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JEE Main 201912 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2019 · 12 January · Evening Shift · Physics · Electromagnetic Induction

A 10m long horizontal wire extends from North East to South West....

A 10m10m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0ms−1,5.0m{s}^{- 1}, at right angles to the horizontal component of the earth's magnetic field of 0.3×10−4Wbm−2.0.3 \times {10}^{- 4}Wb{m}^{- 2} . The value of the induced emf in the wire is:A 10m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0m s ^ - 1 , at right

Options

  1. A0.3×10−3V0.3 \times {10}^{- 3}V
  2. B1.1×10−3V1.1 \times {10}^{- 3}V
  3. C2.5×10−3V2.5 \times {10}^{- 3}V
  4. D1.5×10−3V1.5 \times {10}^{- 3}V

Correct answer

B. 1.1×10−3V1.1 \times {10}^{- 3}V

Step-by-step solution

A 10m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0m s ^ - 1 , at right

The wire makes an angle 45°45 ° with the magnetic field lines. Therefore, the induced emf will be,

ε=Bvlsin450ε = Bvlsin{45}^{0}

=0.3×10−4×5×10×12= 0.3 \times {10}^{- 4} \times 5 \times 10 \times \dfrac{1}{\sqrt{2}}

=1.1×10−3V= 1.1 \times {10}^{- 3}V

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