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A Short Bar Magnet of Magnetic Moment 0.9 J/T is Placed with Its Axis at 30° to a Uniform Magnetic Field. It Experiences a Torque of 0.063 J.(I) Calculate the Magnitude of the Magnetic Field. - Physics

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Question

A short bar magnet of magnetic moment 0.9 J/T is placed with its axis at 30° to a uniform magnetic field. It experiences a torque of 0.063 J.

(i) Calculate the magnitude of the magnetic field.

(ii) In which orientation will the bar magnet be in stable equilibrium in the magnetic field?

Solution

(i) Magnetic moment M = 0.9 J/T

τ = 0.063 J, θ = 30°

We know τ = M × B

= MB sin θ

0.063 = 0.9 × B × sin 30°

`B = (2 xx 0.063)/0.9 = 0.14T`

(ii) Stable equilibrium is position of minimum energy. Since `U = -vecM.vecB`

U = − M B cos θ

Where, U is the energy stored or P.E. of the magnet inside magnetic field B.

So, when θ = 0, U = − MB is the minimum energy.

Thus, when `vecM` and `vecB` are parallel to each other bar magnet is in stable equilibrium.

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