The magnetic field vector of an electromagnetic wave is given by B=B0(^i+^j√2)cos(kz−ωt), where, ^i,^j represents unit vector along x and y−axis respectively. At t=0 s, two electric charges q1 of 4π coulomb and q2 of 2π coulomb located at (0,0,πk) and (0,0,3πk) respectively, have the same velocity of 0.5c ^i, (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is

The magnetic field vector of an electromagnetic wave is given by B=B0(

1.	The magnetic field vector of an electromagnetic wave is given by B=B0(^i+^j√2)cos(kz−ωt), where, ^i,^j represents unit vector along x and y−axis respectively. At t=0 s, two electric charges q1 of 4π coulomb and q2 of 2π coulomb located at (0,0,πk) and (0,0,3πk) respectively, have the same velocity of 0.5c ^i, (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is
Answer» The magnetic field vector of an electromagnetic wave is given by $B = B_{0} (\frac{^i +^j}{\sqrt{2}}) cos (k z - ω t)$ , where, $^i,^j$ represents unit vector along $x$ and $y -$ axis respectively. At $t = 0 s$ , two electric charges $q_{1}$ of $4 π$ coulomb and $q_{2}$ of $2 π$ coulomb located at $(0, 0, \frac{π}{k})$ and $(0, 0, \frac{3 π}{k})$ respectively, have the same velocity of $0.5 c^i$ , (where $c$ is the velocity of light). The ratio of the force acting on charge $q_{1}$ to $q_{2}$ is