If the strength of the current passing in the solenoid changes by 50 A per second, then an EMF of self-induction of 0.08 V appears

univerkov | February 11, 2021 | education

If the strength of the current passing in the solenoid changes by 50 A per second, then an EMF of self-induction of 0.08 V appears at the ends of the solenoid winding. Determine the inductance of the solenoid.

Initial data: ΔI (change in the current in the solenoid) = 50 A; t (duration of current passage through the solenoid) = 1 s; ε (EMF of self-induction arising in the solenoid) = 0.08 V.

The inductance of the solenoid is expressed from the formula: ε = -L * ΔI / Δt, whence L = ε * Δt / ΔI.

Let’s make the calculation: L = 0.08 * 1/50 = 0.0016 H (1.6 mH).

Answer: The inductance of the solenoid is 1.6 mH.

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