Magnetic Properties of Matter: Practice Problem & Solution
15. A bar magnet having a magnetic moment of $2 times 10^4text{ JT}^{-1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 times 10^{-4}text{ T}$ 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 is given by $W = MB(1 - costheta)$. Substituting the values: $W = (2 times 10^4) times (6 times 10^{-4}) times (1 - cos 60^{circ}) = 12 times 0.5 = 6text{ J}$.
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