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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2019 · 8 April · Evening Shift · Physics · Electromagnetic Waves

The magnetic field of an electromagnetic wave is given by: → B =...

The magnetic field of an electromagnetic wave is given by:
\overset{\to}{B} = 1.6 \times {10}^{- 6}\cos \left(2 \times {10}^{7}z + 6 \times {10}^{15}t\right)\left(2\overset{^}{i} + \overset{^}{j}\right)\dfrac{Wb}{{m}^{2}}
The associated electric field will be:

Options

  1. AE→=4.8×102cos\overset{\to}{E} = 4.8 \times {10}^{2}cos\left(2 \times {10}^{7}z - 6 \times {10}^{15}t\right)\left(- 2\overset{^}{j} + \overset{^}{i}\right)\dfrac{V}{m}
  2. BE→=4.8×102cos\overset{\to}{E} = 4.8 \times {10}^{2}cos\left(2 \times {10}^{7}z + 6 \times {10}^{15}t\right)\left(\overset{^}{i} - 2\overset{^}{j}\right)\dfrac{V}{m}
  3. CE→=4.8×102cos\overset{\to}{E} = 4.8 \times {10}^{2}cos\left(2 \times {10}^{7}z + 6 \times {10}^{15}t\right)\left(- \overset{^}{i} + 2\overset{^}{j}\right)\dfrac{V}{m}
  4. DE→=4.8×102cos\overset{\to}{E} = 4.8 \times {10}^{2}cos\left(2 \times {10}^{7}z - 6 \times {10}^{15}t\right)\left(2\overset{^}{i} + \overset{^}{j}\right)\dfrac{V}{m}

Correct answer

C. E→=4.8×102cos\overset{\to}{E} = 4.8 \times {10}^{2}cos\left(2 \times {10}^{7}z + 6 \times {10}^{15}t\right)\left(- \overset{^}{i} + 2\overset{^}{j}\right)\dfrac{V}{m}

Step-by-step solution

Given magnetic field, \overset{\to}{B} = 1.6 \times {10}^{- 6}\cos \left(2 \times {10}^{7}z + 6 \times {10}^{15}t\right)\left(2\overset{^}{i} + \overset{^}{j}\right)\dfrac{Wb}{{m}^{2}}
The relation between magnetic field and electric field is E=cB=(3×108)×(1.6×10−6×5)=4.8×102×5E = cB = ( 3 \times {10}^{8} ) \times ( 1.6 \times {10}^{- 6} \times \sqrt{5} ) = 4.8 \times {10}^{2} \times \sqrt{5}
The direction of light propagation will be along E→×B→\overset{\to}{E} \times \overset{\to}{B} .
E→=4.8×102cos\overset{\to}{E} = 4.8 \times {10}^{2}cos\left(2 \times {10}^{7}z + 6 \times {10}^{15}t\right)\left(- \overset{^}{i} + 2\overset{^}{j}\right)\dfrac{V}{m}
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