A bar magnet of magnetic moment 6 J/T is aligned at 60° 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).

Solution

M = 6J/T

θ = 60^o

B = 0.44T

$τ = mB sinθ τ = 6 x 0.44 \sin 60^{o} = 6 x 0.44 x \frac{\sqrt{3}}{2} = 3 \sqrt{3} x 0.44 = 2.836 dw \int_{θ = 60^{0}}^{90^{0}} τ . dθ = - mB [\cos 90^{0} - \cos 60^{0}] = 6 x 0.44 x [- \frac{1}{2}] = 3 x 0.44 = 1.32 J (ii) dw = \int_{θ = 60^{0}}^{180^{0}} mB \sin θ . dθ = - mB [\cos 180^{0} - \cos 60^{0}] = 6 x 0.44 x [- 1 - \frac{1}{2}] = - 6 x 0.44 [- \frac{3}{2}] = 9 x 0.44 = 39.6 J (b) \vec{τ} = \vec{m} x \vec{B} τ = m Bsin θ = 6 x 0.44 x \sin 180^{0} τ = 0$

Some More Questions From Moving Charges And Magnetism Chapter

Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω N₁ = 30,
A₁ = 3.6 x 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω; N₂ = 42,
A₂ = 1.8 x 10^–3 m², B₂ = 0.50 T
(The spring constants k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M_1.

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Moving Charges And Magnetism

Some More Questions From Moving Charges And Magnetism Chapter

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