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A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil?
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A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.T...
Given:
Resistance of the coil (R) = 150 ohms
Number of turns in the coil (N) = 500
Resistance of the galvanometer (Rg) = 450 ohms
Initial magnetic field (B1) = 0.1 mWb
Final magnetic field (B2) = 0.3 mWb
Time taken to move from B1 to B2 (t) = 0.1 s
Calculating the induced emf:
The induced emf (ε) can be calculated using Faraday's law of electromagnetic induction.
ε = -N(dΦ/dt)
Where N is the number of turns in the coil and (dΦ/dt) is the rate of change of magnetic flux.
The change in magnetic flux (dΦ) can be calculated as the difference between the final and initial magnetic flux.
dΦ = B2 - B1
Substituting the given values:
dΦ = 0.3 mWb - 0.1 mWb
dΦ = 0.2 mWb
Converting mWb to Wb:
dΦ = 0.2 mWb * 10^-3
dΦ = 2 * 10^-4 Wb
Now, we can calculate the induced emf:
ε = -N(dΦ/dt)
ε = -500 * (2 * 10^-4 Wb / 0.1 s)
ε = -500 * 2 * 10^-3 V
ε = -1 V
The average induced emf (ε_avg) can be calculated as:
ε_avg = ε / 2
ε_avg = -1 V / 2
ε_avg = -0.5 V
Calculating the average current:
The average current (I_avg) can be calculated using Ohm's law.
I_avg = ε_avg / (R + Rg)
I_avg = -0.5 V / (150 ohms + 450 ohms)
I_avg = -0.5 V / 600 ohms
I_avg = -0.000833 A
As the direction of the induced emf is opposite to the direction of the current flow, the negative sign indicates the direction of the current.
Answer:
The average induced emf is -0.5 V and the average current through the coil is -0.000833 A. The negative sign indicates the direction of the current.
Resistance of the coil (R) = 150 ohms
Number of turns in the coil (N) = 500
Resistance of the galvanometer (Rg) = 450 ohms
Initial magnetic field (B1) = 0.1 mWb
Final magnetic field (B2) = 0.3 mWb
Time taken to move from B1 to B2 (t) = 0.1 s
Calculating the induced emf:
The induced emf (ε) can be calculated using Faraday's law of electromagnetic induction.
ε = -N(dΦ/dt)
Where N is the number of turns in the coil and (dΦ/dt) is the rate of change of magnetic flux.
The change in magnetic flux (dΦ) can be calculated as the difference between the final and initial magnetic flux.
dΦ = B2 - B1
Substituting the given values:
dΦ = 0.3 mWb - 0.1 mWb
dΦ = 0.2 mWb
Converting mWb to Wb:
dΦ = 0.2 mWb * 10^-3
dΦ = 2 * 10^-4 Wb
Now, we can calculate the induced emf:
ε = -N(dΦ/dt)
ε = -500 * (2 * 10^-4 Wb / 0.1 s)
ε = -500 * 2 * 10^-3 V
ε = -1 V
The average induced emf (ε_avg) can be calculated as:
ε_avg = ε / 2
ε_avg = -1 V / 2
ε_avg = -0.5 V
Calculating the average current:
The average current (I_avg) can be calculated using Ohm's law.
I_avg = ε_avg / (R + Rg)
I_avg = -0.5 V / (150 ohms + 450 ohms)
I_avg = -0.5 V / 600 ohms
I_avg = -0.000833 A
As the direction of the induced emf is opposite to the direction of the current flow, the negative sign indicates the direction of the current.
Answer:
The average induced emf is -0.5 V and the average current through the coil is -0.000833 A. The negative sign indicates the direction of the current.
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A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil?
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A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil?.
A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A coil of resistance 150ohms is placed in a magnetic field of 0.1mWb.The coil has 500turns and a galvanometer of 450 ohms is connected in series with it. The coil is moved in 0.1s from the given field to another field of 0. 3mWb.Find the average induced emf and the average current through the coil?.
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