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A pivoted magnetic needle of length $2 l$ and pole strength $' m'$ is at rest in magnetic meridian. It is held in equilibrium at an angle $θ$ to a magnetic induction field $B_{H}$ 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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F=(2l)mBHtanθr

F=r2lmBHtanθ

F=mBHtanθ2r

F=2rmBHtanθ

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detailed solution

Correct option is A

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=?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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