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Question

Physics Question on Electromagnetic waves

The self inductance of an inductor coil having 100100 turns is 20 mH20\, mH. The magnetic flux through the cross-section of the coil corresponding to a current of 4 mA4 \,mA is

A

2×10−5 Wb2\times 10^{-5}\,Wb

B

4×10−7 Wb4\times 10^{-7}\,Wb

C

8×10−7 Wb8\times 10^{-7}\,Wb

D

8×10−5 Wb8\times 10^{-5}\,Wb

Answer

8×10−7 Wb8\times 10^{-7}\,Wb

Explanation

Solution

Total magnetic flux, Nϕ=LI=20×10−3×4×10−3N\phi=LI=20\times10^{-3}\times4\times10^{-3} =8×10−5 Wb=8\times10^{-5}\,Wb Magnetic flux through the cross-section of the coil, ϕ=8×10−5N=8×10−5100\phi=\frac{8\times10^{-5}}{N}=\frac{8\times10^{-5}}{100} =8×10−7 Wb=8\times10^{-7}\,Wb