Question

A horizontal rod of mass 0.01 kg and length 10 cm is placed on a frictionless plane inclined at an angle 60° with the horizontal and with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field is applied ‘Vertically downwards’. The current through the rod is 1.7 A, then the value of magnetic field induction \( B \) for which the rod remains stationary in the inclined plane is

• A. \( 1 \, \text{T} \)
• B. \( 3 \, \text{T} \)
• C. \( 2 \, \text{T} \)
• D. \( 4 \, \text{T} \)

Hint:

The force on a current-carrying conductor in a magnetic field depends on the length of the conductor, the current, and the angle between the magnetic field and the conductor.

Answer 1

The Correct Option is C

Using the relation for the force on a current-carrying wire in a magnetic field, \( F = BIL \sin(\theta) \), and equating it with the gravitational force, we can calculate the required magnetic field as \( 2 \, \text{T} \).