Electromagnetic Waves

JEE Main level, single correct, medium

124 of 165

Question 124 of 165

An electromagnetic wave travelling in the x−x- direction has frequency of 2×1014Hz2\times {10}^{14}Hz and electric field amplitude of 27Vm–127V{m}^{–1} oscillates in Y−Y-direction. From the options given below, which one describes the magnetic field for this wave?

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Answer: (D) B⃗\vec{B}(x,t)=(9×10−8T)k^sin2π[(x1.5×10−6)−2×1014t]\left(x,t\right)=\left(9\times {10}^{-8}T\right)\hat{k}sin2\pi \left[\left(\dfrac{x}{1.5\times {10}^{-6}}\right)-2\times {10}^{14}t\right]

When undefined

Then B=B0sin⁡⁡B={B}_{0}\sin ⁡(kx−ωt)(kx-\omega t)

Of light in travelling along i^\hat{i} then B⃗\vec{B} in either along j^\hat{j} or k^\hat{k}.

∴∴ speed of light C=E0B0⇒B0=E0CC=\dfrac{{E}_{0}}{{B}_{0}}\Rightarrow {B}_{0}=\dfrac{{E}_{0}}{C}

⇒B0=273×108=9×10−8T\Rightarrow {B}_{0}=\dfrac{27}{3\times {10}^{8}}=9\times {10}^{-8}T

also, ω=2πf=2π×2×1014=4π×1014\omega =2\pi f=2\pi \times 2\times {10}^{14}=4\pi \times {10}^{14}

k=ωc=4π×10143×108k=\dfrac{\omega }{c}=\dfrac{4\pi \times {10}^{14}}{3\times {10}^{8}}

=(2π1.5×10−6)=(\dfrac{2\pi }{1.5\times {10}^{-6}})

Looking into the option the correct

Answer is B⃗=9×10−8sin⁡⁡2π(x1.5×10−6−2×1014t)k^\vec{B}=9\times {10}^{-8}\sin ⁡2\pi \left(\dfrac{x}{1.5\times {10}^{-6}}-2\times {10}^{14}t\right)\hat{k}

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