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A coil of conducting wire carries a current i. In a time interval of Δt = 0.530 s, the current goes from i1 = 3.60 A to i2 = 2.20 A. The average emf induced in the coil is e m f = 16.0 mV. Assuming the current does not change direction, calculate the coil's inductance (in mH).

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Answer:6.1mH

Explanation: The current - voltage relationship of an inductor is given below as

v = L × Δi/Δt

Where induced emf (v) = 16mv = 0.016v

Inductance of inductor (L) =?

Δi = I1 - I2 = 3.6 - 2.2 = 1.4

Δt = 0.53

Δi/Δt = 1.4/0.53 = 2.642.

By substituting the parameters, we have that

0.016 = L * 2.462

L = 0.016/ 2.462

L = 0.0061 H = 6.1mH

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