Question

A large research solenoid has a self-inductance of 25.0 H. What induced emf opposes shutting it off when 100 A of current through it is switched off in 80.0 ms?

a) 0.2 V
b) 2.0 V
c) 20.0 V
d) 200.0 V

Answer 1

The induced emf when a 100 A current is switched off in a 25.0 H solenoid in 80.0 ms is 31.25 kV, calculated using Faraday's law of electromagnetic induction.

Step-by-step explanation:

The induced emf that opposes shutting off a large research solenoid with a self-inductance of 25.0 H when 100 A of current through it is switched off in 80.0 ms can be calculated using Faraday's law of electromagnetic induction.

The formula to calculate the induced emf is E = -L (ΔI/Δt), where E is the induced emf, L is the self-inductance, ΔI is the change in current, and Δt is the change in time.

Plugging in the given values: E = -25.0 H * (100 A / 0.080 s) = -31250 V/s. Since the direction of emf opposes the current shutdown, we consider the magnitude and find that the induced emf is 31250 V or 31.25 kV.