Question

A short bar magnet placed with its axis at \(45^\circ\) with a uniform external magnetic field of \(28.3\times 10^{-3}\,T\) experiences a torque of magnitude equal to \(3.6\times 10^{-5}\,J\). The magnitude of magnetic moment of the magnet is nearly

• A. \(1.8\times 10^{-3}\,J\,T^{-1}\)
• B. \(1.2\times 10^{-3}\,J\,T^{-1}\)
• C. \(2.4\times 10^{-3}\,J\,T^{-1}\)
• D. \(1.6\times 10^{-3}\,J\,T^{-1}\)

Hint:

For a magnetic dipole in a uniform magnetic field, torque is given by \[ \tau = MB\sin\theta \] So, magnetic moment can be found using \[ M=\frac{\tau}{B\sin\theta} \]

Answer 1

The Correct Option is A

Step 1: Write the formula for torque on a magnetic dipole.
When a magnetic dipole of magnetic moment \(M\) is placed in a uniform magnetic field \(B\), the torque experienced by it is \[ \tau = MB\sin\theta \] where \(\theta\) is the angle between the magnetic moment and the magnetic field.

Step 2: Substitute the given values.
Given, \[ \tau = 3.6\times 10^{-5}\,J \] \[ B = 28.3\times 10^{-3}\,T \] \[ \theta = 45^\circ \] Also, \[ \sin 45^\circ = \frac{1}{\sqrt{2}} \] Using \[ \tau = MB\sin\theta \] we get \[ 3.6\times 10^{-5}=M(28.3\times 10^{-3})\sin 45^\circ \]

Step 3: Find the magnetic moment.
\[ M=\frac{3.6\times 10^{-5}}{(28.3\times 10^{-3})\sin 45^\circ} \] \[ M=\frac{3.6\times 10^{-5}}{(28.3\times 10^{-3})\times \frac{1}{\sqrt{2}}} \] \[ M=\frac{3.6\times 10^{-5}\times \sqrt{2}}{28.3\times 10^{-3}} \] Using \[ \sqrt{2}\approx 1.414 \] \[ M=\frac{3.6\times 1.414\times 10^{-5}}{28.3\times 10^{-3}} \] \[ M=\frac{5.0904\times 10^{-5}}{28.3\times 10^{-3}} \] \[ M\approx 1.8\times 10^{-3}\,J\,T^{-1} \]

Step 4: Final conclusion.
Therefore, the magnetic moment of the magnet is nearly \[ \boxed{1.8\times 10^{-3}\,J\,T^{-1}} \]