JEE Main 2021 (31 Aug, Shift 2) Physics Q13: The magnetic…

Question

The magnetic field vector of an electromagnetic wave is given by

B = B_0\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 :

Official answer

Option 3

NTA final key.

Accepted Answer

Option 3: $2:1$

Answer

$\sqrt{2}:1$

Answer

$1:\sqrt{2}$

Answer

$2:1$

Answer

$2\sqrt{2}:1$

JEE Main 31 August 2021, Shift 2, Physics Q13