A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is W. The torque on the magnetic needle at this position is:

2 \sqrt{3} W

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\sqrt{3} W

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\frac{\sqrt{3}}{2} W

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\frac{\sqrt{3}}{4} W

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Solution

The correct option is B $\sqrt{3} W$
Work done in moving the needle in a uniform magnetic field from $θ_{1}$ to $θ_{2} = M B (c o s θ_{1} - c o s θ_{2})$
$⟹ W = M B (1 - c o s 60^{\circ}) = \frac{M B}{2}$
Torque on magnetic needle at an angle $θ$ with magnetic field, $τ = M B s i n θ = M B s i n 60^{\circ} = M B \frac{\sqrt{3}}{2} = \sqrt{3} W$

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A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 60∘ and the work done is W. The torque on the magnetic needle at this position is:

The correct option is B √3WWork done in moving the needle in a uniform magnetic field from θ1 to θ2=MB(cosθ1−cosθ2)⟹W=MB(1−cos60∘)=MB2Torque on magnetic needle at an angle θ with magnetic field, τ=MBsinθ=MBsin60∘=MB√32=√3W

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is W. The torque on the magnetic needle at this position is: