Electromagnetic Waves

JEE Main level, single correct, medium

10 of 165

Question 10 of 165

A plane electromagnetic wave is moving in free space with velocity c=3×108 m/sc=3 \times 10^{8} \mathrm{~m} / \mathrm{s} and its electric field is given as E⃗=54sin⁡(kz−ωt)j^ V/m\vec{E}=54 \sin (k z-\omega t) \hat{j} \mathrm{~V} / \mathrm{m}, where j^\hat{j} is the unit vector along yy-axis. The magnetic field vector B⃗\vec{B} of the wave is:

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Answer: (C) −1.8×10−7sin⁡(kz−ωt)i^ T-1.8 \times 10^{-7} \sin (k z-\omega t) \hat{i} \mathrm{~T}

For an electromagnetic wave, the relationship between electric and magnetic field amplitudes is B0=E0cB_0 = \frac{E_0}{c}.
Here E0=54E_0 = 54 V/m and c=3×108c = 3 \times 10^8 m/s.
So, B0=543×108=1.8×10−7B_0 = \frac{54}{3 \times 10^8} = 1.8 \times 10^{-7} T.
The electric field is along the j^\hat{j} direction.
Since the wave travels in the +z+z direction, the magnetic field direction is determined by E⃗×B⃗\vec{E} \times \vec{B} pointing in the +z+z direction.
Using the right-hand rule: j^×(−i^)=k^\hat{j} \times (-\hat{i}) = \hat{k} (positive zz direction), confirming the magnetic field is in the −i^-\hat{i} direction.
Therefore, B⃗=−1.8×10−7sin⁡(kz−ωt)i^\vec{B} = -1.8 \times 10^{-7}\sin(kz - \omega t)\hat{i} T.

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