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JEE Main 201910 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2019 · 10 January · Morning Shift · Physics · Electromagnetic Induction

A solid metal cube of edge length 2cm is moving in the positive y...

A solid metal cube of edge length 2cm2cm is moving in the positive yy-direction, at a constant speed of 6ms−16m{s}^{- 1}. There is a uniform magnetic field of 0.1T\text{0.1}T in the positive zz-direction. The potential difference between the two faces of the cube, perpendicular to the xx-axis, is

Options

  1. A12mV12mV
  2. B1mV1mV
  3. C2mV2mV
  4. D6mV6mV

Correct answer

A. 12mV12mV

Step-by-step solution

emf developed is given by, e=Bvle = Bvl

∴e=0.1(2×10−2)×6⇒e=12×10−3=12mV∴ e = \text{0.1}\left(2 \times {10}^{- 2}\right) \times 6 ⇒ e = 12 \times {10}^{- 3} = 12mV

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