Question

The self inductance of a coil having 200 turns is 10 mH. Complete the total flux linked with the coil. Also, determine the magnetic flux through the cross section of the coil, corresponding to current of 4 mA.

Answer 1

Here, `N = 200, L = 10 mH = 10 xx 10^(-3) H`,
`I = 4 mA = 4 xx 10^(-3) A , phi = ?`
Total magnetic flux linked with the coil `phi = `
`NLI = 200 xx (10 xx 10^(-3)) xx 4 xx 10^(-3) = 8 xx 10^(-3) Wb`
Magnetic flux through the cross-section of the coil = magnetic flux linked with each turn
`=(phi)/(N) = (8 xx 10^(-3))/(200) = 4 xx 10^(-5) Wb`