Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

262 of 276

Question 262 of 276

A short bar magnet placed with its axis at 30∘30^{\circ} with an external field of 800 Gauss, experiences a torque of 0.016 N.m0.016 \mathrm{~N}. \mathrm{m}. The work done in moving it from most stable to most unstable position is α×10−3 J\alpha \times 10^{-3} \mathrm{~J}. The value of α\alpha is ____\_\_\_\_.

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Answer: 64

For a magnetic dipole in an external field, torque is τ=mBsin⁡θ\tau = mB\sin\theta
At 30°: 0.016=m×0.08×sin⁡30°0.016 = m \times 0.08 \times \sin 30°
0.016=m×0.08×0.50.016 = m \times 0.08 \times 0.5
m=0.4m = 0.4 A·m2^2
Work done from most stable (aligned, 0°) to most unstable (anti-aligned, 180°) position equals the change in potential energy:
W=U0°−U180°=−mBcos⁡0°−(−mBcos⁡180°)=−mB−mB=2mBW = U_0° - U_{180°} = -mB\cos 0° - (-mB\cos 180°) = -mB - mB = 2mB
W=2×0.4×0.08=0.064W = 2 \times 0.4 \times 0.08 = 0.064 J =64×10−3= 64 \times 10^{-3} J
Therefore, α=64\alpha = 64

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