Electromagnetic Waves

JEE Main level, single correct, medium

34 of 165

Question 34 of 165

The electric field in an electromagnetic wave is given by E→=i^40cos⁡ω(t−z/c)NC−1\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}. The magnetic field induction of this wave is (in SI unit) :

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Answer: (D) B→=j^40Ccos⁡ω(t−z/c)\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{C}} \cos \omega(\mathrm{t}-z / \mathrm{c})

E→=i^40cos⁡ω(t−zc)\overrightarrow{\mathrm{E}}=\hat{\mathrm{i}} 40 \cos \omega\left(\mathrm{t}-\frac{\mathrm{z}}{\mathrm{c}}\right)
E→\overrightarrow{\mathrm{E}} is along +x+\mathrm{x} direction
v→\overrightarrow{\mathrm{v}} is along +z+\mathrm{z} direction
So direction of B⃗\vec{B} will be along +y and magnitude of BB will be Ec\frac{E}{c}
So answer is 40ccos⁡ω(t−zc)j^\frac{40}{c} \cos \omega\left(t-\frac{z}{c}\right) \hat{j}

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