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JEE Main 201510 AprilOnline
Exam
JEE Main
Year
2015
Date
10 April
Mode
Online
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2015 · 10 April · Online · Physics · Electromagnetic Waves

An electromagnetic wave travelling in the x - direction has...

An electromagnetic wave travelling in the x−x - direction has frequency of 2×1014Hz2 \times {10}^{14}Hz and electric field amplitude of 27Vm–127V{m}^{– 1} oscillates in Y−Y -direction. From the options given below, which one describes the magnetic field for this wave?

Options

  1. A\overset{\to}{B}\left(x, t\right) = ( 9 \times {10}^{- 8}T ) \overset{^}{j} \sin \left[1.5 \times {10}^{- 6}x - 2 \times {10}^{14}t\right]
  2. B\overset{\to}{B}\left(x, t\right) = ( 9 \times {10}^{- 8}T ) \overset{^}{i}\sin \left[2\pi\left(1.5 \times {10}^{- 8}x - 2 \times {10}^{14}t\right)\right]
  3. CB→\overset{\to}{B}\left(x, t\right) = \left(3 \times {10}^{- 8}T\right)\overset{^}{j}\sin 2 \pi\left[\left(\dfrac{x}{1.5 \times {10}^{- 8}}\right) - 2 \times {10}^{14}t\right]
  4. DB→\overset{\to}{B}\left(x, t\right) = \left(9 \times {10}^{- 8}T\right)\overset{\text{^}}{k}\sin 2 \pi\left[\left(\dfrac{x}{1.5 \times {10}^{- 6}}\right) - 2 \times {10}^{14}t\right]

Correct answer

D. B→\overset{\to}{B}\left(x, t\right) = \left(9 \times {10}^{- 8}T\right)\overset{\text{^}}{k}\sin 2 \pi\left[\left(\dfrac{x}{1.5 \times {10}^{- 6}}\right) - 2 \times {10}^{14}t\right]

Step-by-step solution

When undefined

Then B=B0sinB = {B}_{0}sin(kx−\omegat)( kx - \omegat )

Of light in travelling along \overset{^}{i} then\overset{\to}{B}ineitheralongin either along\overset{^}{j}oror\overset{^}{k}.<br/><br/>.<br /><br />∴speedoflightspeed of lightC = \dfrac{{E}_{0}}{{B}_{0}} ⇒ {B}_{0} = \dfrac{{E}_{0}}{C}<br/><br/><br /><br />⇒ {B}_{0} = \dfrac{27}{3 \times {10}^{8}} = 9 \times {10}^{- 8}T<br/><br/>also,<br /><br />also,\omega = 2\pif = 2\pi \times 2 \times {10}^{14} = 4\pi \times {10}^{14}<br/><br/><br /><br />k = \dfrac{\omega}{c} = \dfrac{4\pi \times {10}^{14}}{3 \times {10}^{8}}<br/><br/><br /><br />= ( \dfrac{2\pi}{1.5 \times {10}^{- 6}} )<br/><br/>Lookingintotheoptionthecorrect<br/><br/>Answeris<br /><br />Looking into the option the correct<br /><br />Answer is\overset{\to}{B} = 9 \times {10}^{- 8}\sin 2\pi\left(\dfrac{x}{1.5 \times {10}^{- 6}} - 2 \times {10}^{14}t\right)\overset{^}{k}
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