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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

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