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JEE Main 20211 SeptemberEvening Shift
Exam
JEE Main
Year
2021
Date
1 September
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2021 · 1 September · Evening Shift · Physics · Electromagnetic Induction

A square loop of side 20cm and resistance 1Ω is moved towards right...

A square loop of side 20cm20cm and resistance 1Ω1Ω is moved towards right with a constant speed v0.{v}_{0} . The right arm of the loop is in a uniform magnetic field of 5T.5T . The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4Ω.4Ω . What should be the value of v0{v}_{0} so that a steady current of 2mA2mA flows in the loop ?

A square loop of side 20cm and resistance 1Ω is moved towards right with a constant speed v _ 0 . The right arm of the

Options

  1. A10−2cms−1{10}^{- 2}cm{s}^{- 1}
  2. B1ms−11m{s}^{- 1}
  3. C1cms−11cm{s}^{- 1}
  4. D102ms−1{10}^{2}m{s}^{- 1}

Correct answer

C. 1cms−11cm{s}^{- 1}

Step-by-step solution

The EMF& is induced in the coil when it is moving& in a uniform magnetic field,
e=Bvle = Bvl~and the current,& i=eR⇒BlvR=0.002Ai = \dfrac{e}{R}⇒\dfrac{Blv}{R} = 0 . 002A
⇒v=0.002×RBl=0.002×(4+1)5×0.2=0.01ms−1=1cms−1⇒ v = \dfrac{0 . 002 \times R}{Bl}= \dfrac{0 . 002 \times \left(4 + 1\right)}{5 \times 0 . 2} = 0 . 01ms^{- 1} = 1cms^{- 1}.

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