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Electromagnetic Induction question

2025 · 28 Jan · Shift 2 · Q54
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  5. /2025 · 28 Jan · Shift 2 · Q54

Electromagnetic Induction question

2025 · 28 Jan · Shift 2 · Q54

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10 π\piπ rad s-1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (π=3.14)(\pi=3.14)(π=3.14)
  1. A
    0.5024 V
  2. B
    0.0628 V
  3. C
    0.2512 V
  4. D
    0.1256 V
View written solutionFree

Correct answer: C

  1. Identify the concept

A rotating conducting disc in a uniform magnetic field develops a potential difference between its center and rim. This is the case of a rotating disc generator (Faraday disc).

The formula for the emf between the center and the rim is:

E=12BωR2\mathcal{E} = \frac{1}{2} B \omega R^2E=21​BωR2

where:

  • B=0.4 TB = 0.4\,\text{T}B=0.4T
  • ω=10π rad s−1\omega = 10\pi\,\text{rad s}^{-1}ω=10πrad s−1
  • R=20 cm=0.2 mR = 20\,\text{cm} = 0.2\,\text{m}R=20cm=0.2m
  1. Substitute the values
E=12(0.4)(10π)(0.2)2\mathcal{E} = \frac{1}{2}(0.4)(10\pi)(0.2)^2E=21​(0.4)(10π)(0.2)2
  1. Calculate step by step

First,

(0.2)2=0.04(0.2)^2 = 0.04(0.2)2=0.04

So,

E=12×0.4×10π×0.04\mathcal{E} = \frac{1}{2} \times 0.4 \times 10\pi \times 0.04E=21​×0.4×10π×0.04 =0.2×10π×0.04= 0.2 \times 10\pi \times 0.04=0.2×10π×0.04 =2π×0.04= 2\pi \times 0.04=2π×0.04 =0.08π= 0.08\pi=0.08π

Using π=3.14\pi = 3.14π=3.14,

E=0.08×3.14=0.2512 V\mathcal{E} = 0.08 \times 3.14 = 0.2512\,\text{V}E=0.08×3.14=0.2512V
  1. Match with the options
0.2512 V0.2512\,\text{V}0.2512V

So the correct option is:

C: 0.2512 V0.2512\,\text{V}0.2512V

  1. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They match.

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