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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2023 · 8 April · Evening Shift · Physics · Electromagnetic Induction

An emf of 0 . 08V is induced in a metal rod of length 10cm held...

An emf of 0.08V0 . 08V is induced in a metal rod of length 10cm10cm held normal to a uniform magnetic field of 0.4T0 . 4T, when move with a velocity of:

Options

  1. A0.5ms−10 . 5m{s}^{- 1}
  2. B20ms−120m{s}^{- 1}
  3. C3.2ms−13 . 2m{s}^{- 1}
  4. D2ms−12m{s}^{- 1}

Correct answer

D. 2ms−12m{s}^{- 1}

Step-by-step solution

The formula to calculate the motional emf induced in a moving conductor in a magnetic field is given by

ε=Blv...(1)ε = Blv . . . \left(1\right)

Substitute the values of the known parameters into equation (1) and solve to calculate the required velocity of the rod.

{0.08V=0.4T×(10cm×1m100cm)×v⇒v=0.08V0.4T×10cm×1m100cm=2ms−1\begin{cases}0 . 08V & = & 0 . 4T \times \left(10cm \times \dfrac{1m}{100cm}\right) \times v \\ & ⇒ & v = \dfrac{0 . 08V}{0 . 4T \times 10cm \times \dfrac{1m}{100cm}} \\ & = & 2m{s}^{- 1}\end{cases}

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