A bar magnet of dipole moment 3 Am2 rests with its centre on a frictionless point. A force F is applied at right angles to the axis of the magnet, 10 cm from the point. It is observed that an external magnetic field of 0.25 T is required to hold the magnet in equilibrium at an angle of 30° with the field. The value of F is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.75 N
The magnet can only rotate about the pivot, so it is in equilibrium when the torques balance: the torque of the field on the dipole equals the torque of the force F about the pivot.
\( \displaystyle \tau_{\mathrm{field}} = mB\sin\theta \)
\( \displaystyle \tau_{\mathrm{field}} = 3\,\mathrm{A\,m^{2}} \times 0.25\,\mathrm{T} \times \sin 30^\circ = 0.375\,\mathrm{N\,m} \)
F acts at right angles to the axis at \( r = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \), so its torque is Fr.
\( \displaystyle F \times 0.1\,\mathrm{m} = 0.375\,\mathrm{N\,m} \)
\( \displaystyle F = 3.75\,\mathrm{N} \)