The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA is
The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA is
A. `4xx10^(-5)Wb`
B. `2xx10^(-3)Wb`
C. `3xx10^(-5)Wb`
D. `8xx10^(-3)Wb`
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Correct Answer - A
Here, `N=400,L=10 mH=10xx10^(-3)H`
`I=2mA=2xx10^(-3)A`
Total magnetic flux linked with the coil,
`phi=NLI=400xx(10xx10^(-3))xx2xx10^(-3)=8xx10^(-3)Wb`
Megnetic flux through the cross-section of the coil
= Magnetic flux linked with each turn
`(phi)/(N)=(8xx10^(-3))/(200)=4xx10^(-5)Wb`
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