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Asked in GUJCET 2011 · Self-inductance and self-induced EMF
Given: Φ = 10 μWb = 10×10⁻⁶ Wb, I = 2 mA = 2×10⁻³ A.
Idea: the flux linked with a coil is proportional to its own current, Φ = LI, so L = Φ/I.
L = (10×10⁻⁶)/(2×10⁻³) = 5×10⁻³ H.
L = 5 mH.
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