A cylindrical solenoid 37 cm long with a radius of 2 mm has 200 tightly-wound tu

Question

A cylindrical solenoid 37 cm long with a radius of 2 mm has 200 tightly-wound turns of wire uniformly distributed along its length (see the figure). Around the middle of the solenoid is a two-turn rectangular loop 3 cm by 2 cm made of resistive wire having a resistance of 125 ohms. One microsecond after connecting the loose wire to the battery to form a series circuit with the battery and a 20 resistor, what is the magnitude of the current in the rectangular loop and its direction (clockwise or counter-clockwise in the diagram)? (The battery has an emf of 9 V.)

Explanation / Answer

At first we must find inductance of the solemoid:

L = u0 N^2 A/l

==> L = (4*3.16416e-7) *(200*200) * (3.1416*2e-3*2e-3)/0.37

==> L = 1.7163e-6 H

magnetic field:

B = u i N/l = (4*3.16416e-7)*(200)/0.37 = (0.0006841) * i

now we must find current:

i = i0 (1 - e^(-tR/L))

i0 = V/R = 9/(20) = 0.45 A

==> i = 0.45 * (1 - e^(-t*20/1.7163e-6))

==> i = 0.45 * (1 - e^(-t*1169506.43))

==> di/dt = 0.45 * (1169506.43 * e^(-t*1169506.43))

---------------------------------

magnetic flux:

emf = N A dB/dt

==> emf= 2 * (3.1416*2e-3*2e-3) * (1.7163e-6) * (di/dt)

==> emf= 2 * (3.1416*2e-3*2e-3) * (1.7163e-6) * (0.45 * (1169506.43 * e^(-t*1169506.43)))

==> at t=1

==> emf= 2 * (3.1416*2e-3*2e-3) * (1.7163e-6) * (0.45 * (1169506.43 * e^(-t*1169506.43)))

==> emf =

--------------------------------

current of the loop:

i = emf/R =

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