Question

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 \(30^{\circ}\) with the field. The value of F is:

• A. 2.75 N
• B. 2.93 N
• C. 3.75 N
• D. 4.08 N

Hint:

Balance torques: \(F d = MB\sin\theta\).

Answer 1

The Correct Option is C

Step 1: Understanding the concept.
The field tries to turn the magnet with a torque. The applied force F gives an opposite torque about the pivot. For equilibrium the two torques are equal.

Step 2: Torque by the field.
\[ \tau_B = MB\sin\theta = 3 \times 0.25 \times \sin 30^{\circ} = 3 \times 0.25 \times 0.5 = 0.375 \text{ N m} \]

Step 3: Torque by the force.
F acts at right angles to the axis at distance 0.10 m from the pivot, so \[ \tau_F = F \times 0.10 \]

Step 4: Balance the torques.
\[ F \times 0.10 = 0.375 \Rightarrow F = 3.75 \text{ N} \]

Step 5: Check the options.
2.75 N, 2.93 N and 4.08 N do not satisfy the balance. 3.75 N is option 3.

Final Answer:
F = 3.75 N. \[ \boxed{3.75 \text{ N}} \]