Question

A bar magnet of magnetic moment $1.5A\,m^{2}$ is initially placed in the direction of a uniform magnetic field of $18\times10^{-2}$ T. The work to be done to rotate the magnet so that its magnetic moment becomes perpendicular to the direction of the magnetic field is}

• A. 540 mJ
• B. 135 mJ
• C. 270 mJ
• D. 90 mJ

Hint:

For magnetic dipoles, \[ U=-MB\cos\theta \] and \[ W=\Delta U \] Always calculate the change in potential energy between initial and final orientations.

Answer 1

The Correct Option is C

Concept: The potential energy of a magnetic dipole in a uniform magnetic field is \[ U=-MB\cos\theta \] where \[ M=\text{magnetic moment} \] \[ B=\text{magnetic field} \] \[ \theta=\text{angle between } \vec M \text{ and } \vec B \] The work done in rotating the dipole is equal to the increase in potential energy.

Step 1: Calculate the initial potential energy.
Initially the magnet is along the magnetic field. Hence, \[ \theta_1=0^\circ \] \[ U_1=-MB\cos0^\circ \] \[ U_1=-MB \]

Step 2: Calculate the final potential energy.
Finally the magnet becomes perpendicular to the field. \[ \theta_2=90^\circ \] \[ U_2=-MB\cos90^\circ \] \[ U_2=0 \]

Step 3: Find the work done.
\[ W=U_2-U_1 \] \[ W=0-(-MB) \] \[ W=MB \] Substituting, \[ M=1.5\,A\,m^2 \] \[ B=18\times10^{-2}\,T \] \[ W=1.5\times18\times10^{-2} \] \[ W=27\times10^{-2} \] \[ W=0.27J \]

Step 4: Convert into millijoules.
\[ 0.27J=270mJ \] \[ \boxed{270mJ} \]