along the x-axis
• B. move along the y-axis
• C. move along the z-axis
• D. remain stationary

Hint:

When considering the motion of a current-carrying conductor in a magnetic field, always check for symmetry and the directions of \( \vec{B} \), \( \vec{dl} \), and the resultant \( \vec{dl} \times \vec{B} \). Symmetry often leads to cancellation of forces, especially in uniform fields.

Answer 1

The Correct Option is D

Step 1: Analyzing the forces on the loop.
A current-carrying loop in a magnetic field experiences a force given by \( \vec{F} = I(\vec{dl} \times \vec{B}) \), where \( \vec{dl} \) is an element of the loop and \( \vec{B} \) is the magnetic field. In this scenario, the magnetic field \( \vec{B} \) is uniform and points along the +z-axis, and the loop lies in the xy-plane. Step 2: Calculating the net force.
Since the loop is symmetrical and the field is uniform, the forces on opposite sides of the loop will be equal in magnitude but opposite in direction, leading to a cancellation of all horizontal (x and y directions) forces. The vertical (z-axis) force components also cancel out due to the symmetry and the orientation of the magnetic field and current direction. Step 3: Determining the resultant motion.
With all forces canceling out, there is no resultant force on the loop. Additionally, if there is no additional torque acting to tilt or twist the loop out of the xy-plane, the loop will not experience any translational or rotational motion.