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: Use the torque formula for a magnetic dipole.
The torque on a magnetic needle is \[ \tau = MB\sin\theta, \] where \[ M=\text{magnetic dipole moment}, \] \[ B=\text{magnetic field}, \] and \[ \theta=\text{angle between } \vec{M} \text{ and } \vec{B}. \]

Step 2: Determine the angle in the first case.
The magnetic field is along \[ \hat{i}. \] The needle is directed along \[ (\sqrt3\hat{i}+\hat{j}). \] Therefore, \[ \tan\theta_1=\frac{1}{\sqrt3}. \] Hence, \[ \theta_1=30^\circ. \] Given torque, \[ \tau_1=0.06\ \text{N m}. \] Thus, \[ 0.06=MB\sin30^\circ. \] \[ 0.06=MB\left(\frac12\right). \] \[ MB=0.12. \]

Step 3: Determine the angle in the second case.
The magnetic field is along \[ \hat{j}. \] The needle is directed along \[ (\hat{i}+\sqrt3\hat{j}). \] Now, \[ \tan\theta_2=\frac{1}{\sqrt3}. \] Hence, \[ \theta_2=30^\circ. \] The magnetic field magnitude is \[ 2B. \] Therefore, \[ \tau_2=M(2B)\sin30^\circ. \] \[ \tau_2=2MB\left(\frac12\right). \] \[ \tau_2=MB. \] Using \[ MB=0.12, \] we obtain \[ \tau_2=0.12\ \text{N m}. \]

Step 4: Final conclusion.
Hence, the torque experienced by the magnetic needle is \[ \boxed{0.12\ \text{N m}} \] Therefore, the correct option is \[ \boxed{(1)} \]