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 bar magnet (magnetic dipole of moment $M$) is placed in an external field $B$ at angle $\theta$:
\[ \tau = MB\sin\theta \]

Step 2: Compute $B\sin\theta$ first for simplification.
$B = 28.3 \times 10^{-3}\,\text{T}$ and $\theta = 45^\circ$, so $\sin 45^\circ = 1/\sqrt{2}$:
\[ B\sin\theta = 28.3 \times 10^{-3} \times \frac{1}{\sqrt{2}} = \frac{28.3 \times 10^{-3}}{1.414} \approx 20 \times 10^{-3}\,\text{T} \]

Step 3: Rearrange the torque formula to solve for $M$.
\[ M = \frac{\tau}{B\sin\theta} \]

Step 4: Substitute the values.
\[ M = \frac{3.6 \times 10^{-5}}{20 \times 10^{-3}} = \frac{3.6 \times 10^{-5}}{2 \times 10^{-2}} = 1.8 \times 10^{-3}\,\text{J\,T}^{-1} \]

Step 5: Check units.
$[\tau/B] = \text{J}/\text{T} = \text{A\,m}^2$, which is the unit of magnetic moment. Correct.

Step 6: State the answer.
\[ \boxed{1.8 \times 10^{-3}\,\text{J\,T}^{-1}} \]