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Electromagnetic Induction question

2022 · 26 Jun · Shift 2 · Q60
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  5. /2022 · 26 Jun · Shift 2 · Q60

Electromagnetic Induction question

2022 · 26 Jun · Shift 2 · Q60

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s −-− 1. If the horizontal component of earth's magnetic field is 0.2 ×\times× 10 −-− 4 T, then emf induced between the two ends of the conductor is :
  1. A
    5 μ\muμ V
  2. B
    50 μ\muμ V
  3. C
    5 mV
  4. D
    50 mv
View written solutionFree

Correct answer: B

  1. Given data

    • Length of rod: l=1 ml = 1\,\text{m}l=1m
    • Angular velocity: ω=5 rad s−1\omega = 5\,\text{rad s}^{-1}ω=5rad s−1
    • Horizontal component of earth's magnetic field: BH=0.2×10−4 T=2×10−5 TB_H = 0.2 \times 10^{-4}\,\text{T} = 2 \times 10^{-5}\,\text{T}BH​=0.2×10−4T=2×10−5T

  2. Relevant concept

    For a rod of length lll rotating about one end in a magnetic field component perpendicular to the plane of rotation, the induced emf between its ends is E=12Bωl2\mathcal{E} = \frac{1}{2} B \omega l^2E=21​Bωl2

  3. Which component of earth's field is effective?

    The rod rotates in a vertical plane parallel to east-west direction.

    • A vertical plane parallel to east-west has a normal along north-south direction.
    • The horizontal component of earth's magnetic field BHB_HBH​ is directed approximately along north-south.
    • Therefore, BHB_HBH​ is perpendicular to the plane of rotation, so it is the effective magnetic field.

    Hence, B=BH=2×10−5 TB = B_H = 2 \times 10^{-5}\,\text{T}B=BH​=2×10−5T

  4. Calculate the induced emf

    E=12Bωl2\mathcal{E} = \frac{1}{2} B \omega l^2E=21​Bωl2 E=12(2×10−5)(5)(1)2\mathcal{E} = \frac{1}{2}(2 \times 10^{-5})(5)(1)^2E=21​(2×10−5)(5)(1)2 E=5×10−5 V\mathcal{E} = 5 \times 10^{-5}\,\text{V}E=5×10−5V

  5. Convert into microvolts

    Since 1 μV=10−6 V1\,\mu\text{V} = 10^{-6}\,\text{V}1μV=10−6V we get 5×10−5 V=50×10−6 V=50 μV5 \times 10^{-5}\,\text{V} = 50 \times 10^{-6}\,\text{V} = 50\,\mu\text{V}5×10−5V=50×10−6V=50μV

  6. Check options

    • A: 5 μV5\,\mu\text{V}5μV ❌
    • B: 50 μV50\,\mu\text{V}50μV ✅
    • C: 5 mV5\,\text{mV}5mV ❌
    • D: 50 mV50\,\text{mV}50mV ❌

Final Answer: 50 μV\boxed{50\,\mu\text{V}}50μV​

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