The magnetic field vector of an electromagnetic wave is given by B = B0\frac{\hat{i} + \hat{j}}{\sqrt2}cos (kz - ωt); where \hat{i}, \hat{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 \Big(0, 0, \frac{\pi}{k}\Big) and \Big(0, 0, \frac{3 \pi}{k}\Big), 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 q1 to q2 is
(1) 2√2 : 1
(2) 1 : √2
(3) 2 : 1
(4) √2 : 1
Correct answer: (3) 2 : 1
Explanation:
\vec{F} = q(\vec{V} \times \vec{B})
\vec{F_1} = 4π\Big[0.5c \hat{i} \times B_0 \Big(\frac{\hat{i} + \hat{j}}{2}\Big)cos \Big(K.\frac{\pi}{K}-0 \Big) \Big]
\vec{F_2} = 2π\Big[0.5c \hat{i} \times B_0 \Big(\frac{\hat{i} + \hat{j}}{2}\Big)cos \Big(K.\frac{3\pi}{K}-0 \Big) \Big]
cosπ = -1, cos3π = -1
∴ \frac{F_1}{F_2} = 2