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Electromagnetic Induction question

2023 · 25 Jan · Shift 2 · Q51
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  5. /2023 · 25 Jan · Shift 2 · Q51

Electromagnetic Induction question

2023 · 25 Jan · Shift 2 · Q51

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be ‾\underline{\hspace{2cm}}​.
  1. A
    20 V
  2. B
    8 V
  3. C
    16 V
  4. D
    12 V
View written solutionFree

Correct answer: C

  1. Use motional emf formula
    For a wire of length lll moving with speed vvv perpendicular to a magnetic field BBB, the induced emf is ε=Blv\varepsilon = B l vε=Blv

  2. Substitute the given values
    B=2 T,l=1 m,v=8 m/sB = 2\,\text{T}, \quad l = 1\,\text{m}, \quad v = 8\,\text{m/s}B=2T,l=1m,v=8m/s

    So, ε=2×1×8=16 V\varepsilon = 2 \times 1 \times 8 = 16\,\text{V}ε=2×1×8=16V

  3. Match with the options
    16 V16\,\text{V}16V corresponds to Option C.

  4. Comparison with stored answer
    Stored correct answer is C, which matches the derived answer.

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