First slide

Electromagnetic waves and its nature

Question

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 is:

Difficult

A

zero
B

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

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

Parallel to $\hat{k}$

Solution

$\begin{array}{l} A t t = 0, x = K \\ E l e c t r i c f i e l d = \vec{E} = E_{0} \frac{(\hat{i} + \hat{j})}{\sqrt{2}} \cos (π) = - \frac{(\hat{i} + \hat{j})}{\sqrt{2}} E_{0} \end{array}$
$∴$ $F o r c e d u e t o e l e c t r i c f i e l d i s i n d i r e c t i o n - \frac{(\hat{i} + \hat{j})}{\sqrt{2}}$
$\begin{array}{l} D i r e c t i o n o f e l e c t r o m a g n e t i c w a v e i s i n (- z) d i r e c t i o n \\ ∴ \hat{C} = - \hat{k} i s w a v e d i r e c t i o n . \\ L e t \hat{B} i s t h e u n i t v e c t o r a l o n g t h e d i r e c t i o n o f m a g n e t i c f i e l d . \\ \hat{B} = \hat{C} \times \hat{E} = - \hat{k} \times - \frac{(\hat{i} + \hat{j})}{\sqrt{2}} = \frac{(- \hat{i} + \hat{j})}{\sqrt{2}} \end{array}$
Now, Force due to magnetic field is in direction $q (\vec{v} \times \vec{B})$ and $\vec{v} | | \hat{k}$
$F o r c e d u e t o m a g n e t i c f i e l d i s \hat{F} = \hat{k} \times (\frac{- \hat{i} + \hat{j}}{\sqrt{2}}) = (\frac{- \hat{i} - \hat{j}}{\sqrt{2}}) = - \frac{(\hat{i} + \hat{j})}{\sqrt{2}}$
$W e c a n s e e t h a t d i r e c t i o n o f b o t h e l e c t r i c a n d m a g n e t i c f o r c e i s s a m e .$
$∴$ Net force is antiparallel to $\frac{(\hat{i} + \hat{j})}{\sqrt{2}}$

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