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A long solenoid has `500` turns. When a current of `2A` is passed through it, the resulting magnetic flux linked with each turn of the solenoid is `4xx10^(-3)Wb` . The self-inductance of the solenoid isA. 1.0 henryB. 4.0 henryC. 2.5 henryD. 2.0 henry |
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Answer» Correct Answer - A `"Flux linkage "="Flux through each turn"xx"number of turn"` `phi=500xx4xx10^(-3)=2Wb` since `phi=Li` `rArr L=(phi)/(i)=(2)/(2)=1H` |
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