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A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring is
- a)zero, only if θ=180º
- b)proportional to (180º-θ)
- c)zero for all values of θ
- d)inversely proportional to r
Correct answer is option 'C'. Can you explain this answer?
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A battery is connected between two points A and B on the circumference...
Θ at the center of the ring. The circuit also includes a resistor of resistance r connected between points A and B.
We can start by applying Kirchhoff's laws to the circuit. Kirchhoff's first law, also known as the law of conservation of charge, states that the sum of currents entering a node (or junction) is equal to the sum of currents leaving the node. In this case, there is only one node, which is the point where the battery is connected to the ring. Therefore, the current I flowing through the ring must be equal to the current flowing through the resistor, which we can call I'.
Kirchhoff's second law, also known as the law of conservation of energy, states that the sum of potential differences around any closed loop in a circuit must be zero. We can choose the loop that goes from A to B through the ring, and back to A through the resistor. Along this loop, there are three potential differences: V1 across the battery, V2 across the ring, and V3 across the resistor. We can write:
V1 - V2 - V3 = 0
Since the ring is uniform, the potential difference across any arc of the ring is proportional to the length of the arc. Therefore, we can write:
V2 = (θ/2π) V1
where θ/2π is the fraction of the circumference that is subtended by the arc AB.
The potential difference across the resistor is simply:
V3 = I' R
where R is the resistance of the resistor.
Substituting these expressions into Kirchhoff's second law, we get:
V1 - (θ/2π) V1 - I' R = 0
Solving for I', we get:
I' = (1 - θ/2π) (V1/R)
This is the current flowing through the resistor. The power dissipated by the resistor is given by:
P = I'^2 R = (1 - θ/2π)^2 (V1^2/R)
This expression shows that the power dissipated by the resistor decreases as the angle θ increases, since the term (1 - θ/2π)^2 is a decreasing function of θ. Intuitively, this makes sense, since if the angle θ is very small, most of the current will flow through the resistor, and if the angle θ is very close to 2π, almost no current will flow through the resistor.
Note that if the resistor is removed from the circuit (i.e. if R goes to infinity), then I' goes to zero and the power dissipated by the resistor also goes to zero. In this case, all the current flows through the ring, and the power dissipated by the ring is given by:
P = I^2 r = (V1^2/r)
This is the maximum power that can be dissipated by the ring, and it occurs when all the current flows through the ring.
We can start by applying Kirchhoff's laws to the circuit. Kirchhoff's first law, also known as the law of conservation of charge, states that the sum of currents entering a node (or junction) is equal to the sum of currents leaving the node. In this case, there is only one node, which is the point where the battery is connected to the ring. Therefore, the current I flowing through the ring must be equal to the current flowing through the resistor, which we can call I'.
Kirchhoff's second law, also known as the law of conservation of energy, states that the sum of potential differences around any closed loop in a circuit must be zero. We can choose the loop that goes from A to B through the ring, and back to A through the resistor. Along this loop, there are three potential differences: V1 across the battery, V2 across the ring, and V3 across the resistor. We can write:
V1 - V2 - V3 = 0
Since the ring is uniform, the potential difference across any arc of the ring is proportional to the length of the arc. Therefore, we can write:
V2 = (θ/2π) V1
where θ/2π is the fraction of the circumference that is subtended by the arc AB.
The potential difference across the resistor is simply:
V3 = I' R
where R is the resistance of the resistor.
Substituting these expressions into Kirchhoff's second law, we get:
V1 - (θ/2π) V1 - I' R = 0
Solving for I', we get:
I' = (1 - θ/2π) (V1/R)
This is the current flowing through the resistor. The power dissipated by the resistor is given by:
P = I'^2 R = (1 - θ/2π)^2 (V1^2/R)
This expression shows that the power dissipated by the resistor decreases as the angle θ increases, since the term (1 - θ/2π)^2 is a decreasing function of θ. Intuitively, this makes sense, since if the angle θ is very small, most of the current will flow through the resistor, and if the angle θ is very close to 2π, almost no current will flow through the resistor.
Note that if the resistor is removed from the circuit (i.e. if R goes to infinity), then I' goes to zero and the power dissipated by the resistor also goes to zero. In this case, all the current flows through the ring, and the power dissipated by the ring is given by:
P = I^2 r = (V1^2/r)
This is the maximum power that can be dissipated by the ring, and it occurs when all the current flows through the ring.
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A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring isa)zero, only if θ=180ºb)proportional to (180º-θ)c)zero for all values of θd)inversely proportional to rCorrect answer is option 'C'. Can you explain this answer?
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A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring isa)zero, only if θ=180ºb)proportional to (180º-θ)c)zero for all values of θd)inversely proportional to rCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring isa)zero, only if θ=180ºb)proportional to (180º-θ)c)zero for all values of θd)inversely proportional to rCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring isa)zero, only if θ=180ºb)proportional to (180º-θ)c)zero for all values of θd)inversely proportional to rCorrect answer is option 'C'. Can you explain this answer?.
A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring isa)zero, only if θ=180ºb)proportional to (180º-θ)c)zero for all values of θd)inversely proportional to rCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring isa)zero, only if θ=180ºb)proportional to (180º-θ)c)zero for all values of θd)inversely proportional to rCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius r and resistance R. One of the arcs AB of the ring subtends an angle θ at the center . The value of the magnetic induction at the center due to the current in the ring isa)zero, only if θ=180ºb)proportional to (180º-θ)c)zero for all values of θd)inversely proportional to rCorrect answer is option 'C'. Can you explain this answer?.
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