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NEET PHYSICSELECTROMAGNETIC INDUCTIONMedium

Question

A wheel with 20 metallic spokes, each 1 m long, is rotated with a speed of 120 rpm in a plane perpendicular to a magnetic field of 0.4 G. The induced emf between the axle and rim of the wheel will be: (1 G = 10⁻⁴ T)

A

2.51 × 10⁻⁴ V

B

2.51 × 10⁻⁵ V

C

4.0 × 10⁻⁵ V

D

2.51 V

Step-by-Step Solution

  1. Formula: The induced electromotive force (emf) ε\varepsilon between the center (axle) and the rim of a rotating rod (spoke) is given by ε=12BωR2\varepsilon = \frac{1}{2} B \omega R^2, where BB is the magnetic field, ω\omega is the angular velocity, and RR is the length of the spoke .
  2. Parallel Combination: Although there are 20 spokes, they are all connected in parallel between the axle and the rim. In a parallel combination of identical cells (spokes), the equivalent emf is equal to the emf of a single cell. Therefore, the number of spokes does not affect the magnitude of the induced voltage .
  3. Unit Conversions: Magnetic Field (BB): 0.4 G=0.4×104 T0.4 \text{ G} = 0.4 \times 10^{-4} \text{ T}. Angular Velocity (ω\omega): 120 rpm=120×2π60 rad/s=4π rad/s120 \text{ rpm} = \frac{120 \times 2\pi}{60} \text{ rad/s} = 4\pi \text{ rad/s}.
  • Length (RR): 1 m1 \text{ m}.
  1. Calculation: ε=12×(0.4×104)×(4π)×(1)2\varepsilon = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4\pi) \times (1)^2 ε=0.2×104×4×3.14\varepsilon = 0.2 \times 10^{-4} \times 4 \times 3.14 ε=0.8×3.14×104\varepsilon = 0.8 \times 3.14 \times 10^{-4} ε2.512×104 V\varepsilon \approx 2.512 \times 10^{-4} \text{ V}.

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 INDUCTIONmetallicspokesrotatedperpendicularmagnetic

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