The electric field of an electromagnetic wave in free space is E to =57cos[7.5 \!\!×\!\! 10 6 t-5 \!\!×\!\! 10 (-3) (3x+4y)](4 i wedge -3 j wedge )N/C. The associated magnetic field in Tesla is-

B to =-frac 57 3× 10 8 cos [ 7.5× 10 6 t -5× 10 -3 ( 3 x +4 y ) ] ( 5overset mathop k \, )

B to =frac 57 3× 10 8 cos [ 7.5× 10 6 t -5× 10 -3 ( 3 x +4 y ) ] ( 5overset mathop k \, )

B to =57/(3 \!\!×\!\! 10 8 )cos[7.5 \!\!×\!\! 10 6 t-5 \!\!×\!\! 10 (-3) (3x+4y)]( k wedge )

B to =-frac 57 3× 10 8 cos [ 7.5× 10 6 t -5× 10 -3 ( 3 x +4 y ) ] ( k )

Physics · Electromagnetic Waves

JEE Main 2025 — 23 January, Morning Shift — Question 62

The electric field of an electromagnetic wave in free space is ${{E}^{\to }}=57cos[7.5\text{ }\!\!\times\!\!\text{ }{{10}^{6}}t-5\text{ }\!\!\times\!\!\text{ }{{10}^{(-3)}}(3x+4y)](4{{i}^{\wedge }}-3{{j}^{\wedge }})N/C.$ The associated magnetic field in Tesla is-

Option A:
${{B}^{\to }}=\frac{57}{3\times {{10}^{8}}}\text{cos}\left[ 7.5\times {{10}^{6}}\text{t}-5\times {{10}^{-3}}\left( 3\text{x}+4\text{y} \right) \right]\left( 5\overset{\text{}}{\mathop{\text{k}}}\, \right)$
Option B:
${{B}^{\to }}=57/(3\text{ }\!\!\times\!\!\text{ }{{10}^{8}})cos[7.5\text{ }\!\!\times\!\!\text{ }{{10}^{6}}t-5\text{ }\!\!\times\!\!\text{ }{{10}^{(-3)}}(3x+4y)]({{k}^{\wedge }}\text{ })$
Option C:
${{B}^{\to }}=-\frac{57}{3\times {{10}^{8}}}\text{cos}\left[ 7.5\times {{10}^{6}}\text{t}-5\times {{10}^{-3}}\left( 3\text{x}+4\text{y} \right) \right]\left( 5\overset{\text{}}{\mathop{\text{k}}}\, \right)$
Correct
Option D:
${{B}^{\to }}=-\frac{57}{3\times {{10}^{8}}}\text{cos}\left[ 7.5\times {{10}^{6}}\text{t}-5\times {{10}^{-3}}\left( 3\text{x}+4\text{y} \right) \right]\left( \text{ }k\text{ } \right)$

Answer: C

Step-by-step solution

${{\text{K}}^{\to }}=3\overset{\text{}}{\mathop{\text{i}}}\,+4\overset{\text{}}{\mathop{\text{j}}}\,$

$\overset{\text{}}{\mathop{\text{K}}}\,=\frac{3\overset{\text{}}{\mathop{\text{i}}}\,+4\overset{\text{}}{\mathop{\text{j}}}\,}{5}$

$\overset{}{\mathop{B}}\,=\overset{\text{}}{\mathop{\text{K}}}\,\times \overset{\text{}}{\mathop{\text{E}}}\,$

${{\text{B}}_{0}}=\frac{{{\text{E}}_{0}}}{\text{C}}=\frac{57}{3\times {{10}^{8}}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves
Topic: Displacement Current and Equation of EM Waves