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A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap?
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A perfectly conducting filament is formed into a circular ring of radi...
Answer:
Part (a)
Applying Faraday's Law:
Faraday's law of electromagnetic induction states that the electromotive force (EMF) induced in any closed loop is equal to the negative rate of change of the magnetic flux enclosed by the loop. Mathematically, it can be represented as:
EMF = -dΦ/dt
where EMF is the electromotive force induced, Φ is the magnetic flux enclosed by the loop, and t is time.
Calculating Magnetic Flux:
The magnetic flux through a closed loop can be calculated using the formula:
Φ = ∫B.dA
where Φ is the magnetic flux, B is the magnetic field, and dA is the differential area element of the loop.
Applying Faraday's Law to the Given Circuit:
In the given circuit, the resistance R and the battery of voltage Vo are inserted at different points in the circular ring. The loop current itself produces a negligible magnetic field.
We can apply Faraday's law to the loop of the circuit as follows:
EMF = -dΦ/dt
where EMF is the electromotive force induced in the loop, and Φ is the magnetic flux enclosed by the loop.
Calculating EMF:
The electromotive force (EMF) induced in the loop can be calculated using Ohm's law:
EMF = IR
where I is the current flowing through the circuit.
Calculating Magnetic Flux:
The magnetic flux through the loop can be calculated as follows:
Φ = ∫B.dA
where B is the magnetic field, and dA is the differential area element of the loop.
Since the loop current itself produces a negligible magnetic field, the magnetic field B is due to the battery of voltage Vo.
The magnetic field due to the battery of voltage Vo is given by:
B = μ0I/2a
where μ0 is the permeability of free space, I is the current flowing through the circuit, and a is the radius of the circular ring.
Since the current flowing through the circuit is constant, the magnetic flux through the loop is also constant.
Therefore, dΦ/dt = 0.
Conclusion:
Substituting the values of EMF and dΦ/dt in Faraday's law, we get:
IR = 0
This shows that the electromotive force induced in the loop is equal to the voltage drop across the resistance R.
Part (b)
Applying Faraday's Law:
Faraday's law of electromagnetic induction states that the electromotive force (EMF) induced in any closed loop is equal to the negative rate of change of the magnetic flux enclosed by the loop. Mathematically, it can be represented as:
EMF = -dΦ/dt
where EMF is the electromotive force induced, Φ is the magnetic flux enclosed by the loop, and t is time.
Calculating Magnetic Flux:
The magnetic flux through a closed loop can be calculated using the formula:
Φ = ∫B.dA
where Φ is the magnetic flux, B is the magnetic field, and dA is the differential area element of the loop.
Applying Faraday's Law to the Given Circuit
Part (a)
Applying Faraday's Law:
Faraday's law of electromagnetic induction states that the electromotive force (EMF) induced in any closed loop is equal to the negative rate of change of the magnetic flux enclosed by the loop. Mathematically, it can be represented as:
EMF = -dΦ/dt
where EMF is the electromotive force induced, Φ is the magnetic flux enclosed by the loop, and t is time.
Calculating Magnetic Flux:
The magnetic flux through a closed loop can be calculated using the formula:
Φ = ∫B.dA
where Φ is the magnetic flux, B is the magnetic field, and dA is the differential area element of the loop.
Applying Faraday's Law to the Given Circuit:
In the given circuit, the resistance R and the battery of voltage Vo are inserted at different points in the circular ring. The loop current itself produces a negligible magnetic field.
We can apply Faraday's law to the loop of the circuit as follows:
EMF = -dΦ/dt
where EMF is the electromotive force induced in the loop, and Φ is the magnetic flux enclosed by the loop.
Calculating EMF:
The electromotive force (EMF) induced in the loop can be calculated using Ohm's law:
EMF = IR
where I is the current flowing through the circuit.
Calculating Magnetic Flux:
The magnetic flux through the loop can be calculated as follows:
Φ = ∫B.dA
where B is the magnetic field, and dA is the differential area element of the loop.
Since the loop current itself produces a negligible magnetic field, the magnetic field B is due to the battery of voltage Vo.
The magnetic field due to the battery of voltage Vo is given by:
B = μ0I/2a
where μ0 is the permeability of free space, I is the current flowing through the circuit, and a is the radius of the circular ring.
Since the current flowing through the circuit is constant, the magnetic flux through the loop is also constant.
Therefore, dΦ/dt = 0.
Conclusion:
Substituting the values of EMF and dΦ/dt in Faraday's law, we get:
IR = 0
This shows that the electromotive force induced in the loop is equal to the voltage drop across the resistance R.
Part (b)
Applying Faraday's Law:
Faraday's law of electromagnetic induction states that the electromotive force (EMF) induced in any closed loop is equal to the negative rate of change of the magnetic flux enclosed by the loop. Mathematically, it can be represented as:
EMF = -dΦ/dt
where EMF is the electromotive force induced, Φ is the magnetic flux enclosed by the loop, and t is time.
Calculating Magnetic Flux:
The magnetic flux through a closed loop can be calculated using the formula:
Φ = ∫B.dA
where Φ is the magnetic flux, B is the magnetic field, and dA is the differential area element of the loop.
Applying Faraday's Law to the Given Circuit
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A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap?
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A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap?.
A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap?.
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A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap?, a detailed solution for A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap? has been provided alongside types of A perfectly conducting filament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage Vo is inserted. Assume that the loop current itself produces negligible magnetic field. (a) Apply Faraday's law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B field is ap? theory, EduRev gives you an
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