Question

A current carrying circular loop of magnetic moment \( \vec{M} \) is suspended in a vertical plane in an external magnetic field \( \vec{B} \) such that its plane is normal to \( \vec{B} \). The work done in rotating this loop by 45° about an axis perpendicular to \( \vec{B} \) is closest to:

• A. \(-0.3 MB\)
• B. \(0.3 MB\)
• C. \(-1.7 MB\)
• D. \(1.7 MB\)

Hint:

Remember, the work done on the system in magnetic fields involves changes in potential energy that reflect the orientation of the magnetic moment relative to the magnetic field direction.

Answer 1

The Correct Option is B

Step 1: Determine initial magnetic potential energy. The magnetic potential energy \( U \) of a magnetic dipole in a magnetic field is \( U = -\vec{M} \cdot \vec{B} \). When the loop's plane is perpendicular to \( \vec{B} \), \( \vec{M} \) aligns with \( \vec{B} \), resulting in initial potential energy \( U_i = -M B \). Step 2: Calculate final magnetic potential energy. After rotating the loop by 45 degrees, the angle \( \theta \) between \( \vec{M} \) and \( \vec{B} \) becomes \( 45^\circ \). Since \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \), the final potential energy \( U_f \) is: \[ U_f = -MB \cos(45^\circ) = -MB \frac{\sqrt{2}}{2} \] Step 3: Compute work done. The work done \( W \) in rotating the dipole equals the change in potential energy: \[ W = U_f - U_i = \left(-MB \frac{\sqrt{2}}{2}\right) - (-MB) \] \[ W = MB \left(1 - \frac{\sqrt{2}}{2}\right) \] Using \( \sqrt{2} \approx 1.414 \), this simplifies to: \[ W = MB \left(1 - 0.707\right) \] \[ W = MB (0.293) \] For simplification and clarity in answer choices, this can be approximated as: \[ W \approx 0.3 MB \]