The electric field of a plane electromagnetic wave is given by $\vec{E} = E_{0} \frac{\hat{i} + \hat{j}}{\sqrt{2}} \cos (k z + ω t) .$ At t = 0, a positively charged particle is at the point $(x, y, z) = (0, 0, \frac{π}{k})$ . If its instantaneous velocity at t = 0 is $v_{0} \hat{k}$ , the force acting on it due to the wave at t = 0 is

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Parallel to $\hat{k}$

Anti-parallel to $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$

Parallel to $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$

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answer is A.

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Detailed Solution

Given, $\bar{E} = E_{0} \frac{(\hat{i} + \hat{j})}{\sqrt{2}} \cos (k z + ω t)$

At t = 0 positively charged particle is at, $(x, y, z) = (0, 0, π / 4)$

$∴ \bar{E} = E_{0} \frac{(\hat{i} + \hat{j})}{\sqrt{2}} \cos π = - E_{0} \frac{(\hat{i} + \hat{j})}{\sqrt{2}}$

Magnetic field at t = 0 and at point $(x, y, z) = (0, 0, \frac{π}{k})$ will be, $N o w, n e t f o r c e, \vec{F} = q \vec{E} + q (\vec{v} \times \vec{B})$

So, force due to magnetic field will be : ${\vec{F}}_{B} = q v_{0} \hat{k} \times (- \frac{E_{0}}{c}) (\frac{\hat{i} - \hat{j}}{\sqrt{2}}) = - \frac{q E_{0} v_{0}}{c} (\frac{\hat{i} + \hat{j}}{\sqrt{2}})$

So direction of force due to magnetic field parallel to that of $\vec{E}$ Net force acting on the charged particle due to the wave is anti-parallel to $\frac{\hat{i} + \hat{j}}{\sqrt{2}}$

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