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  • 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:Option: 1</stro

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:

Option: 1

\mathrm{4 \times 10^{-5} \mathrm{~Wb}}


Option: 2

\mathrm{2 \times 10^{-3} \mathrm{~Wb}}


Option: 3

\mathrm{3 \times 10^{-5} \mathrm{~Wb}}


Option: 4

\mathrm{8 \times 10^{-3} \mathrm{~Wb}}


Answers (1)

best_answer

\text { Here, } \mathrm{N}=400, \mathrm{~L}=10 \mathrm{mH}=10 \times 10^{-3} \mathrm{H}, \mathrm{I}=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A}

Total magnetic flux linked with the coil,

\phi=\mathrm{NLI}=400 \times\left(10 \times 10^{-3}\right) \times 2 \times 10^{-3}=8 \times 10^{-3} \mathrm{~Wb}

Magnetic flux through the cross – section of the coil = Magnetic flux linked with each turn 

=\frac{\phi}{\mathrm{N}}=\frac{8 \times 10^{-3}}{200}=4 \times 10^{-5} \mathrm{~Wb}

Posted by

Irshad Anwar

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