Two plane electromagnetic waves are moving in vacuum in whose electric field vectors are given byE1-E0-coskx-wt andE2-E0k-cosky-wtAt t-0- a charge q is at origin with velocity v-0-8c- c is the speed of light in vacuum-The instantaneous force on this charge is- all data are in SI units

Step 1: Given dataGiven that the electric field vectors are,E_1=E_0ĵcos(kx wt) E_2=E_0k̂cos(ky wt)velocity,v=0.8cĵAt t=0, charge is at origin, i.e., x=y=0Step 2 ...

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Standard XII

Physics

Magnetic Flux and Faraday's Law

Two plane ele...

Question

Two plane electromagnetic waves are moving in vacuum in whose electric field vectors are given by
$\vec{E_{1}} = E_{0} \hat{j} \cos (k x - w t)$ and
$\vec{E_{2}} = E_{0} \hat{k} \cos (k y - w t)$
At $t = 0$ , a charge $q$ is at origin with velocity $\vec{v} = 0.8 c \hat{j}$ ( $c$ is the speed of light in vacuum).The instantaneous force on this charge is? (all data are in SI units)

$q E_{0} (0.4 \hat{i} - 3 \hat{j} + 0.8 \hat{k})$

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$q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$

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$q E_{0} (0.8 \hat{i} - \hat{j} + 0.4 \hat{k})$

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$q E_{0} (- 0.8 \hat{i} + \hat{j} + \hat{k})$

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Solution

The correct option is B
$q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$

Step 1: Given data
Given that the electric field vectors are,
$\vec{E_{1}} = E_{0} \hat{j} \cos (k x - w t)$
$\vec{E_{2}} = E_{0} \hat{k} \cos (k y - w t)$
velocity, $\vec{v} = 0.8 c \hat{j}$
At $t = 0$ , charge is at origin, i.e., $x = y = 0$
Step 2: Finding the magnetic field
From the equation we can see that the variation of the vectors $\vec{E_{1}}$ and $\vec{E_{2}}$ are along $x$ and $y$ respectively.
Since $\vec{E} \times \vec{B}$ gives the direction of propagation and $\frac{E_{0}}{B_{0}} = c$
Magnetic field vectors are given by,
$\vec{B_{1}} = \frac{E_{0}}{c} \hat{k} \cos (k x - w t)$
$\vec{B_{2}} = \frac{E_{0}}{c} \hat{i} \cos (k y - w t)$
Step 3: Finding the force
$\vec{F} = q \vec{E} + q (\vec{v} \times \vec{B}) \Rightarrow \vec{F} = q (\vec{E_{1}} + \vec{E_{2}}) + q (\vec{v} \times (\vec{B_{1}} + \vec{B_{2}}))$
By substituting the values for $\vec{E_{1}}, \vec{E_{2}}, \vec{B_{1}}$ and $\vec{B_{2}}$ , we get
$\vec{F} = q (E_{0} \hat{j} \cos (k x - w t) + E_{0} \hat{k} \cos (k y - w t)) + q (\vec{v} \times (\frac{E_{0}}{c} \hat{i} \cos (k y - w t) + \frac{E_{0}}{c} \hat{k} \cos (k x - w t)))$
Finding $\vec{v} \times \vec{B}$ by cross multiplication,
$\vec{v} \times \vec{B} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0.8 c & 0 \\ \frac{E_{0}}{c} \cos (k y - w t) & u & \frac{E_{0}}{c} \cos (k x - w t) \end{matrix}| \vec{v} \times \vec{B} = \hat{i} (0.8 E_{0} \cos (k x - w t) - 0) - \hat{j} (0) + \hat{k} (0 - 0.8 E_{0} \cos (k y - w t) \vec{v} \times \vec{B} = E_{0} (0.8 \cos (k x - w t) \hat{i} - 0.8 E_{0} \cos (k y - w t) \hat{k})$
So, the force becomes,
$\vec{F} = q E_{0} (\cos (k x - w t) \hat{j} + \cos (k y - w t) \hat{k}) + q E_{0} (0.8 \cos (k x - w t) \hat{i} - 0.8 E_{0} \cos (k y - w t) \hat{k}) \vec{F} = q E_{0} [\cos (k x - w t) \hat{j} + \cos (k y - w t) \hat{k} + 0.8 \cos (k x - w t) \hat{i} - 0.8 E_{0} \cos (k y - w t) \hat{k}] \vec{F} = q E_{0} [\cos (k x - w t) \hat{j} + 0.2 \cos (k y - w t) \hat{k} + 0.8 \cos (k x - w t) \hat{i}]$
We are asked to find the instantaneous force, so when $t = 0$ and $x = y = 0$ , we get
$\vec{F} = q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$
Therefore, the instantaneous force is $\vec{F} = q E_{0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k})$
Hence, option B is the correct answer.

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