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Question

Question: A short bar magnet placed with its axis at \({30^o}\) with a uniform external magnetic field of \(0....

A short bar magnet placed with its axis at 30o{30^o} with a uniform external magnetic field of 0.16T0.16T experiences a torque of magnitude 0.032Nm0.032Nm . 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 - 0.064J, + 0.064J
B. 0.032J,+0.032J - 0.032J, + 0.032J
C. +0.064J,0.128J + 0.064J, - 0.128J
D. 0.032J,0.032J0.032J, - 0.032J

Explanation

Solution

Hint First 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 : 30o{30^o}
The value of magnetic field : 0.16T0.16T
And the amount of torque on the magnetic field : 0.032Nm0.032Nm
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=mBsinθz = mB\sin \theta
Where zz is torque, mm is magnetic moment and BB 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=mBsinθ 0.032=m×0.16×12 m=0.4Am  z = mB\sin \theta \\\ 0.032 = m \times 0.16 \times \dfrac{1}{2} \\\ m = 0.4Am \\\
Therefore, the value for m=0.4Amm = 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:
Umax=mB Umax=0.4×0.16 Umax=0.064T Umin=mB Umin=0.4×0.16 Umin=0.064T  {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=mBSinθ so, Umax=mB Umin=mB  U = mB\operatorname{Sin} \theta \\\ so, \\\ {U_{\max }} = mB \\\ {U_{min}} = - mB \\\
Therefore the correct answer would be option A, 0.064J,+0.064J - 0.064J, + 0.064J .

Note:- The value for the min and max potential energies is derived by the help of Sinθ\operatorname{Sin} \theta . The value of Sinθ\operatorname{Sin} \theta varies between the 11 and 1 - 1 , therefore the min and max potential energies would be Umax=mB & Umin=mB{U_{\max }} = mB{\text{ \& }}{U_{min}} = - mB .