Question

A bar magnet having a magnetic movement $2 \times {10^ - }^4J{T^ - }^1$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times {10^ - }^4T$ exists in space. The work done in taking the magnet slowly from a direction parallel to the field to a direction 60 degree from the field is:
A. \[2{\text{ }}J\]
B. \[0.6{\text{ }}J\]
C. \[12{\text{ }}J\]
D. \[6{\text{ }}J\]

Answer 1

Hint: The magnetic dipole moment is known as the Magnetic moment.
The object's tendency which is aligned with the magnetic field is measured by the magnetic moment. The magnetic moments' direction is the south to the north pole

Complete step by step solution:
Here, the initial direction of the magnet is equal to the parallel field and they are zero degrees.
From the given data magnet’s final direction is $60^\circ $
The magnetic field is $B = 6 \times {10^{ - 4}}$
The magnetic movement is, $T = 2 \times {10^4}J{T^{ - 1}}$
When rotating a magnet in the magnetic field the expression is given by,
$W = B.T\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$
Now substitute the given details in the above equation we get,
$W = 6 \times {10^{ - 4}} \times 2 \times {10^4}\left( {1 - 0.5} \right)$
Now by solving the above equation we get work done,
$W = 6J$
Hence, it requires an $6J$of work done to rotate the magnet from the $0^\circ $ to $60^\circ $
So, the correct answer is an option (D) $6J$

Note:
The motion of the electric charge and the spin angular momentum is used to generate the magnetic momentum.
The magnet's orientation and the magnet's strength are defined as the magnetic moment measured by the magnetometers.
The magnetic moment is directly proportional to the magnetic field which is created by the magnet.