Verified Solution • 2025–26 Session moderate

QJEEM19Ph109d: JEE Main 2019 Physics - Class 12 JEEM19Ph109d Solution: A plane electromagnetic wave of frequency 50 MHz travels in…

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, E→=6.3j^ V/m.\overrightarrow E = 6.3\widehat j\,V/m. The corresponding magnetic field B,→\overrightarrow {B,} at that point will be :

Multiple Choice Options

  • A 18.9 ×\times 10 −- 8 k^\widehat k T
  • B 2.1 ×\times 10 −- 8 k^\widehat k T
  • C 6.3 ×\times 10 −- 8 k^\widehat k T
  • D 18.9 ×\times 108 k^\widehat k T

Step-by-Step Verified Solution

Given, E→=6.3j^\overrightarrow E = 6.3\widehat j

∴\therefore     ∣E→∣=(6.3)2=6.3\left| {\overrightarrow E } \right| = \sqrt {{{\left( {6.3} \right)}^2}} = 6.3

∴\therefore     ∣B→∣=∣E→∣C\left| {\overrightarrow B } \right| = {{\left| {\overrightarrow E } \right|} \over C}

=6.33×108= {{6.3} \over {3 \times {{10}^8}}}

=2.1×10−8 = 2.1 \times {10^{ - 8}}\, T

Now we have to find the direction of B→\overrightarrow B .

We know, E^×B^=C^\widehat E \times \widehat B = \widehat C and given E→\overrightarrow E is in y-direction and wave moving in positive x-direction.

∴\therefore     J^×B^=i^\widehat J \times \widehat B = \widehat i

⇒\Rightarrow     B^=K^\widehat B = \widehat K

∴\therefore     B→=∣B→∣B^\overrightarrow B = \left| {\overrightarrow B } \right|\widehat B

=2.1×10−8 K^ T= 2.1 \times {10^{ - 8}}\,\widehat K\,T

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring