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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil. What is the ratio of the potential difference applied across them so that the magnetic field at their centres is the same?

A

3

B

4

C

6

D

2

Step-by-Step Solution

  1. Magnetic Field Condition: The magnetic field BB at the center of a circular coil of radius rr carrying current II is given by B=μ0I2rB = \frac{\mu_0 I}{2r} . For the fields to be equal (B1=B2B_1 = B_2): μ0I12r1=μ0I22r2  ⟹  I1r1=I2r2  ⟹  I1I2=r1r2\frac{\mu_0 I_1}{2r_1} = \frac{\mu_0 I_2}{2r_2} \implies \frac{I_1}{r_1} = \frac{I_2}{r_2} \implies \frac{I_1}{I_2} = \frac{r_1}{r_2}
  2. Resistance Relationship: The resistance RR of a wire is R=ρLAR = \rho \frac{L}{A}. Since the coils are made of the same wire, resistivity ρ\rho and cross-sectional area AA are constant. The length LL of the wire forming the coil is proportional to the radius (L=2πr×NL = 2\pi r \times N). Assuming single-turn coils (or equal turns), resistance is proportional to radius: R∝rR \propto r. R1R2=r1r2\frac{R_1}{R_2} = \frac{r_1}{r_2}
  3. Potential Difference: According to Ohm's Law, V=IRV = IR . V1V2=I1R1I2R2=(I1I2)×(R1R2)\frac{V_1}{V_2} = \frac{I_1 R_1}{I_2 R_2} = \left( \frac{I_1}{I_2} \right) \times \left( \frac{R_1}{R_2} \right)
  4. Calculation: Substituting the ratios derived above (given r1=2r2r_1 = 2r_2, so r1/r2=2r_1/r_2 = 2): V1V2=(2)×(2)=4\frac{V_1}{V_2} = (2) \times (2) = 4 Therefore, the ratio of potential differences is 4.

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 MAGNETISMcircularradiuspotentialdifferenceapplied

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