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MCQEasyJEE 2024Faraday's Laws of EMI

JEE Physics 2024 Question with Solution

A horizontal straight wire 5 m5 \, \text{m} long extending from east to west falls freely at a right angle to the horizontal component of Earth’s magnetic field 0.60×10−4 Wb/m20.60 \times 10^{-4} \, \text{Wb/m}^2. The instantaneous value of emf induced in the wire when its velocity is 10 m/s10 \, \text{m/s} is:

  • A

    3×10−3 V3 \times 10^{-3} \, \text{V}

  • B

    6×10−3 V6 \times 10^{-3} \, \text{V}

  • C

    9×10−3 V9 \times 10^{-3} \, \text{V}

  • D

    2×10−3 V2 \times 10^{-3} \, \text{V}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: magnetic field component B=0.60×10−4 Wb/m2B = 0.60 \times 10^{-4} \, \text{Wb/m}^2, wire length l=5 ml = 5 \, \text{m}, and speed v=10 m/sv = 10 \, \text{m/s}.

Find: the instantaneous induced emf.

For a conductor moving perpendicular to the magnetic field, the motional emf is

emf=Blv\text{emf} = Blv

Substituting the given values,

emf=(0.60×10−4)×5×10\text{emf} = \left(0.60 \times 10^{-4}\right) \times 5 \times 10 emf=3.0×10−3 V\text{emf} = 3.0 \times 10^{-3} \, \text{V}

Therefore, the induced emf is 3×10−3 V3 \times 10^{-3} \, \text{V} and the correct option is A.

The solution text contains an arithmetic inconsistency, but its listed correct option is 3, which matches 3×10−3 V3 \times 10^{-3} \, \text{V} in the options.

Common mistakes

  • Using the full Earth’s magnetic field instead of the horizontal component is incorrect because the question explicitly states that the wire falls at right angle to the horizontal component. Use only the given component 0.60×10−4 Wb/m20.60 \times 10^{-4} \, \text{Wb/m}^2.

  • Making an arithmetic error in evaluating BlvBlv leads to a wrong emf. Compute 5×10=505 \times 10 = 50 first, then 0.60×50×10−4=30×10−4=3×10−30.60 \times 50 \times 10^{-4} = 30 \times 10^{-4} = 3 \times 10^{-3}.

  • Confusing Wb/m2\text{Wb/m}^2 with another unit and not recognizing it as tesla can cause formula misuse. Treat the given magnetic field in standard SI form and apply emf=Blv\text{emf} = Blv directly.

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