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When a current carrying straight conductor is placed in a uniform magnetic field, making an angle of $30^{0}$ with the direction of field, force experienced by the conductor is 10 N. when the same current carrying conductor is placed in the same magnetic field making an angle of $60^{0}$ with the direction of field, force experienced by the conductor will be

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$5 \sqrt{3} N$

$10 \sqrt{3} N$

$\frac{20}{\sqrt{3}} N$

$15 \sqrt{3} N .$

answer is D.

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Detailed Solution

$F = ilB sinθ ∴ \frac{F_{2}}{F_{1}} = \frac{{sinθ}_{2}}{{sinθ}_{1}} \Rightarrow \frac{F_{2}}{10} = \frac{\sin 60^{0}}{\sin 30^{0}} \Rightarrow F_{2} = 10 \sqrt{3} N$

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