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A magnetic needle suspended parallel to a magnetic field requires √3J of work

A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60⁰. The torque needed to maintain the needle in this position will be:

Options

(a) 2 √3 J
(b) 3 J
(c) √3 J
(d) 3/2 J

Correct Answer:

3 J

Explanation:

According to work energy theorem
W = U final – U initial = MB (cos 0 – cos 60⁰)
W = MB/2 = √3J …(i)
? = M⃗ x B⃗ = MB sin 60⁰ = (MB √3 / 2) …(ii)
From eq (i) and (ii)
? = 2 √3 x √3 / 2 = 3 J

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