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A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200 turns, a diameter of 2.0 cm, and is 7.5 cm long. The resistance of the loop is 0.032 Ω. If the current in the solenoid is increased by 1.2 A in 30 ms, what is the induced current in the loop?

A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid-example-1

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Step-by-step explanation:

To find the induced current in the loop, we need to use Faraday's law of electromagnetic induction, which states that the electromotive force (EMF) induced in a closed loop is equal to the rate of change of the magnetic flux passing through the loop.

Firstly, let's find the change in magnetic flux passing through the loop. The magnetic flux through a loop is given by the equation:

Φ = B*A

where Φ is the magnetic flux, B is the magnetic field, and A is the area of the loop. Since the solenoid passes through the center of the loop, the magnetic field is constant and equal to the magnetic field inside the solenoid.

The magnetic field inside a solenoid is given by the equation:

B = μ₀*n*I

where B is the magnetic field, μ₀ is the permeability of free space (4π x 10^-7 Tm/A), n is the number of turns per unit length (turns/m), and I is the current.

The number of turns per unit length (n) is given by the equation:

n = N/L

where N is the total number of turns in the solenoid and L is the length of the solenoid.

Using the given values, we can calculate the magnetic field inside the solenoid:

n = N/L = 1200 turns / 0.075 m = 16000 turns/m

B = μ₀*n*I = (4π x 10^-7 Tm/A)(16000 turns/m)(1.2 A) ≈ 0.024 T

Next, let's calculate the area of the loop. Since the solenoid passes through the center of the loop, the loop has the same diameter as the solenoid. The area of a circle is given by the equation:

A = π*r²

where A is the area and r is the radius. In this case, the radius is half the diameter of the solenoid.

r = 2.0 cm / 2 = 1.0 cm = 0.01 m

A = π*(0.01 m)² ≈ 0.000314 m²

Now we can calculate the change in magnetic flux passing through the loop:

Φ = B*A = (0.024 T)(0.000314 m²) ≈ 7.536 x 10^-6 Tm²

The change in magnetic flux with time is equal to the EMF induced in the loop, which can be written as:

EMF = ΔΦ/Δt

where EMF is the electromotive force, ΔΦ is the change in magnetic flux, and Δt is the change in time.

Using the given values, we can calculate the change in time:

Δt = 30 ms = 0.03 s

Now we can calculate the induced EMF:

EMF = ΔΦ/Δt = (7.536 x 10^-6 Tm²)/(0.03 s) ≈ 0.000251 V

Finally, we can calculate the induced current in the loop using Ohm's law:

EMF = I_loop * R

where I_loop is the induced current in the loop and R is the resistance of the loop.

Using the given resistance, we can calculate the induced current in the loop:

I_loop = EMF / R = (0.000251 V) / (0.032 Ω) ≈ 0.00784 A

Therefore, the induced current in the loop is approximately 0.00784 A.

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