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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2020 · 5 September · Morning Shift · Physics · Magnetic Effects of Current

An electron is constrained to move along the y -axis with a speed...

An electron is constrained to move along the yy-axis with a speed of 0.1c0 . 1c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is \overset{\to}{E} = 30\overset{^}{j}\sin \left(1 . 5 \times {10}^{7}t - 5 \times {10}^{- 2}x\right)V{m}^{- 1} . where tt in in seconds and xx is im meters.The maximum magnetic force experienced by the electron will be: (given c=3×108ms−1c = 3 \times {10}^{8}m{s}^{- 1} and electron charge =1.6×10−19Coloumbs= 1 . 6 \times {10}^{- 19}Coloumbs

Options

  1. A3.2×10−18N3 . 2 \times {10}^{- 18}N
  2. B2.4×10−18N2 . 4 \times {10}^{- 18}N
  3. C4.8×10−19N4 . 8 \times {10}^{- 19}N
  4. D1.6×10−19N1 . 6 \times {10}^{- 19}N

Correct answer

C. 4.8×10−19N4 . 8 \times {10}^{- 19}N

Step-by-step solution

In electromagnetic wave is E0B0=C\dfrac{{E}_{0}}{{B}_{0}} = C
so maximum value of magnetic field is

B0=E0C{B}_{0} = \dfrac{{E}_{0}}{C}

{F}_\max = qV{B}_\max \cdot \sin 90 °

=qV0E0CN= \dfrac{q{V}_{0}{E}_{0}}{C}N

1.6×10−19×0.1×3×108×303×108=4.8×10−19N\dfrac{1 . 6 \times {10}^{- 19} \times 0 . 1 \times 3 \times {10}^{8} \times 30}{3 \times {10}^{8}} = 4 . 8 \times {10}^{- 19}N

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