Question

If a magnetic dipole of moment M situated in the direction of a magnetic field B is rotated by \(180^\circ\), then the amount of work done is

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

Hint:

Rotating a dipole from stable equilibrium (\(0^\circ\)) to unstable equilibrium (\(180^\circ\)) requires the maximum amount of work, which is exactly \(2 MB\). If rotated only to \(90^\circ\), the work would be \(MB\).

Answer 1

The Correct Option is B

Concept: The potential energy (\(U\)) of a magnetic dipole in a uniform magnetic field is given by: \[ U = -MB \cos \theta \] where \(\theta\) is the angle between the magnetic moment \(\vec{M}\) and the magnetic field \(\vec{B}\).
• Work Done (\(W\)): The work done by an external agent to rotate the dipole is equal to the change in its potential energy: \[ W = \Delta U = U_{final} - U_{initial} \] \[ W = -MB \cos \theta_2 - (-MB \cos \theta_1) = MB(\cos \theta_1 - \cos \theta_2) \]

Step 1: Identify the initial and final angles.
- The dipole is initially situated "in the direction of the field," meaning \(\theta_1 = 0^\circ\). - It is rotated by \(180^\circ\), so the final angle \(\theta_2 = 180^\circ\).

Step 2: Calculate the change in energy.
\[ W = MB(\cos 0^\circ - \cos 180^\circ) \] Since \(\cos 0^\circ = 1\) and \(\cos 180^\circ = -1\): \[ W = MB(1 - (-1)) \] \[ W = MB(1 + 1) = 2 MB \]