Question

A bar magnet of magnetic moment $1.5J.{{T}^{-1}}$ lies aligned with the direction of a uniform magnetic field of $0.22T$.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in the cases (i) and (ii)?

Answer 1

This problem can be solved by using the direct formula for the work done by an external torque to rotate a bar magnet in a magnetic field through an angle. The second part of this problem can be solved by using the direct formula for the torque on a bar magnet placed in a magnetic field.

Formula used:
${{W}_{ext}}=MB\left( \cos {{\theta }_{1}}-\cos {{\theta }_{2}} \right)$
$\tau =MB\sin \theta$

Complete step by step solution:
We will use the direct formula for the work done by an external torque to rotate a bar magnetic in a magnetic field through an angle and also the formula for the torque on a bar magnet in a magnetic field.
The work ${{W}_{ext}}$ done by an external torque to rotate a magnetic dipole of magnetic moment $M$ in a magnetic field $B$ is given by
${{W}_{ext}}=MB\left( \cos {{\theta }_{1}}-\cos {{\theta }_{2}} \right)$ --(1)
Where ${{\theta }_{1}},{{\theta }_{2}}$ are the angles made by the axis of the magnetic dipole with the magnetic field in the initial and final positions of the magnetic dipole.
Also, the torque $\tau$ experienced by a magnetic dipole of magnetic moment $M$ when placed in a magnetic field of magnitude $B$ is given by
$\tau =MB\sin \theta$ --(2)
Where $\theta$ is the angle made by the axis of the dipole with the direction of the magnetic field.
Now, let us analyze the question.
The bar magnet can be considered to be a magnetic dipole.
The dipole moment of this bar magnet is $M=1.5J.{{T}^{-1}}$.
The magnitude of the magnetic field is $B=0.22T$.
(a) (i) Magnet is rotated to normal to the direction of the magnetic field.
Let the angle made by the direction of the magnetic field with the axis of the dipole in the initial position be ${{\theta }_{1}}$.
Let the angle made by the direction of the magnetic field with the axis of the dipole in the final position be ${{\theta }_{2}}$.
It is given that initially, the bar magnet is aligned with the direction of the magnetic field. Therefore, ${{\theta }_{1}}={{0}^{0}}$.
It is also given that the bar magnet is rotated and finally aligned normal to the direction of the magnetic field. Therefore, ${{\theta }_{2}}={{90}^{0}}$.
Let the required amount of work done by the external torque in this case be ${{W}_{1}}$.
Therefore, using (1), we get,
${{W}_{1}}=MB\left( \cos {{0}^{0}}-\cos {{90}^{0}} \right)$
$\therefore {{W}_{1}}=1.5\times 0.22\times \left( 1-0 \right)=1.5\times 0.22\times 1=0.33J$ $\left( \because \cos {{0}^{0}}=1,\cos {{90}^{0}}=0 \right)$
Therefore, the required amount of work done by the external torque in this case is $0.33J$.
(a)(ii) ) Magnet is rotated opposite to the direction of the magnetic field.
Let the angle made by the direction of the magnetic field with the axis of the dipole in the initial position be ${{\theta }_{1}}$.
Let the angle made by the direction of the magnetic field with the axis of the dipole in the final position be ${{\theta }_{2}}$.
It is given that initially, the bar magnet is aligned with the direction of the magnetic field. Therefore, ${{\theta }_{1}}={{0}^{0}}$.
It is also given that the bar magnet is rotated and finally aligned normal to the direction of the magnetic field. Therefore, ${{\theta }_{2}}=-{{180}^{0}}$.
Let the required amount of work done by the external torque in this case be ${{W}_{2}}$.
Therefore, using (1), we get,
${{W}_{2}}=MB\left( \cos {{0}^{0}}-\cos {{180}^{0}} \right)$
$\therefore {{W}_{2}}=1.5\times 0.22\times \left( 1-\left( -1 \right) \right)=1.5\times 0.22\times \left( 1+1 \right)=1.5\times 0.22\times 2=0.66J$ $\left( \because \cos {{0}^{0}}=1,\cos {{180}^{0}}=-1 \right)$
Therefore, the required amount of work done by the external torque in this case is $0.66J$.

(b)(i) Magnetic dipole (bar magnet) is placed normal to the magnetic field.
Let the angle made by the bar magnet with the direction of the magnetic field be $\theta$.
Since, the bar magnet is placed normal to the magnetic field, $\theta ={{90}^{0}}$.
Le the torque acting upon the bar magnet in this case be ${{\tau }_{1}}$.
Using (2), we get,
${{\tau }_{1}}=MB\sin \theta$
$\therefore {{\tau }_{1}}=1.5\times 0.22\times \left( \sin {{90}^{0}} \right)=1.5\times 0.22\times 1=0.33N.m$ $\left( \because \sin {{90}^{0}}=1 \right)$
Hence, the required torque on the bar magnet in this position is $0.33N.m$.

(b)(ii) Magnetic dipole (bar magnet) is placed in the direction opposite to the magnetic field.
Let the angle made by the bar magnet with the direction of the magnetic field be $\theta$.
Since, the bar magnet is placed opposite to the direction of the magnetic field, $\theta ={{180}^{0}}$.
Le the torque acting upon the bar magnet in this case be ${{\tau }_{2}}$.
Using (2), we get,
${{\tau }_{2}}=MB\sin \theta$
$\therefore {{\tau }_{2}}=1.5\times 0.22\times \left( \sin {{180}^{0}} \right)=1.5\times 0.22\times 0=0N.m$ $\left( \because \sin {{180}^{0}}=0 \right)$
Hence, the required torque on the bar magnet in this position is $0N.m$.

Note: Students must note that when the bar magnet is placed opposite to the direction of the magnetic field, its potential energy is maximum and it is in a state of unstable equilibrium. Even if a slight deflection is given to the bar magnet at this point, it will move away from this point. However, when the bar magnet is aligned in the direction of the magnetic field, it is in stable equilibrium and if a slight deflection is given to the bar magnet at this point, it tries to come back to the original alignment. This is an important difference between these two alignments since the torque on the magnet in both of these alignments is zero.