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Electromagnetic Induction question

2022 · 26 Jun · Shift 1 · Q52
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  5. /2022 · 26 Jun · Shift 1 · Q52

Electromagnetic Induction question

2022 · 26 Jun · Shift 1 · Q52

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
The magnetic flux through a coil perpendicular to its plane is varying according to the relation ϕ=(5t3+4t2+2t−5)\phi = (5{t^3} + 4{t^2} + 2t - 5)ϕ=(5t3+4t2+2t−5) Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,
  1. A
    15.6 A
  2. B
    16.6 A
  3. C
    17.6 A
  4. D
    18.6 A
View written solutionFree

Correct answer: A

  1. Use Faraday’s law

    The induced emf in the coil is e=−dϕdte = -\frac{d\phi}{dt}e=−dtdϕ​

    Given magnetic flux: ϕ=5t3+4t2+2t−5\phi = 5t^3 + 4t^2 + 2t - 5ϕ=5t3+4t2+2t−5

  2. Differentiate flux with respect to time

    dϕdt=15t2+8t+2\frac{d\phi}{dt} = 15t^2 + 8t + 2dtdϕ​=15t2+8t+2

    So magnitude of induced emf is ∣e∣=15t2+8t+2|e| = 15t^2 + 8t + 2∣e∣=15t2+8t+2

  3. Evaluate at t=2 st=2\,\text{s}t=2s

    ∣e∣=15(2)2+8(2)+2|e| = 15(2)^2 + 8(2) + 2∣e∣=15(2)2+8(2)+2 =15⋅4+16+2= 15\cdot 4 + 16 + 2=15⋅4+16+2 =60+16+2=78 V= 60 + 16 + 2 = 78\,\text{V}=60+16+2=78V

  4. Use Ohm’s law to find induced current

    Resistance of coil: R=5 ΩR = 5\,\OmegaR=5Ω

    I=∣e∣R=785=15.6 AI = \frac{|e|}{R} = \frac{78}{5} = 15.6\,\text{A}I=R∣e∣​=578​=15.6A

  5. Match with options

    15.6 A15.6\,\text{A}15.6A corresponds to Option A.

Final Answer: A: 15.6 A15.6\,\text{A}15.6A

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