Electromagnetic Waves

JEE Main level, single correct, medium

72 of 165

Question 72 of 165

The magnetic field vector of an electromagnetic wave is given by B=B0i^+j^2cos⁡(kz−ωt)B={B}_{0}\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\cos \left(kz-\omega t\right) where i^,j^\hat{i},\hat{j} represents unit vector along xx and yy-axis respectively. At t=0s,t=0s, two electric charges q1{q}_{1} of 4π4\pi coulomb and q2{q}_{2} of 2π2\pi coulomb located at (0,0,πk)\left(0,0,\dfrac{\pi }{k}\right) and (0,0,3πk),\left(0,0,\dfrac{3\pi }{k}\right), respectively, have the same velocity of 0.5ci^,0.5c\hat{i}, (where cc is the velocity of light ). The ratio of the force acting on charge q1{q}_{1} to q2{q}_{2} is :

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Answer: (D) 2:12:1

At t=0t=0
BB at (0,0,πk)=B0(i^+j^2)cos⁡(π)\left(0,0,\dfrac{\pi }{k}\right)={B}_{0}\left(\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\cos \left(\pi \right)
BB at (0,0,3πk)=B0(i^+j^2)cos⁡(3π)\left(0,0,\dfrac{3\pi }{k}\right)={B}_{0}\left(\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\cos \left(3\pi \right)
Force on charged particle q1{q}_{1}
F1=q1V⃗1×B⃗1{F}_{1}={q}_{1}{\vec{V}}_{1}\times {\vec{B}}_{1}
=4π(0.5ci^)×[−B0(i^+j^2)]=4πB0c22(−k^)=4\pi \left(0.5c\hat{i}\right)\times \left[-{B}_{0}\left(\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\right]=\dfrac{4\pi {B}_{0}c}{2\sqrt{2}}\left(-\hat{k}\right)
Force on charged particle q2{q}_{2}
F2=q2V⃗2×B⃗2{F}_{2}={q}_{2}{\vec{V}}_{2}\times {\vec{B}}_{2}
=2π(0.5ci^)×[−B0(i^+j^2)]=2πB0c22(−k^)⇒F1F2=2=2\pi \left(0.5c\hat{i}\right)\times \left[-{B}_{0}\left(\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\right]=\dfrac{2\pi {B}_{0}c}{2\sqrt{2}}\left(-\hat{k}\right)\Rightarrow \dfrac{{F}_{1}}{{F}_{2}}=2

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