# 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) $4 \times 10^{-5} Wb$ (b) $2 \times 10^{-3} Wb$ (c) $3 \times 10^{-5} Wb$ (d) $8 \times 10^{-3} Wb$

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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 (a) $4 \times 10^{-5} Wb$ (b) $2 \times 10^{-3} Wb$ (c) $3 \times 10^{-5} Wb$ (d) $8 \times 10^{-3} Wb$

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