An electromagnetic coil has dΦ/dt = 3 V. The induced emf is ε = -N dΦ/dt. What is ε for N = 1 turn? What about N turns?

• A. ε = -N dΦ/dt; For N=1, 3 V; For N turns, 3N V
• B. ε = -N dΦ/dt; For N=1, -3 V; For N turns, -3N V
• C. ε = N dΦ/dt; For N=1, 3 V; For N turns, 3N V
• D. ε = N dΦ/dt; For N=1, -3 V; For N turns, -3N V

Answer: (A)

The key idea is Faraday’s law with Lenz’s law: the induced emf around a coil opposes the change in magnetic flux. This gives the relationship ε = -N dΦ/dt, where N is the number of turns.

Here dΦ/dt is 3 V (that’s the rate of change of flux, equivalent to 3 Weber per second). So the emf becomes ε = -N × 3 V = -3N V. For a single turn, ε = -3 V; for N turns, ε = -3N V.

The magnitude of the induced emf is 3 V per turn, but the sign is negative due to Lenz’s law, indicating the direction of the induced emf opposes the change in flux.