Modern Physics 7 Question 6
6. 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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Solution:
- 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,
$$ \begin{array}{rlrl} & \nabla \times \mathbf{E} & =-\frac{\partial \mathbf{B}}{\partial t} \\ & \text { Here, } & \frac{\partial \mathbf{E}}{\partial Z} & =-\frac{\partial B}{\partial t} \\ & \text { and } & B _0 & =E _0 / C \\ & \text { Given, } & \mathbf{E} & =E _0 \hat{\mathbf{i}} \cos k z \cos \omega t \\ \Rightarrow & \frac{-\partial \mathbf{E}}{\partial Z} & =k E _0 \sin k z \cos \omega t \end{array} $$
$\therefore$ 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
$$ \begin{aligned} & \Rightarrow \quad \mathbf{B}=\frac{k}{\omega} E _0 \sin k z \sin \omega t=\frac{E _0}{C} \sin k z \sin \omega t \\ & \Rightarrow \quad \mathbf{B}=\frac{E _0}{C} \sin (k z) \sin (\omega t) \hat{\mathbf{j}} \end{aligned} $$
