Modern Physics Ques 310
- The electric field of a plane electromagnetic wave is given by
$ \mathbf{E}=E _0 \hat{\mathbf{i}} \cos (k z) \cos (\omega t) $
The corresponding magnetic field $\mathbf{B}$ is then given by
(Main 2019, 10 April I)
(a) $\mathbf{B}=\frac{E _0}{c} \hat{\mathbf{j}} \sin (k z) \sin (\omega t)$
(b) $\mathbf{B}=\frac{E _0}{c} \hat{\mathbf{j}} \sin (k z) \cos (\omega t)$
(c) $\mathbf{B}=\frac{E _0}{c} \hat{\mathbf{k}} \sin (k z) \cos (\omega t)$
(d) $\mathbf{B}=\frac{E _0}{c} \hat{\mathbf{j}} \cos (k z) \sin (\omega t)$
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Answer:
Correct Answer: 310.(a)
Solution:
Formula:
Relation Between The Magnetic Field Vector And The Electric Field Vector:
- Key Idea: For an electromagnetic wave, its electric field vector (E) and magnetic field vector (B) is mutually perpendicular to each other and also to its direction of propagation.
We know that, $\mathbf{E} \times \mathbf{B}$ represents direction of propagation of an electromagnetic wave
$ \Rightarrow \quad(\mathbf{E} \times \mathbf{B}) || v $
$\therefore$ From the given electric field, we can state that direction of propagation is along $Z$-axis and direction of $\mathbf{E}$ is along $X$-axis.
Thus, from the above discussion, direction of $\mathbf{B}$ must be $Y$-axis.
From Maxwell’s equation,
$\nabla \times \mathbf{E} =-\frac{\partial \mathbf{B}}{\partial t} $
$\text { Here, } \frac{\partial \mathbf{E}}{\partial Z} =-\frac{\partial B}{\partial t} \quad …….(i) $
$\text { and } B _0 =E _0 / C \quad …….(ii)$
$\text { Given, } \mathbf{E} =E _0 \hat{\mathbf{i}} \cos k z \cos \omega t $
$\Rightarrow \quad \frac{-\partial \mathbf{E}}{\partial Z} =k E _0 \sin k z \cos \omega t$
$\therefore$ Using Eq. (i), we get
$ \frac{\partial \mathbf{B}}{\partial t}=k E _0 \sin k z \cos \omega t $
Integrating both sides of the above equation w.r.t. $t$, we get
$ \begin{aligned} & \Rightarrow \quad \mathbf{B}=\frac{k}{\omega} E _0 \sin k z \sin \omega t=\frac{E _0}{C} \sin k z \sin \omega t \\ & \Rightarrow \quad \mathbf{B}=\frac{E _0}{C} \sin (k z) \sin (\omega t) \hat{\mathbf{j}} \end{aligned} $
BETA
