Question

A bar magnet of magnetic moment 2$A{m^2}$ is free to rotate about a vertical axis passing through its center. The magnet is released from rest from the east - west position. Then the KE of the magnet as it takes N-S position is (${B_H}$​=25μT)
A. 25 μ J
B. 50 μ J
C. 100 μ J
D. 12.5 μ J

Answer 1

Hint: In order to solve this numerical we should know the formula of torque produced by the magnetic field such that we can calculate the kinetic energy of the magnet.

Complete step by step answer:
From the given data
The magnetic moment is given by
$m = 2A{m^2}$
${B_H} = 25\mu T$
Let us take the initial and final angle
${\theta _1} = \dfrac{\pi }{2},{\theta _2} = 0$
By using the formula, in both the cases
When it making an angle with ${90^0}$
${\mu _1} = - mB\cos \theta $
$\
  {\mu _1} = - mB\cos \dfrac{\pi }{2} \\
  \Rightarrow {\mu _1} = 0 \\
\ $
When it is making an angle with ${0^0}$
$\
  {\mu _2} = - mB\cos \theta \\
  \Rightarrow {\mu _2} = - mB\cos {0^0} \\
\Rightarrow {\mu _2} = - 2 \times 25 \\
\Rightarrow {\mu _2} = - 50H \\
\ $
By using the kinetic energy due to the magnet is given by
${K_1} = 0$
$\
  {K_2} - {K_1} = {U_1} - {U_2} \\
\Rightarrow {K_2} = {U_1} - {U_2} + {K_1} \\
\Rightarrow {K_2} = 50\mu J \\
\ $
Hence the correct option is B

Note:The magnetic energy of a magnet is the work done by the magnetic force usually it can be taken as magnetic torque produced by the magnet. Electrostatic potential and magnetic energy both are related by Maxwell's equations.