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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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प्रश्न

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?

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उत्तर

(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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2011-2012 (March) Foreign Set 1

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