Q154 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Physics
The magnetic field in a plane electromagnetic wave is given as
$\mathbf{B}=\left(3 \times 10^{-7} \mathrm{~T}\right) \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{j}$
The electric field of this wave is given as
- A$90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{i} \mathrm{Vm}^{-1}$
- Correct answerB$90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1}$
- C$45 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{i} \mathrm{Vm}^{-1}$
- D$45 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1}$
Solution
Given, Magnetic field in a plane EM wave,
$B=\left(3 \times 10^{-7} \mathrm{~T}\right) \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \mathrm{T}$
Here, $B_0=3 \times 10^{-7} \mathrm{~T}$
Amplitude of electric field,
$\begin{aligned} E_0 & =B_0 c \\ & =3 \times 10^{-7} \times 3 \times 10^8\end{aligned}$
$\Rightarrow \quad E_0=90 \mathrm{Vm}^{-1}$
From the given equation of $B$, it is clear that $E M$ wave is travelling along negative direction of
$X$ - axis i.e.
$\therefore$ If $\hat{\mathbf{n}}$ be the unit vector along the direction of electric field, then
$\hat{\mathbf{n}} \times \hat{j}=-\hat{i}$
Clearly, $\hat{\mathbf{n}}=\hat{k}$
Hence, $E=E_0 \sin (k x+\omega t) \hat{k}$
$=90 \sin \left(3 \times 10^4 x+9 \times 10^{12} t\right) \hat{k} \mathrm{Vm}^{-1}$
Hence, the correct answer is option: 'B'.