Find the rate of change of the magnetic flux in a 2000-turn solenoid when an induction emf of 120 V is excited in it.
| September 16, 2021 | education
Initial data: N (the number of turns that the solenoid consists of) = 2000 turns; ε (EMF that occurred with the solenoid) = 120 V.
The rate of change of the magnetic flux in the solenoid can be expressed from the formula: ε = | ΔF / Δt | * N, whence | ΔФ / Δt | = ε / N.
Let’s calculate: | ΔФ / Δt | = 120/2000 = 0.06 Wb / s.
Answer: The rate of change of the magnetic flux in the solenoid is 0.06 Wb / s.
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