Modern Physics Ques 269
- The electric field of a plane polarised electromagnetic wave in free space at time $t=0$ is given by an expression.
$ \mathbf{E}(x, y)=10 \hat{\mathbf{j}} \cos [(6 x+8 z)] $
The magnetic field $\mathbf{B}(x, z, t)$ is given by (where, $c$ is the velocity of light)
(Main 2019, 10 Jan II)
(a) $\frac{1}{c}(6 \hat{\mathbf{k}}-8 \hat{\mathbf{i}}) \cos [(6 x+8 z+10 c t)]$
(b) $\frac{1}{c}(6 \hat{\mathbf{k}}-8 \hat{\mathbf{i}}) \cos [(6 x+8 z-10 c t)]$
(c) $\frac{1}{c}(6 \hat{\mathbf{k}}+8 \hat{\mathbf{i}}) \cos [(6 x-8 z+10 c t)]$
(d) $\frac{1}{c}(6 \hat{\mathbf{k}}+8 \hat{\mathbf{i}}) \cos [(6 x+8 z-10 c t)]$
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Answer:
Correct Answer: 269.(b)
Solution:
Formula:
Relation Between The Magnetic Field Vector And The Electric Field Vector:
- We are given electric field as
$ \mathbf{E}=10 \hat{\mathbf{j}} \cos (6 x+8 z) \quad …….(i) $
where, phase angle is independent of time, i.e., phase angle at $t=0$ is $\phi=6 x+8 z$.
Phase angle for $\mathbf{B}$ will also be $6 x+8 z$ because for an electromagnetic wave $\mathbf{E}$ and $\mathbf{B}$ oscillate in same phase.
Thus, direction of wave propagation
$ =\frac{6 \hat{\mathbf{i}}+8 \hat{\mathbf{k}}}{\sqrt{6^{2}+8^{2}}}=\frac{6 \hat{\mathbf{i}}+8 \hat{\mathbf{k}}}{10} \quad …….(ii) $
Let magnetic field vector,
$\mathbf{B}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+d \hat{\mathbf{k}}$, then direction of wave propagation is given by
$ \begin{aligned} \frac{\mathbf{E} \times \mathbf{B}}{\left|\mathbf{E}^{\prime}\right||\mathbf{B}|} & =\frac{10 \hat{\mathbf{j}} \times(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+d \hat{\mathbf{k}})}{10 \times\left(a^{2}+b^{2}+d^{2}\right)^{1 / 2}} \\ & =\frac{-10 a \hat{\mathbf{k}}+10 d \hat{\mathbf{i}}}{10\left(a^{2}+b^{2}+d^{2}\right)^{1 / 2}} \quad …….(iii) \end{aligned} $
As, $\quad|\mathbf{B}|=\frac{|\mathbf{E}|}{c}=\frac{10}{c}$
We get, $\quad|\mathbf{B}|=\sqrt{a^{2}+b^{2}+d^{2}}=10 / c$
By putting this value in Eqs. (iii) and (ii), we get direction of propagation
$\Rightarrow \quad \frac{c 10(d \hat{\mathbf{i}}-a \hat{\mathbf{k}})}{10 \times 10}=\frac{6 \hat{\mathbf{i}} \times 8 \hat{\mathbf{k}}}{10} $
$\Rightarrow \quad d=6 / c \text { and } a=-8 / c$
Hence, $\mathbf{B}=\frac{6}{c} \hat{\mathbf{k}}-\frac{8}{c} \hat{\mathbf{i}}=\frac{1}{c}(6 \hat{\mathbf{k}}-8 \hat{\mathbf{i}})$
As the general equation of magnetic field of an EM wave propagating in positive $y$-direction is given as,
$\left.B=B _0 \cos (R y-\omega t)\right)$ $\therefore \mathbf{B}=\frac{1}{c}(6 \hat{\mathbf{k}}-8 \hat{\mathbf{i}}) \cos (6 x+8 z-10 c t)$
Alternate method
Given, electric field is $E(x, y)$, i.e. electric field is in $x y$-plane which is given as
$\because \mathbf{E}=10 \hat{\mathbf{j}} \cos (6 x+8 z)$
Since, the magnetic field given is $\mathbf{B}(x, z, t)$, this means $\mathbf{B}$ is in $x z$-plane.
$\therefore$ Propagation of wave is in $y$-direction.
$[\because$ for an electromagnetic wave, $\mathbf{E} \perp \mathbf{B} \perp$ propagation direction]
As Poynting vector suggests that $\mathbf{E} \times \mathbf{B}$ is parallel to $(6 \hat{\mathbf{i}}+8 \hat{\mathbf{k}})$.
$ \begin{aligned} & \text { Let } \quad \mathbf{B}=(x \hat{\mathbf{i}}+z \hat{\mathbf{k}}) \text {, } \\ & \text { then } \quad \mathbf{E} \times \mathbf{B}=\hat{\mathbf{j}} \times(x \hat{\mathbf{i}}+z \hat{\mathbf{k}})=6 \hat{\mathbf{i}}+8 \hat{\mathbf{k}} \\ & \text { or } \quad-x \hat{\mathbf{k}}+z \hat{\mathbf{i}}=6 \hat{\mathbf{i}}+8 \hat{\mathbf{k}} \\ & \text { or } \quad x=-8 \text { and } z=6 \\ & \therefore \quad \mathbf{B}=\frac{1}{c}(6 \hat{\mathbf{k}}-8 \hat{\mathbf{i}}) \cos (6 x+8 z-10 c t) \quad [\because \frac{|\mathbf{E}|}{|\mathbf{B}|}=c] \end{aligned} $
BETA
