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Step-by-Step Solution
Step 1: Understand the Physical Situation
A bar magnet with a magnetic moment $M$ is placed in a uniform horizontal magnetic field $B$. The magnet is initially oriented parallel to the field (i.e., the angle between the magnet’s axis and the field is $0^\circ$), and it is slowly rotated to $60^\circ$ from the field direction. We need to find the work done in this process.
Step 2: Recall the Formula for Potential Energy of a Magnetic Dipole
The potential energy $U$ of a magnetic dipole of moment $M$ placed in a uniform magnetic field $B$, oriented at an angle $\theta$ to the field, is given by:
$U = - M B \cos \theta$
Step 3: Work Done in Changing the Orientation
The work done in rotating the dipole from an initial angle $\theta_1$ to a final angle $\theta_2$ is given by the change in potential energy:
$W = U_{\text{final}} - U_{\text{initial}} = -M B \cos \theta_2 - \bigl(-M B \cos \theta_1\bigr)$
$W = M B \bigl(\cos \theta_1 - \cos \theta_2\bigr)$
Step 4: Substitute the Given Values
Magnetic moment, $M = 2 \times 10^4 \,\text{J}\,\text{T}^{-1}$
Magnetic field, $B = 6 \times 10^{-4}\,\text{T}$
Initial angle, $\theta_1 = 0^\circ$ (parallel to field, $\cos 0^\circ = 1$)
Final angle, $\theta_2 = 60^\circ$ (so, $\cos 60^\circ = \tfrac{1}{2}$)
Step 5: Calculate the Work Done
$W = M B \left(\cos 0^\circ - \cos 60^\circ\right)$
$W = (2 \times 10^4) \times (6 \times 10^{-4}) \left(1 - \tfrac{1}{2}\right)$
$W = (2 \times 10^4) \times (6 \times 10^{-4}) \times \tfrac{1}{2}
= 12 \times 10^0 \times \tfrac{1}{2}
= 6 \,\text{J}
$
Step 6: Final Answer
The work done in taking the magnet from parallel orientation to $60^\circ$ orientation is 6 J.
