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The work done in turning a magnet of magneticmoment M by an angle of 90° from the meridian,is n times the corresponding work done to turnit through an angle of 60°. The value of n isgiven by      [1995]
  • a)
    2
  • b)
    1
  • c)
    0.5
  • d)
    0.25
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The work done in turning a magnet of magneticmoment M by an angle of 9...
Magnetic moment = M; Initial angle through which magnet is turned (θ1) = 90º and final angle through which magnet is turned (θ2)= 60º.
Work done in turning the magnet
through 90º(W1) = MB (cos 0º – cos  0º)
= MB (1–0) = MB.
Similarly, W2 = MB (cos 0º – cos 60º)
∴ W1 = 2W2 or n = 2.
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Most Upvoted Answer
The work done in turning a magnet of magneticmoment M by an angle of 9...
Degrees is given by:

W = MΔθ

where W is the work done, M is the magnetic moment of the magnet, and Δθ is the angle through which the magnet is turned.

If the angle of rotation is 90 degrees, then Δθ = 90 degrees. Therefore, the work done in turning the magnet by 90 degrees is:

W = M × 90

W = 90M

So, the work done is directly proportional to the magnetic moment of the magnet. The larger the magnetic moment, the more work is required to turn the magnet.
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The work done in turning a magnet of magneticmoment M by an angle of 9...
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