Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

A bar magnet experiment torque 0.018 Nm when placed in uniform magnetic field, B = 0.06 T and makes 30° angle with the magnetic field as shown in figure. find out work done by the external force if magnet rotates from minimum potential energy to maximum potential energy.

Solution : torque, τ = 0.018 Nm

magnetic field, B = 0.06 T

angle between magnet and magnetic field, θ = 30°

now find magnetic moment, M using formula τ = MBsinθ

⇒0.018 = M × 0.06 × sin30°

⇒M = 0.6 Am²

now work done by the external force = change in potential energy

= Uf - Ui

= maximum potential energy - minimum potential energy

= (-MBcos180°) - (-MBcos0°)

= MB + MB

= 2MB

= 2 × 0.6 × 0.06

= 0.072 J

Therefore the workdone by the external force is 0.072 J