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  • The magnetic field vector of an EM wave is given by <img alt="B=B_0 \frac{i+\hat{\jmath}}{\sqrt{2}} \cos (k z-\omega t)" src="https://entrancecorner.oncodecogs.com/gif.latex?B%3DB_0%20%5Cfrac%7Bia

The magnetic field vector of an EM wave is given by B=B_0 \frac{i+\hat{\jmath}}{\sqrt{2}} \cos (k z-\omega t) At t=0, two electric charges q_1 of 4 \pi coulomb and q_2 of 2 \pi coulomb located at \left(0,0, \frac{\pi}{k}\right) and \left(0,0, \frac{3 \pi}{k}\right) respectively, have the same velocity of 0.5 \mathrm{C} \hat{i}, The ratio of the force acting on charge q_1 to q_2 is:-

Option: 1

2 V2: 1


Option: 2

1:{\sqrt{2}}


Option: 3

2: 1


Option: 4

\sqrt{2}: 1


Answers (1)

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\begin{aligned} \overrightarrow{F_1} & =q(\vec{V} \times \vec{B}) \\ & =4 \pi\left[0.5 c \hat{\imath} \times B_0 \frac{(\hat{\imath}+\hat{\jmath})}{\sqrt{2}} \cos \left(k \cdot \frac{\pi}{k}-0\right)\right] \\ \overrightarrow{F_2} & =2 \pi\left[0.5 c \hat{\imath} \times B_0 \frac{(\hat{\imath}+\hat{\jmath})}{\sqrt{2}} \cos \left(k \cdot \frac{3 \pi}{k}-0\right)\right] \\ & \cos \pi=-1 \& \cos 3 \pi=-1 . \\ \Rightarrow & \frac{F_1}{F_2}=2: 1 . \end{aligned}

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himanshu.meshram

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