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NEET PHYSICSELECTROMAGNETIC INDUCTIONHard

Question

PQ is an infinite current carrying conductor carrying current II. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity vv as shown. The conductor EF is perpendicular to PQ and its ends are at distances aa and bb from the wire. The force needed to maintain constant speed of EF is:

A

1vR[μ0Iv2πln(ba)]2\frac{1}{vR} \left[ \frac{\mu_0 I v}{2\pi} \ln \left( \frac{b}{a} \right) \right]^2

B

1vR[μ0Iv2ln(ba)]2\frac{1}{vR} \left[ \frac{\mu_0 I v}{2} \ln \left( \frac{b}{a} \right) \right]^2

C

vR[μ0I2ln(ba)]2\frac{v}{R} \left[ \frac{\mu_0 I}{2} \ln \left( \frac{b}{a} \right) \right]^2

D

vR[μ0I2πln(ba)]2\frac{v}{R} \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right]^2

Step-by-Step Solution

  1. Magnetic Field: The magnetic field BB at a distance xx from an infinite wire carrying current II is B=μ0I2πxB = \frac{\mu_0 I}{2\pi x}.

  2. Motional EMF: The conductor EF moves with velocity vv parallel to the wire PQ. A small element of length dxdx at distance xx induces an emf dε=vBdxd\varepsilon = v B dx. The total emf ε\varepsilon across EF (from aa to bb) is: ε=abvμ0I2πxdx=μ0Iv2π[lnx]ab=μ0Iv2πln(ba)\varepsilon = \int_a^b v \frac{\mu_0 I}{2\pi x} dx = \frac{\mu_0 I v}{2\pi} [\ln x]_a^b = \frac{\mu_0 I v}{2\pi} \ln \left( \frac{b}{a} \right)

  3. Induced Current: The current ii in the loop is i=εRi = \frac{\varepsilon}{R}.

  4. Magnetic Force: The magnetic force dFdF on the element dxdx carrying induced current ii is dF=iBdxdF = i B dx. The total force opposing motion is: Fmag=abiBdx=iabμ0I2πxdx=i[μ0I2πln(ba)]F_{mag} = \int_a^b i B dx = i \int_a^b \frac{\mu_0 I}{2\pi x} dx = i \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right]

  5. External Force: To maintain constant velocity, the external force FF must balance FmagF_{mag}. Substituting i=ε/Ri = \varepsilon/R: F=εR[μ0I2πln(ba)]=1R[μ0Iv2πln(ba)][μ0I2πln(ba)]F = \frac{\varepsilon}{R} \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right] = \frac{1}{R} \left[ \frac{\mu_0 I v}{2\pi} \ln \left( \frac{b}{a} \right) \right] \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right] F=vR[μ0I2πln(ba)]2F = \frac{v}{R} \left[ \frac{\mu_0 I}{2\pi} \ln \left( \frac{b}{a} \right) \right]^2

This expression can be rewritten to match Option A by multiplying numerator and denominator by vv: F=1vR[μ0Iv2πln(ba)]2F = \frac{1}{vR} \left[ \frac{\mu_0 I v}{2\pi} \ln \left( \frac{b}{a} \right) \right]^2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROMAGNETIC INDUCTION. 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.

PHYSICSELECTROMAGNETIC INDUCTIONinfinitecurrentcarryingconductorcarrying

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