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A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ?
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Calculating the induced emf in the loop
To calculate the induced emf in the loop, we can use Faraday's law of electromagnetic induction, which states that the induced emf in a closed loop is equal to the rate of change of magnetic flux through the loop.
Calculating the magnetic flux through the loop
The magnetic flux through the loop is given by the product of the magnetic field and the area of the loop. Since the loop is inside a long solenoid with 15 turns per cm, the magnetic field is given by:
B = μ0nI
where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current in the solenoid. Substituting the given values, we get:
B = μ0 * 15 * 2 * 10^-2 * 2
B = 1.88 * 10^-4 T
The area of the loop is given as 2.0 cm^2, which is equal to 2.0 * 10^-4 m^2. Therefore, the magnetic flux through the loop is:
Φ = B * A
Φ = 1.88 * 10^-4 * 2.0 * 10^-4
Φ = 3.76 * 10^-8 Wb
Calculating the rate of change of magnetic flux
The rate of change of magnetic flux is equal to the change in magnetic flux divided by the time taken for the change. In this case, the current in the solenoid changes from 2.0A to 4.0A in 0.1s. Therefore, the change in magnetic flux is:
ΔΦ = B * A * ΔI/Δt
ΔΦ = 1.88 * 10^-4 * 2.0 * 10^-4 * (4.0 - 2.0)/0.1
ΔΦ = 7.52 * 10^-8 Wb
The time taken for the change is 0.1s. Therefore, the rate of change of magnetic flux is:
dΦ/dt = ΔΦ/Δt
dΦ/dt = 7.52 * 10^-8/0.1
dΦ/dt = 7.52 * 10^-7 V
Calculating the induced emf in the loop
Finally, the induced emf in the loop is equal to the rate of change of magnetic flux, which is:
emf = -dΦ/dt
emf = -7.52 * 10^-7 V
The negative sign indicates that the emf is induced in a direction that opposes the change in magnetic flux.
To calculate the induced emf in the loop, we can use Faraday's law of electromagnetic induction, which states that the induced emf in a closed loop is equal to the rate of change of magnetic flux through the loop.
Calculating the magnetic flux through the loop
The magnetic flux through the loop is given by the product of the magnetic field and the area of the loop. Since the loop is inside a long solenoid with 15 turns per cm, the magnetic field is given by:
B = μ0nI
where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current in the solenoid. Substituting the given values, we get:
B = μ0 * 15 * 2 * 10^-2 * 2
B = 1.88 * 10^-4 T
The area of the loop is given as 2.0 cm^2, which is equal to 2.0 * 10^-4 m^2. Therefore, the magnetic flux through the loop is:
Φ = B * A
Φ = 1.88 * 10^-4 * 2.0 * 10^-4
Φ = 3.76 * 10^-8 Wb
Calculating the rate of change of magnetic flux
The rate of change of magnetic flux is equal to the change in magnetic flux divided by the time taken for the change. In this case, the current in the solenoid changes from 2.0A to 4.0A in 0.1s. Therefore, the change in magnetic flux is:
ΔΦ = B * A * ΔI/Δt
ΔΦ = 1.88 * 10^-4 * 2.0 * 10^-4 * (4.0 - 2.0)/0.1
ΔΦ = 7.52 * 10^-8 Wb
The time taken for the change is 0.1s. Therefore, the rate of change of magnetic flux is:
dΦ/dt = ΔΦ/Δt
dΦ/dt = 7.52 * 10^-8/0.1
dΦ/dt = 7.52 * 10^-7 V
Calculating the induced emf in the loop
Finally, the induced emf in the loop is equal to the rate of change of magnetic flux, which is:
emf = -dΦ/dt
emf = -7.52 * 10^-7 V
The negative sign indicates that the emf is induced in a direction that opposes the change in magnetic flux.
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A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ?
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A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ?.
A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long solenoid with 15turns per cm has a small loop of area 2.0cm2 placed inside the solenoid normal to it's axis. If the current carried by the solenoid changes steadily from 2.0A to 4.0A in 0.1s, what is the induced emf in the loop while the current is changing ?.
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