The magnetic field vector of an electromagnetic wave is given by {\B}=\B_{0} {{{i}}^{\^}+{{j}}^{\^}}/{\\sqrt{2} \cos (k z-\\omegat) \T where {{i}}^{\^}, {{j}}^{\^} represents unit vector along X and Y-axis respectively. At t=0, two electric charges q_{1}of 4 \\pi {\C} and q_{2} of 2 \\pi C located at (0,0, \frac{\\pi}{k}) and (0,0, \frac{3 \\pi}{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 [31 Aug 2021 Shift 2]

JEE Main 31-Aug-2021 Shift 2 Solved Paper

Section: Physics

Question : 12 of 90

Marks: +1, -0

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

√2

‌cos(kz−ωt)‌T where

represents unit vector along X and Y-axis respectively. At t=0, two electric charges q1of 4πC and q2 of 2π C located at (0,0,

) and (0,0,

3π

) respectively, have the same velocity of 0.5c

.
(where, c is the velocity of light). The ratio of the force acting on charge q1 to q2 is

[31 Aug 2021 Shift 2]

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