Question

The work done in rotating a magnet of magnetic moment $2A - {m^2}$ in a magnetic field of $5 \times {10^{ - 3}}T$ from the direction along the magnetic field to opposite direction to the magnetic field, is

(A) Zero
(B) $2 \times {10^{ - 2}}J$$$
(C) ${10^{ - 2}}J$
(D) $10J$

Answer 1

Hint:
Here first start with the given information provided in the question such as the value of the magnetic moment is 2, value of the magnetic field is $5 \times {10^{ - 3}}$ and the angle of rotation will be ${180^0}$( as the direction is opposite). Then use the formula for the work done in terms of magnetic moment, magnetic field and angle of rotation.




Complete step by step solution:
Let start with the given information in the question:
Magnetic moment of the magnet rotating in the magnetic field is $2A{m^2}$.
Magnetic field strength is$5 \times {10^{ - 3}}T$.
Also the magnetic is rotated at an angle opposite to the direction of magnetic field.
Therefore, $\theta = {180^0}$

Now, we know that work done W in rotating a magnet of magnetic moment M in a magnetic field B with the angle of rotation $\theta $ is given as follows:
$W = MB\left( {1 - \cos \theta } \right)$
Now, putting the value of magnetic moment, magnetic field and angle of rotation from the question, we get;
$W = MB\left( {1 - \cos {{180}^0}} \right)$
$W = 2 \times 5 \times {10^{ - 3}}\left( {1 - \left( { - 1} \right)} \right)$
$W = 2 \times 5 \times {10^{ - 3}} \times 2$
By solving, we get;
$W = 2 \times {10^{ - 2}}J$

Hence the correct answer is Option(B).







Note:
Here the value of the magnetic moment, magnetic field and the angle of rotation was already given in the question so it was easy to find the work done just by simply putting in the formula of work done. Also the angle of rotation was ${180^0}$ as the direction was given opposite to the direction of magnetic field.