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A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the efficient of self-induction of the coil will be …… |
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Answer» 25 MH `B=(mu_0 NI)/(2a)`,where a is the radius of a coil. `therefore` The flux LINKED with coil `phi`=ABN `therefore phi=(mu_0N^2AI)/(2a)=(mu_0N^2 xx pia^2I)/(2a)` `therefore phi=(mu_0N^2 piaI)/2` `therefore L=phi/I=(mu_0N^2pia)/2` `=(4xxpixx10^(-7)xx(500)^2xxpixx5xx10^(-2))/2` `=pi^2 xx 250000xx10^(-8)` `=10xx250000xx10^(-8)` [`because` taking `pi^2`=10] `=25xx10^(5-8)` `=25xx10^(-3)` `therefore` L=25 mH |
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