Question

The electric field of a radio wave is given by $\overrightarrow{E}=Esin(Kz-\omega t)(\hat{i}+\hat{j})$ . Give a unit vector
in the direction of the magnetic field at a place and time where sin$(kz-\omega t)$ is positive.

• A. $\left({\hat{i}}^{+}\hat{k}\right)/\sqrt{2}$
• B. $(-\hat{i}-\hat{k}/\sqrt{2}$
• C. $-\hat{i}+\hat{j}/\sqrt{2}$
• D. $\hat{i}-\hat{j}/\sqrt{2}$

Answer 1

The correct option is C $-\hat{i}+\hat{j}/\sqrt{2}$

$\overrightarrow{E}=E\sin (Kz-\omega t)(\hat{i}+\hat{j})$

$\hat{E}=\frac{\hat{i}+\hat{j}}{\sqrt{1^2+1^2}}$

$\hat{E}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

unit vector in the direction of the magnetic field where $\sin (kz-\omega t)$ is positive will be in same plane and perpendicular to $\hat{E}$

unit vector $=\frac{1}{\sqrt{x_1^2+y_1^2}}(x_1\hat{i}+y_1\hat{j})$

Both vector are perpendicular to each other hence dot product will be zero.

$\frac{1}{\sqrt{2}}\ast \frac{x_1}{\sqrt{x_1^2+y_1^2}}+\frac{1}{\sqrt{2}}\ast \frac{y_1}{\sqrt{x_1^2+y_1^2}}=0$

$x_1=-y_1$

or $y_1=-x_1$

unit vector $=\frac{1}{\sqrt{x_1^2+{-x_1}^2}}(x_1\hat{i}-x_1\hat{j})or\frac{1}{\sqrt{y_1^2+{-y_1}^2}}(-y_1\hat{i}+y_1\hat{j})=\frac{1}{\sqrt{2x_1^2}}\left(x_1\right(\hat{i}-\hat{j}\left)\right)or=\frac{1}{\sqrt{2y_1^2}}\left(y_1\right(\widehat{-i}+\hat{j}\left)\right)=\frac{1}{\sqrt{2}}(\hat{i}-\hat{j})or=\frac{1}{\sqrt{2}}(-\hat{i}+\hat{j})$

But for $\frac{\hat{i}-\hat{j}}{\sqrt{2}},\sin =-ve$ so this will not possible

Hence $\frac{-\hat{i}+\hat{j}}{\sqrt{2}}$ is correct answer.