CBSE Class 12 · 2018 · Magnetic Properties of MatterShort Answer

A bar magnet of magnetic moment

6 J T^{- 1}

is aligned at

6 0^{\circ}

with a uniform external magnetic field of

0.44 T

. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

Answer: (a)(i) $1.32 J$ ; (a)(ii) $3.96 J$ ; (b) $0$ .

Solution

The work done in turning a dipole from

θ_{1}

θ_{2}

in a uniform field is

W = M B (cos θ_{1} - cos θ_{2}) .

Here

M = 6 J T^{- 1}

B = 0.44 T

and the initial angle

θ_{1} = 6 0^{\circ}

.
(a)(i) Turning to normal (

θ_{2} = 9 0^{\circ}

W = 6 \times 0.44 (cos 6 0^{\circ} - cos 9 0^{\circ}) = 6 \times 0.44 \times \frac{1}{2} = 1.32 J .

(a)(ii) Turning to opposite (

θ_{2} = 18 0^{\circ}

W = 6 \times 0.44 (cos 6 0^{\circ} - cos 18 0^{\circ}) = 6 \times 0.44 \times \frac{3}{2} = 3.96 J .

(b) In the final orientation of case (ii) the magnet is anti-parallel to

B

(

θ = 18 0^{\circ}

), so the torque is

τ = M B sin 18 0^{\circ} = 0.

W_{(i)} = 1.32 J, W_{(ii)} = 3.96 J, τ = 0

CBSE Class 12 Board · 2018 · Subjective

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CBSE Class 12 · 2018 · Magnetic Properties of MatterShort Answer

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