Question

A short bar magnet placed with its axis at ${30^o}$ with a uniform external magnetic field of $0.16T$ experiences a torque of magnitude $0.032Nm$ . If the bar magnet is free to rotate, its potential energies when it is in stable and unstable equilibrium are respectively
A. $ - 0.064J, + 0.064J$
B. $ - 0.032J, + 0.032J$
C. $ + 0.064J, - 0.128J$
D. $0.032J, - 0.032J$

Answer 1

HintFirst step is to figure out the known quantities ( in this case angle, magnetic field and torque). After that we have to derive the formula for the torque in order to find the much needed magnetic moment. After that we can put the given values in the equation and get the magnetic moment. Then we can use the value of the magnetic moment to find the value of the min and max potential energies

 Complete step-by-step solution:
Firstly we have to sort out what we have got :
Angle of the axis of magnet with magnetic field : ${30^o}$
The value of magnetic field : $0.16T$
And the amount of torque on the magnetic field : $0.032Nm$
Step1:firstly we have to find the magnetic moment for the given magnet. For that we have to give the proper formula for it :
$z = mB\sin \theta $
Where $z$ is torque, $m$ is magnetic moment and $B$ is magnetic field
Step 2: Therefore, we have to put all the given quantities in the formula and find the answer for the given question:
$
  z = mB\sin \theta \\
  0.032 = m \times 0.16 \times \dfrac{1}{2} \\
  m = 0.4Am \\
 $
Therefore, the value for $m = 0.4Am$
Step3: now we have to put the found magnetic moment in formula of potential energies and then we can find the value for its max and min:
$
  {U_{\max }} = mB \\
  {U_{\max }} = 0.4 \times 0.16 \\
  {U_{\max }} = 0.064T \\
  {U_{min}} = - mB \\
  {U_{min}} = - 0.4 \times 0.16 \\
  {U_{min}} = - 0.064T \\
 $
Step4: we just have to put the values in the equation and find the correct answer.
$
  U = mB\operatorname{Sin} \theta \\
  so, \\
  {U_{\max }} = mB \\
  {U_{min}} = - mB \\
 $
Therefore the correct answer would be option A, $ - 0.064J, + 0.064J$ .

Note:- The value for the min and max potential energies is derived by the help of $\operatorname{Sin} \theta $. The value of $\operatorname{Sin} \theta $ varies between the $1$ and $ - 1$ , therefore the min and max potential energies would be ${U_{\max }} = mB{\text{ \& }}{U_{min}} = - mB$ .