The electric field of a plane electromagnetic wave is given by

$\vec{E} = E_{0} \frac{\hat{i} + \hat{j}}{\sqrt[]{2}} c o s (k z + ω t)$

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 is:

Option 1 -

Zero

Option 2 -

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

Option 3 -

Parallel to $\hat{k}$

Option 4 -

Antiparallel to $\frac{\hat{i} + \hat{j}}{\sqrt[]{2}}$

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

Electric field at $t = 0$ & $(x, y, z) = (0,0, \frac{π}{k})$ is

$\begin{array}{r} \overset{?}{E} = - (\frac{\hat{i} + \hat{j}}{\sqrt[]{2}}) E_{0} and \overset{?}{F} = \overset{?}{E} q \\ ∴ \overset{?}{F} = - (\frac{\hat{i} + \hat{j}}{\sqrt[]{2}}) q \end{array}$

Which is antiparallel to $(\frac{\hat{i} + \hat{j}}{\sqrt[]{2}})$

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The electric field of a plane electromagnetic wave is given by E ⃗ = E 0 i ˆ + j ˆ 2 c o s ⁡ ( k z + ω t ) At t = 0 , a positively charged particle is at the point ( x , y , z ) = 0,0 , π k . If its instantaneous velocity at ( t = 0 ) is v 0 k ˆ , the force acting on it due to the wave is: