Question

In a uniform magnetic field \(\vec{B}\), a bar magnet of magnetic moment \(M\) is kept suspended at an angle of \(60^\circ\) with respect to \(\vec{B}\). The work done to turn it from \(60^\circ\) to \(90^\circ\) with respect to the field is

• A. $MB$
• B. $\sqrt{2}MB$
• C. $0$
• D. $2MB$
• E. $\frac{MB}{2}$

Hint:

Magnetic dipole energy: - $U = -MB\cos\theta$ - Work = change in potential energy

Answer 1

Answer: (E)

Concept: Potential energy of magnetic dipole in magnetic field: \[ U = -MB\cos\theta \] Work done: \[ W = U_2 - U_1 \]

Step 1: Find initial energy.
\[ U_1 = -MB\cos 60^\circ = -MB \cdot \frac{1}{2} = -\frac{MB}{2} \]

Step 2: Find final energy.
\[ U_2 = -MB\cos 90^\circ = 0 \]

Step 3: Calculate work done.
\[ W = U_2 - U_1 = 0 - \left(-\frac{MB}{2}\right) = \frac{MB}{2} \]