Question

The work done in turning a magnetic dipole of magnetic moment \(M\) in a magnetic field \(B\) by an angle \(90^\circ\) from the meridian is \(n\) times the corresponding work done in turning it through an angle of \(60^\circ\) in the same field and from the same initial condition. Find the value of \(n\).

• A. \(\frac14\)
• B. \(\frac12\)
• C. \(2\)
• D. \(4\)

Hint:

For a magnetic dipole: \[ U=-MB\cos\theta. \] Work done equals the change in potential energy.

Answer 1

The Correct Option is C

Concept: Potential energy of a magnetic dipole is \[ U=-MB\cos\theta. \] Work done in rotating the dipole from \(\theta_1\) to \(\theta_2\) is \[ W=MB(\cos\theta_1-\cos\theta_2). \]

Step 1: Calculate work done for rotation through \(90^\circ\). Initially the dipole is along the meridian, \[ \theta_1=0^\circ. \] Final position: \[ \theta_2=90^\circ. \] Hence, \[ W_{90} = MB(\cos0^\circ-\cos90^\circ) \] \[ = MB(1-0) \] \[ = MB. \]

Step 2: Calculate work done for rotation through \(60^\circ\). \[ W_{60} = MB(\cos0^\circ-\cos60^\circ) \] \[ = MB\left(1-\frac12\right) \] \[ = \frac{MB}{2}. \]

Step 3: Find \(n\). \[ n = \frac{W_{90}}{W_{60}} \] \[ = \frac{MB}{MB/2} \] \[ =2. \] \[\begin{aligned} \boxed{2} \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.