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Electromagnetic Induction question

2023 · 8 Apr · Shift 2 · Q43
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  5. /2023 · 8 Apr · Shift 2 · Q43

Electromagnetic Induction question

2023 · 8 Apr · Shift 2 · Q43

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
An emf of 0.08 V0.08 \mathrm{~V}0.08 V is induced in a metal rod of length 10 cm10 \mathrm{~cm}10 cm held normal to a uniform magnetic field of 0.4 T0.4 \mathrm{~T}0.4 T, when moves with a velocity of:
  1. A
    20 ms−120 \mathrm{~ms}^{-1}20 ms−1
  2. B
    2 ms−12 \mathrm{~ms}^{-1}2 ms−1
  3. C
    3.2 ms−13.2 \mathrm{~ms}^{-1}3.2 ms−1
  4. D
    0.5 ms−10.5 \mathrm{~ms}^{-1}0.5 ms−1
View written solutionFree

Correct answer: B

  1. For a rod of length lll moving with speed vvv perpendicular to a magnetic field BBB, the motional emf is

E=Blv\mathcal{E} = B l vE=Blv

  1. Given:

E=0.08 V,l=10 cm=0.1 m,B=0.4 T\mathcal{E} = 0.08\ \text{V}, \quad l = 10\ \text{cm} = 0.1\ \text{m}, \quad B = 0.4\ \text{T}E=0.08 V,l=10 cm=0.1 m,B=0.4 T

  1. Substitute into the formula:

0.08=0.4×0.1×v0.08 = 0.4 \times 0.1 \times v0.08=0.4×0.1×v

0.08=0.04v0.08 = 0.04v0.08=0.04v

  1. Solve for vvv:

v=0.080.04=2 m s−1v = \frac{0.08}{0.04} = 2\ \text{m s}^{-1}v=0.040.08​=2 m s−1

  1. Check options:
  • A: 20 m s−120\ \text{m s}^{-1}20 m s−1
  • B: 2 m s−12\ \text{m s}^{-1}2 m s−1
  • C: 3.2 m s−13.2\ \text{m s}^{-1}3.2 m s−1
  • D: 0.5 m s−10.5\ \text{m s}^{-1}0.5 m s−1

Hence, the correct option is B.

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