Question

A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B . The work done to rotate the loop by 30 about an axis perpendicular to its plane is_____

Answer 1

Step 1: Given data

Potential energy is given by:

$U=-MB\cos \theta$

here,

M = magnetic moment of the current carrying loop

B = magnetic field in which the loop is kept

$\theta$= Angle between M and B

Step 2; Calculation:

Change in potential energy will give work done as initial and final kinetic energy is 0.The rotation of loop by ${30}^0$make no change in angle by axis of loop. Therefore the work done to rotate the loop is zero

The work done to rotate the loop =$W=MB(\cos {\theta }_1-\cos {\theta }_2)$