Question

Two loops of wire are arranged so that a changing current in one will induce a current in the other. If the current in the first is increasing clockwise by one amp every second, the current in the second loop will

Answer 1

Answer:

The current in the second loop will stay constant

Explanation:

Since the induced emf in the second coil, ε due to the changing current i₁ in the first wire loop ε = -Mdi₁/dt where M = mutual inductance of the coils and di₁/dt = rate of change of current in the first coil = + 1 A/s (positive since it is clockwise)

Now ε = i₂R where i₂ = current in second wire loop and R = resistance of second wire loop.

So,  i₂R = -Mdi₁/dt

i₂ = -Mdi₁/dt/R

Since di₁/dt = + 1 A/s,

i₂ = -Mdi₁/dt/R

i₂ = -M × + 1 A/s/R

i₂ = -M/R

Since M and R are constant, this implies that i₂ = constant

So, the current in the second wire loop will stay constant.