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