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The self inductance of an inductance coi...

The self inductance of an inductance coil having 100 turns is 20 mH. Calculate the magnetic flux through the cross section of the coil corresponding to a current of 4 milliampere. Also, find the total flux

A

`2xx10^(-5)Wb`

B

`4xx10^(-7)Wb`

C

`8xx10^(-7)Wb`

D

`8xx10^(-5)Wb`

Text Solution

Verified by Experts

The correct Answer is:
C
Total megnetic flux,
`Nphi=LI=20xx10^(-3)xx4xx10^(-3)=8xx10^(-5)Wb`
Megnetic flux through the corss-section of the coil,
`phi=(8xx10^(-5))/(N)(10xx^(-5))/(100)=8xx10^(-1)Wb`
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