Verified Solution • 2025–26 Session moderate

QJEEM20Ph285f: JEE Main 2020 Physics - Class 12 JEEM20Ph285f Solution: For a plane electromagnetic wave, the magnetic field at a point …

For a plane electromagnetic wave, the magnetic field at a point x and time t is

B→(x,t)\overrightarrow B \left( {x,t} \right) = [1.2×10−7sin⁡(0.5×103x+1.5×1011t)k^]\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right] T

The instantaneous electric field E→\overrightarrow E corresponding to B→\overrightarrow B is :
(speed of light c = 3 × 108 ms–1)

Multiple Choice Options

  • A E→(x,t)=[36sin⁡(1×103x+1.5×1011t)i^]\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat i} \right] Vm{V \over m}
  • B E→(x,t)=[36sin⁡(0.5×103x+1.5×1011t)k^]Vm\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]{V \over m}
  • C E→(x,t)=[36sin⁡(1×103x+0.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}
  • D E→(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}

Step-by-Step Verified Solution

Given,
B→(x,t)\overrightarrow B \left( {x,t} \right) = [1.2×10−7sin⁡(0.5×103x+1.5×1011t)k^]\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right] T

Wave is travelling along (–x) axis and B→\overrightarrow B is along +z axis.

We know, Magnitude of electric field

E = BC

= 1.2 ×\times 10-7 sin ( 0.5 10 + 1.5 ×\times 1011t ) ×\times 3 ×\times 108

= 36 sin (0.5 ×\times 103x+1.5 ×\times 1011t) V/m

Also, s→=E→×B→μ0\overrightarrow s = {{\overrightarrow E \times \overrightarrow B } \over {{\mu _0}}}

⇒\Rightarrow −i^=E→×k^μ0- \widehat i = {{\overrightarrow E \times \widehat k} \over {{\mu _0}}}

∴\therefore Direction of E→=−j^{\overrightarrow E = - \widehat j}

∴\therefore Instantaneous electric field,

E→(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}

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