Q150 / 155 · WBJEE 2021 · Physics
For a plane eletromagnetic wave, the electric field is given by $\overrightarrow{\mathrm{E}}=90 \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \mathrm{k} . \mathrm{v} / \mathrm{m} .$ The corresponding magnetic field $\vec{B}$ will be
- A$\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{i} \mathrm{~T}$
- Correct answerB$\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$
- C$\vec{B}=27 \times 10^{9} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$
- D$\overrightarrow{\mathrm{B}}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{k}} \mathrm{T}$
Solution
$E=90 \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{k} v / m$
Direction of propagation is $-\hat{i}$
$\& E_{0}=90$
$\frac{\mathrm{E}_{0}}{\mathrm{~B}_{0}}=\mathrm{C} \Rightarrow \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}=\frac{90}{3 \times 10^{8}} \mathrm{~T}=3 \times 10^{-7} \mathrm{~T}$
$\overrightarrow{\mathrm{E}}$ is along $\hat{\mathrm{k}} \& \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$ should be along $-\hat{\mathrm{i}} \Rightarrow \overrightarrow{\mathrm{B}}$ is should be along $\hat{\mathrm{j}}$ Hence equation of magnetic field is
$\mathrm{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}} \mathrm{T}$
Direction of propagation is $-\hat{i}$
$\& E_{0}=90$
$\frac{\mathrm{E}_{0}}{\mathrm{~B}_{0}}=\mathrm{C} \Rightarrow \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}=\frac{90}{3 \times 10^{8}} \mathrm{~T}=3 \times 10^{-7} \mathrm{~T}$
$\overrightarrow{\mathrm{E}}$ is along $\hat{\mathrm{k}} \& \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$ should be along $-\hat{\mathrm{i}} \Rightarrow \overrightarrow{\mathrm{B}}$ is should be along $\hat{\mathrm{j}}$ Hence equation of magnetic field is
$\mathrm{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}} \mathrm{T}$
Hence, the correct answer is option: 'B'.