Physics, asked by Ankush254, 1 year ago

A magnet of magnetic moment M is lying along a uniform magnetic field B the work done in rotating the magnet by 90 is

Answers

Answered by shirleywashington
3

Answer : W = M B

Explanation :

It is given that a magnet if magnetic moment M is lying along a uniform magnetic field B. the work done in rotating the magnet by 90^0 is :

W=\int\limits^a_b {MB\ sin\theta} \, d\theta

W=\int\limits^{90^0}_0 {M.Bsin\theta} \, d\theta

W=-MB[cos\ 90^0-cos\ 0]

W=MB

So, the work done in rotating the magnet by 90^0 is MB.

Hence, this is the required solution.

Answered by branta
3

Answer: The work done in rotating the magnet by ninety degree is MB.

Explanation:

In the given problem, a magnet of magnetic moment M is lying along a uniform magnetic field B.

The expression for torque in terms of magnetic moment and magnetic field is as follows;

\tau =MBsin\theta

Here, tau is torque, M is the magnetic moment, B is the magnetic field and theta is the angle.

The expression for the work done in rotating the magnet is as follows;

W = \int_{0}^{\theta} \tau\ d\theta

Put \theta =90\degree and \tau =MBsin\theta.

W = \int_{0}^{90\degree} \MBsin\theta\ d\theta

W = MB\ [-cos\theta]_{0}^{90^{o}}W = MB [-cos90^{o} + cos0^{0}]

W =MB

Therefore, the work done in rotating the magnet by ninety degree is MB.

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