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A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be
- a)2n B
- b)n2 B
- c)nB
- d)2 n2 B
Correct answer is option 'B'. Can you explain this answer?
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A long wire carries a steady current. It is bent into a circle of one ...
KEY CONCEPT : Magentic field at the centre of a circular coil of radius R carrying current i is 
Given : n ´ (2pr ') =2pR
⇒ nr '=R ...(1)
from (1) and (2), 
Most Upvoted Answer
A long wire carries a steady current. It is bent into a circle of one ...
Explanation:
When a long wire carries a steady current, it creates a magnetic field around it. The magnitude of the magnetic field at a point depends on the distance from the wire and the current flowing through it.
Bending the wire into a circle of one turn:
When the wire is bent into a circle of one turn, the magnetic field at the center of the coil can be determined using Ampere's law. According to Ampere's law, the line integral of the magnetic field around a closed loop is equal to the product of the current passing through the loop and the permeability of free space.
Since there is only one turn in the coil, the magnetic field at the center is given by:
B = μ₀I / (2πr₁)
where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r₁ is the radius of the circle.
Bending the wire into a circular loop of n turns:
When the wire is bent into a circular loop of n turns, the magnetic field at the center of the coil can be determined in a similar way. However, now there are n turns in the coil, which means the current passing through the loop is n times the original current.
The radius of the circular loop remains the same as the original circle (r₁), as bending the wire does not change the radius. Therefore, the magnetic field at the center of the coil with n turns is given by:
B' = μ₀(nI) / (2πr₁)
Simplifying this equation, we get:
B' = n(μ₀I / (2πr₁))
B' = nB
Therefore, the magnetic field at the center of the coil with n turns is n times the magnetic field at the center of the coil with one turn.
Conclusion:
Hence, the correct option is (b) n²B, as the magnetic field at the center of the coil with n turns is n times the magnetic field at the center of the coil with one turn.
When a long wire carries a steady current, it creates a magnetic field around it. The magnitude of the magnetic field at a point depends on the distance from the wire and the current flowing through it.
Bending the wire into a circle of one turn:
When the wire is bent into a circle of one turn, the magnetic field at the center of the coil can be determined using Ampere's law. According to Ampere's law, the line integral of the magnetic field around a closed loop is equal to the product of the current passing through the loop and the permeability of free space.
Since there is only one turn in the coil, the magnetic field at the center is given by:
B = μ₀I / (2πr₁)
where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r₁ is the radius of the circle.
Bending the wire into a circular loop of n turns:
When the wire is bent into a circular loop of n turns, the magnetic field at the center of the coil can be determined in a similar way. However, now there are n turns in the coil, which means the current passing through the loop is n times the original current.
The radius of the circular loop remains the same as the original circle (r₁), as bending the wire does not change the radius. Therefore, the magnetic field at the center of the coil with n turns is given by:
B' = μ₀(nI) / (2πr₁)
Simplifying this equation, we get:
B' = n(μ₀I / (2πr₁))
B' = nB
Therefore, the magnetic field at the center of the coil with n turns is n times the magnetic field at the center of the coil with one turn.
Conclusion:
Hence, the correct option is (b) n²B, as the magnetic field at the center of the coil with n turns is n times the magnetic field at the center of the coil with one turn.
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A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer?
Question Description
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer?.
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer?.
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A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will bea)2n Bb)n2 Bc)nBd)2 n2 BCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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