Electromagnetic Waves

JEE Main level, single correct, medium

18 of 165

Question 18 of 165

The electric field in a plane electromagnetic wave is given by :
Ey=69sin⁡[0.6×103x−1.8×1011t]V/mE_{y}=69 \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right] \mathrm{V} / \mathrm{m}.
The expression for magnetic field associated with this electromagnetic wave is ____\_\_\_\_ T.

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Answer: (A) Bz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_{z}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right]

Given Ey=69sin⁡[0.6×103x−1.8×1011t]E_y = 69 \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] V/m
Wave propagates along +x (form kx−ωtkx - \omega t), E⃗\vec{E} is along y.
For EM wave: E⃗×B⃗\vec{E} \times \vec{B} = direction of propagation
j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, so B⃗\vec{B} is along z-direction.
B0=E0c=693×108=2.3×10−7B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 2.3 \times 10^{-7} T
E and B are in phase for plane EM wave.
Bz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] T

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