Magnetic Properties of Matter: Practice Problem & Solution
15. A bar magnet having a magnetic moment of $2 \times 10^4$ J T$^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times 10^{-4}$ T exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is: (2009)
Solution Explained:
To solve this problem, we apply the core principles of Magnetic Properties of Matter. Understanding the underlying formula is key to arriving at the correct answer below:
Work done, $W = MB(\cos \theta_1 - \cos \theta_2) = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - \frac{1}{2}) = \frac{MB}{2}$.
$W = \frac{1}{2} \times (2 \times 10^4) \times (6 \times 10^{-4}) = 6$ J.
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