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The self-inductance of a coil having 200...

The self-inductance of a coil having 200 turns is 10 milli henry. Calculate the magnetic flux through the cross-section of the coil corresponding to current of 4 milliampere. Also determine the total flux linked with each turn.

Text Solution

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Total magnetic flux linked with the coil,
`N phi = LI = 10^(-2) xx 4 xx 10^(-3) = 4 xx 10^(-5) Wb`
` therefore `Flux per turn `phi = (4 xx 10^(-5))/(200) = 2 xx 10^(-7) Wb`
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