To solve this problem, we need to use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.

First, we need to calculate the current in the solenoid. The current I_sol in the solenoid can be calculated using Ohm's law, which states that the current is equal to the voltage divided by the resistance.

I_sol = E / R_sol = 32.43 V / 129.2 Ω = 0.251 A

Next, we need to calculate the magnetic field B inside the solenoid. The magnetic field inside a solenoid is given by the formula B = μ0 * n * I, where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.

n = 1719 turns / 11.43 cm = 150.39 turns/cm = 15039 turns/m

B = μ0 * n * I = (4π * 10^-7 T m/A) * (15039 turns/m) * (0.251 A) = 0.047 T

The magnetic flux Φ through the outer loop is given by the formula Φ = B * A, where A is the area of the loop.

A = π * b^2 = π * (3.37 cm)^2 = 35.65 cm^2 = 0.003565 m^2

Φ = B * A = 0.047 T * 0.003565 m^2 = 0.000167 Wb

The rate of change of the magnetic flux dΦ/dt is then given by the change in flux divided by the change in time. Since the flux was initially zero, the change in flux is just Φ, and the change in time is 2.83 μs = 2.83 * 10^-6 s.

dΦ/dt = Φ / t = 0.000167 Wb / 2.83 * 10^-6 s = 59.01 Wb/s

Finally, the current I_o in the outer loop is given by Faraday's law, which states that the induced EMF is equal to the rate of change of the magnetic flux, and Ohm's law, which states that the current is equal to the voltage divided by the resistance.

I_o = dΦ/dt / R_o = 59.01 Wb/s / 1791.6 Ω = 0.033 mA

So, at a time t = 2.83 μs, the current that flows through the outer loop is 0.033 mA.

Question

To solve this problem, we need to use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.

First, we need to calculate the current in the solenoid. The current I_sol in the solenoid can be calculated using Ohm's law, which states that the current is equal to the voltage divided by the resistance.

I_sol = E / R_sol = 32.43 V / 129.2 Ω = 0.251 A

Next, we need to calculate the magnetic field B inside the solenoid. The magnetic field inside a solenoid is given by the formula B = μ0 * n * I, where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.

n = 1719 turns / 11.43 cm = 150.39 turns/cm = 15039 turns/m

B = μ0 * n * I = (4π * 10^-7 T m/A) * (15039 turns/m) * (0.251 A) = 0.047 T

The magnetic flux Φ through the outer loop is given by the formula Φ = B * A, where A is the area of the loop.

A = π * b^2 = π * (3.37 cm)^2 = 35.65 cm^2 = 0.003565 m^2

Φ = B * A = 0.047 T * 0.003565 m^2 = 0.000167 Wb

The rate of change of the magnetic flux dΦ/dt is then given by the change in flux divided by the change in time. Since the flux was initially zero, the change in flux is just Φ, and the change in time is 2.83 μs = 2.83 * 10^-6 s.

dΦ/dt = Φ / t = 0.000167 Wb / 2.83 * 10^-6 s = 59.01 Wb/s

Finally, the current I_o in the outer loop is given by Faraday's law, which states that the induced EMF is equal to the rate of change of the magnetic flux, and Ohm's law, which states that the current is equal to the voltage divided by the resistance.

I_o = dΦ/dt / R_o = 59.01 Wb/s / 1791.6 Ω = 0.033 mA

So, at a time t = 2.83 μs, the current that flows through the outer loop is 0.033 mA.

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