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 $t=0 s$, 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 $0.5 c \hat{ i }$, (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)

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