Question

A bar magnet having a magnetic movement of $2 \times 10^{4} JT ^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B =6 \times 10^{-4}$ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is :

• A. 12 J
• B. 6 J
• C. 2 J
• D. 0.6 J

Answer 1

The Correct Option is B

To solve this problem, we need to calculate the work done when a bar magnet is rotated in a magnetic field. The formula for the work done by a magnetic field when rotating a magnetic dipole is given by:

W = M \cdot B \cdot (\cos \theta_1 - \cos \theta_2)

where:

• W is the work done.
• M is the magnetic moment of the bar magnet.
• B is the magnetic field strength.
• \theta_1 and \theta_2 are the initial and final angles between the magnetic moment and the magnetic field.

Given:

• Magnetic moment, M = 2 \times 10^{4} \, JT^{-1}
• Magnetic field, B = 6 \times 10^{-4} \, T
• Initial angle, \theta_1 = 0^{\circ} \; (\text{{parallel to the field}})
• Final angle, \theta_2 = 60^{\circ}

Substituting the values into the formula:

W = 2 \times 10^{4} \times 6 \times 10^{-4} \times (\cos 0^{\circ} - \cos 60^{\circ})

Calculate the cosines:

• \cos 0^{\circ} = 1
• \cos 60^{\circ} = 0.5

Now plug the cosine values back into the equation:

W = 2 \times 10^{4} \times 6 \times 10^{-4} \times (1 - 0.5)

W = 2 \times 10^{4} \times 6 \times 10^{-4} \times 0.5

W = 6 \, J

Therefore, the work done in rotating the magnet is 6 \, J.

Thus, the correct answer is 6 J.