1.

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 ……

Answer»

25 MH
`25xx10^(-3)` mH
`50xx10^(-3)` mH
`50xx10^(-3)` H

Solution :The MAGNETIC field at the centre of a circular coil carries CURRENT I is ,
`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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