1.

A coil of 100 turns carries a current of 5 A and creates a magnitude flux of 10^(-5) T m^(2) per turn. The value of its inductance is

Answer»

0.05 mH
0.10 m
0.15 mH
0.20 mH

Solution :Here N=100, l=5A, MAGNETIC flux per turn `phi_(B)=10^(-5) Tm^(2)`
Total magnetic flux `Nphi_(B)=LI`
`RARR L=(N phi_(B))/(I)=(100 xx 10^(-5))/(5)=2 xx 10^(-4) H=0.20mH`


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