Question

A bar magnet of length \(0.1~\text{m}\) and magnetic moment of \(5~\text{Am}^2\) is placed in a uniform magnetic field of intensity \(0.4~\text{T}\), with its axis making an angle of \(60^\circ\) with the field. The torque on the magnet is

• A. \(1.532~\text{Nm}\)
• B. \(1.414~\text{Nm}\)
• C. \(0\)
• D. \(1.732~\text{Nm}\)

Hint:

Use: \(\tau = MB\sin\theta\) for torque on a magnetic dipole in a magnetic field.

Answer 1

The Correct Option is D

Torque on a magnetic dipole in a magnetic field is given by: \[ \tau = MB\sin\theta \] Where:
- \(M = 5~\text{Am}^2\), magnetic moment
- \(B = 0.4~\text{T}\), magnetic field
- \(\theta = 60^\circ\)
\[ \tau = 5 \cdot 0.4 \cdot \sin(60^\circ) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.732~\text{Nm} \]

Answer 2

Step 1: Understand the given data
- Length of bar magnet, \(l = 0.1~\text{m}\) (not needed for torque calculation directly)
- Magnetic moment, \(M = 5~\text{Am}^2\)
- Magnetic field intensity, \(B = 0.4~\text{T}\)
- Angle between magnetic moment and magnetic field, \(\theta = 60^\circ\)

Step 2: Recall the formula for torque on a magnetic dipole
The torque \(\tau\) experienced by a magnetic dipole in a magnetic field is:
\[ \tau = M B \sin \theta \]

Step 3: Calculate the torque
\[ \tau = 5 \times 0.4 \times \sin 60^\circ = 2 \times \frac{\sqrt{3}}{2} = 2 \times 0.866 = 1.732~\text{Nm} \]

Step 4: Final answer
The torque on the magnet is \(1.732~\text{Nm}\).