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If the magnetic field in a plane electromagnetic wave is given by B=3× 10^-8 ⁡ (1.6× 10^3x+48× 10^10t )jT, then what will be expression for electric field ?

E= (9 ⁡ (1.6× 10^3x+48× 10^10t )k(V)/(m) )

E= (60 ⁡ (1.6× 10^3x+48× 10^10t )k(V)/(m) )

E= (3× 10^-8 ⁡ (1.6× 10^3x+48× 10^10t )j(V)/(m) )

E= (3× 10^-8 ⁡ (1.6× 10^3x+48× 10^10t )i(V)/(m) )

Electromagnetic Waves

JEE Main level, single correct, medium

100 of 165

Question 100 of 165

If the magnetic field in a plane electromagnetic wave is given by $\vec{B}=3\times {10}^{-8}\sin ⁡\left(1.6\times {10}^{3}x+48\times {10}^{10}t\right)\hat{j}T,$ then what will be expression for electric field ?

Show answer and solution

Answer: (B) $\vec{E}=\left(9\sin ⁡\left(1.6\times {10}^{3}x+48\times {10}^{10}t\right)\hat{k}\dfrac{V}{m}\right)$

\dfrac{{E}_{0}}{{B}_{0}}=C

(speed of light in vacuum)

{E}_{0}={B}_{0}C=3\times {10}^{-8}\times 3\times {10}^{8}

=9\dfrac{N}{C}

So,

E=9\sin ⁡\left(\omega t+kx+\phi \right)

\vec{E}=\left(9\sin ⁡\left(1.6\times {10}^{3}x+48\times {10}^{10}t\right)\hat{k}\dfrac{V}{m}\right)

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JEE Main 2025 Apr 2 Shift 1 question paper

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