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Q.

A current carrying straight conductor is placed in a uniform magnetic field and the force experienced by it is $10 \sqrt{3}$ N. If the maximum possible force experienced by the conductor, when placed in the same magnetic field is 20 N, then in the first case the angle between the direction of current in the conductor and the magnetic field is

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a

$\tan^{- 1} (\frac{2}{\sqrt{3}})$

b

$\sin^{- 1} (\frac{1}{\sqrt{3}})$

c

30°

d

60°

answer is D.

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$\begin{array}{l} \vec{F_{B}} = I \vec{PQ} \times \vec{B} \\ \Rightarrow F_{B} = I (PQ) B sinθ \\ Also, {(F_{B})}_{\max} = I (PQ) B . \\ ∴ Sinθ = \frac{F_{B}}{{(F_{B})}_{\max}} = \frac{10 \sqrt{3}}{20} = \frac{\sqrt{3}}{2} \Rightarrow θ = 60^{o} \end{array}$

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