Question

The magnetic field vector of an electromagnetic wave is given by $B=B_o\frac{\hat{i}+\hat{j}}{\sqrt{2}}\cos (kz-\omega t)$; where $\hat{i},\hat{j}$ represents unit vector along $x$ and $y$-axis respectively. At $t=0s$, 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 $0.5c\hat{i}$, (where $c$ is the velocity of light). The ratio of the force acting on charge $q_1$ to $q_2$ is :

• A. $2\sqrt{2}:1$
• B. $1:\sqrt{2}$
• C. $2:1$
• D. $\sqrt{2}:1$

Answer 1

Answer: (C)

$\vec{F}=q(\vec{V}\times \vec{B})$
${\vec{F}}_1=4\pi \left[0.5c\hat{i}\times B_0\left(\frac{\hat{i}+\hat{j}}{2}\right)\cos \left(K\cdot \frac{\pi }{K}-0\right)\right]$
${\vec{F}}_2=2\pi \left[0.5c\hat{i}\times B_0\left(\frac{\hat{i}+\hat{j}}{2}\right)\cos \left(K\cdot \frac{3\pi }{K}-0\right)\right]$
$\cos \pi =-1,\cos 3\pi =-1$
$\therefore \frac{F_1}{F_2}=2$