A) 10 mH
B) 50 mH
C) 200 mH
D) 100 mH
Correct Answer: D
Solution :
In resonance condition, maximum current flows in the circuit. Current in the LCR circuit is given by \[\therefore \] For current to be maximum denominator should be minimum \[{{E}_{R}}=\frac{\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[\tau =qEL\sin \theta \] \[=qEL\theta \] \[\because \] \[\theta \] \[\text{I=m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{+m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{=}\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}\] \[\tau =I\alpha \] \[\text{ }\!\!\alpha\!\!\text{ =}\frac{\text{ }\!\!\tau\!\!\text{ }}{\text{I}}\text{=}\frac{\text{qEL }\!\!\theta\!\!\text{ }}{\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}}\] \[\Rightarrow \] \[{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ =}\frac{\text{2qEL }\!\!\theta\!\!\text{ }}{\text{m}{{\text{L}}^{\text{2}}}}\] L = 0.1 H = 100 mH
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