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Question: The self inductance of an inductor coil having 100 turns is 20 mH. The magnetic flux through the cro...

The self inductance of an inductor coil having 100 turns is 20 mH. The magnetic flux through the cross-section of the coil corresponding to a current of 4 mA is

A

2 × 10-5 WI

B

4 × 10-7 Wb

C

8 × 10-7 Wl

D

8 × 10-5 Wb

Answer

8 × 10-7 Wl

Explanation

Solution

Total magnetic flux

Nϕ=LI=20×10−3×4×10−3\mathrm { N } \phi = \mathrm { LI } = 20 \times 10 ^ { - 3 } \times 4 \times 10 ^ { - 3 }

=8×10−5 Wb= 8 \times 10 ^ { - 5 } \mathrm {~Wb}

Magnetic flux through the cross section of the coil,

ϕ=8×10−5 N=8×105100=8×10−7 Wb\phi = \frac { 8 \times 10 ^ { - 5 } } { \mathrm {~N} } = \frac { 8 \times 10 ^ { 5 } } { 100 } = 8 \times 10 ^ { - 7 } \mathrm {~Wb}