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A pivoted magnetic needle of length 2l and pole strength 'm' is at rest in magnetic meridian. It is held in equilibrium at an angle θ to a magnetic induction field BH by applying a force F at a distance 'r' from the pivot along a direction perpendicular to the field, then F=?

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a

F=(2l)mBHtanθr

b

F=r2lmBHtanθ

c

F=mBHtanθ2r

d

F=2rmBHtanθ

answer is A.

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Detailed Solution

In equilibrium, torque due to F is balanced by torque due to BH ∴τF=τBH ⇒ ( Fcosθ)r=MBHsinθ ⇒F=MBHsinθcosθ×1r ⇒F=m(2l)BHtanθr

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A pivoted magnetic needle of length 2l and pole strength 'm' is at rest in magnetic meridian. It is held in equilibrium at an angle θ to a magnetic induction field BH by applying a force F at a distance 'r' from the pivot along a direction perpendicular to the field, then F=?