A short bar magnet is placed in an external magnetic field of $600\,G$ . When its axis makes an angle of ${30^ \circ }$ with the external field, it experiences a torque of $0.012\,Nm$ . What is the magnetic moment of the magnet? What is the work done in moving it from its most stable to most unstable position? The bar magnet is replaced by a solenoid of cross sectional area $2 \times {10^{ - 4}}\,{m^2}$ and $1000$ turns, but having the same magnetic moment. Determine the current flowing through the solenoid.
(A) $m = 0.2\,A{m^2},\,W = 0.048\,J,\,I = 2.0\,A$ (B) $m = 0.4\,A{m^2},\,W = 0.048\,J,\,I = 3.0\,A$
(C) $m = 0.4\,A{m^2},\,W = 0.048\,J,\,I = 2.0\,A$
(D) $m = 0.04\,A{m^2},\,W = 0.048\,J,\,I = 2.0\,A$

Hint:Use the below formula of the magnetic moment and substitute the values known to find the value of the moment. Then substitute this and also the other required values in the work done formula, to find the work done to move the magnet from the stable to unstable position.

Formula used:

(1) The formula of the magnetic moment is given by

$M = \dfrac{T}{{B\sin \theta }}$

Where $M$ is the magnetic moment, $T$ is the torque experienced by the magnet, $B$ is the magnetic field and $\theta $ is the angle between the axis of the magnet with the external magnetic field.

(2) The formula of the work done to move the magnet from a stable to the unstable position is given by

$W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)$

Where $W$ is the work done to move the magnet.

Complete step by step solution:
It is given that the
The magnetic field, $B = 600\,G$
The angle that the axis makes with the magnetic field, $\theta = {30^ \circ }$
The torque experienced by the magnet, $T = 0.012\,Nm$

Using the formula of the magnetic moment,

$M = \dfrac{T}{{B\sin \theta }}$

Substituting the known values in the above formula,

$M = \dfrac{{0.012}}{{600 \times {{10}^{ - 4}} \times \sin {{30}^ \circ }}}$

By performing the various arithmetic an the trigonometric operation, we get

$M = 0.4\,A{m^2}$

Hence the magnetic moment is obtained as $0.4\,A{m^2}$ .

Using the formula (2) to calculate the work done,

$W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)$

Substituting the known parameters in it, we get

$W = - 0.4 \times 600 \times {10^{ - 4}}\left( {\cos {{180}^ \circ } - \cos {0^ \circ }} \right)$

By simplifying the above equation, we get

$W = 0.048\,J$

Hence the work done to move the magnet from the stable to the unstable position is $0.048\,J$ .

Note:In the above solution, the value of the one Gauss is substituted as $1 \times {10^{ - 4}}$ Tesla. This is because the derived $SI$ unit of the magnetic moment and it is derived and formed as the unit of $Wb{m^{ - 2}}$ . The most stable position is taken at ${180^0}$ , this is because the magnet has maximum potential energy at this angle and minimum at ${0^ \circ }$ .