Question

A square loop of wire surrounds a solenoid. The side of the square is 0.1 m, while the radius of the solenoid is 0.025 m. The square loop has a resistance of 30 ohm . The solenoid has 500 turns and is 0.3 m long. The current in the solenoid is increasing at a constant rate of 0.7 A/s. What is the magnitude of the current flowing in the square loop?

Answer 1

I=9.6×e^{-8} A

Step-by-step explanation:

The magnetic field inside the solenoid.

B=I*500*muy0/0.3=2.1×e ^-3×I.

so the total flux go through the square loop.

B×π×r^2=I×2.1×e^-3π×0.025^2

=4.11×e^-6×I

we have that

(flux)'= -U

so differentiating flux we get

so the inducted emf in the loop.

U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)

so, I=2.9×e^{-6}÷30

I=9.6×e^{-8} A