Question

A short magnetic needle is placed in a magnetic field \(B\hat{i}\) in the direction \((\sqrt{3}\hat{i}+\hat{j})\). The needle experiences a torque of \(0.06\ \text{N m}\). If the same magnetic needle is placed in a magnetic field \(2B\hat{j}\) in the direction \((\hat{i}+\sqrt{3}\hat{j})\), the torque experienced by it is

• A. \(0.12\ \text{N m}\)
• B. \(0.84\ \text{N m}\)
• C. \(0.10\ \text{N m}\)
• D. \(0.03\ \text{N m}\)

Hint:

The torque on a magnetic dipole is \[ \tau=MB\sin\theta. \] When the magnetic field doubles and the angle remains the same, the torque also doubles.

Answer 1

The Correct Option is A

Step 1: Find magnetic moment from first configuration.
First field: \( B\hat{i} \), needle direction \( \hat{n}_1 = (\sqrt{3}\hat{i}+\hat{j})/2 \). Torque \( \tau = mB\sin\theta \) where \( \theta \) is angle between needle and field. \( \cos\theta = \sqrt{3}/2 \Rightarrow \theta = 30^\circ \), \( \sin\theta = 1/2 \). \( \tau_1 = mB\times\frac{1}{2} = 0.06 \Rightarrow mB = 0.12 \).

Step 2: Second configuration.
Field: \( 2B\hat{j} \), needle direction \( \hat{n}_2 = (\hat{i}+\sqrt{3}\hat{j})/2 \). Angle between needle and field: \( \cos\phi = \sqrt{3}/2 \Rightarrow \phi = 30^\circ \). \( \tau_2 = m(2B)\sin30^\circ = 0.12\times2\times\frac{1}{2} = 0.12\text{ N m} \).

\[ \boxed{\tau_2 = 0.12\text{ N m}} \]