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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of the loop is BB. It is then bent into a circular coil of nn turns. The magnetic field at the centre of this coil of nn turns will be:

A

nBnB

B

n2Bn^2B

C

2nB2nB

D

2n2B2n^2B

Step-by-Step Solution

  1. Identify the initial magnetic field: The magnetic field BB at the centre of a circular loop of radius RR carrying current II is given by B=μ0I2RB = \frac{\mu_0 I}{2R} .
  2. Relate radius to the number of turns: Let the total length of the wire be LL. For one turn, L=2πRL = 2\pi R. For nn turns of radius rr, the same wire length is used, so L=n(2πr)L = n(2\pi r). Equating the two, we get 2πR=n(2πr)2\pi R = n(2\pi r), which implies r=Rnr = \frac{R}{n}.
  3. Calculate the new magnetic field (B′B'): The magnetic field at the centre of a coil with nn turns is B′=μ0nI2rB' = \frac{\mu_0 n I}{2r} .
  4. Substitute the new radius: B′=μ0nI2(R/n)=n2(μ0I2R)B' = \frac{\mu_0 n I}{2(R/n)} = n^2 \left( \frac{\mu_0 I}{2R} \right)
  5. Conclusion: Since B=μ0I2RB = \frac{\mu_0 I}{2R}, we find that B′=n2BB' = n^2 B.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMcarryingsteadycurrentcircularmagnetic

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