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Q.

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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a

$F = \frac{(2 l) m B_{H} \tan θ}{r}$

b

$F = \frac{r}{2 l m B_{H} \tan θ}$

c

$F = \frac{m B_{H} \tan θ}{2 r}$

d

$F = \frac{2 r}{m B_{H} \tan θ}$

answer is A.

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Detailed Solution

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$\frac{I n equilibrium, torque due to F is balanced by torque due to B_{H}}{} ∴ τ_{F} = τ_{B_{H}} \Rightarrow (F \cos θ) r = M B_{H} \sin θ \Rightarrow F = \frac{M B_{H} \sin θ}{\cos θ} \times \frac{1}{r} \Rightarrow F = \frac{m (2 l) B_{H} \tan θ}{r}$

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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=?