At a given point in space, the electric field associated with an electromagnetic wave is given by $\vec{E} = (2\hat{i} – 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]}$. At the same point, which one among the following is the correct value of the unit vector $(\hat{B})$ of the magnetic field associated with this electromagnetic wave ?
$-hat{k}$
$1.5hat{i} + 2hat{j}$
$1.5hat{i} - 2hat{j}$
$3hat{i} - 4hat{j}$
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC Geoscientist – 2024
The given electric field is $\vec{E} = (2\hat{i} – 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]}$. The term $k_0(3x+4y)$ represents $\vec{k} \cdot \vec{r}$. This indicates that the wave vector $\vec{k}$ is in the direction $3\hat{i} + 4\hat{j}$.
The direction of $\vec{E}$ is given by the amplitude vector $(2\hat{i} – 1.5\hat{j})$. Let this be $\vec{E}_0$.
We need to find a unit vector $\hat{B}$ such that $\vec{E}_0 \times \hat{B}$ is in the direction of $3\hat{i} + 4\hat{j}$.
Let’s check option A: $\hat{B} = -\hat{k}$.
$\vec{E}_0 \times \hat{B} = (2\hat{i} – 1.5\hat{j}) \times (-\hat{k}) = (2\hat{i} \times -\hat{k}) + (-1.5\hat{j} \times -\hat{k}) = 2\hat{j} + 1.5\hat{i} = 1.5\hat{i} + 2\hat{j}$.
The direction of this vector $1.5\hat{i} + 2\hat{j}$ is indeed the same as $3\hat{i} + 4\hat{j}$ (since $1.5\hat{i} + 2\hat{j} = \frac{1}{2}(3\hat{i} + 4\hat{j}) \times 2 = 3\hat{i} + 4\hat{j}$).
Thus, with $\hat{B} = -\hat{k}$, $\vec{E}_0 \times \hat{B}$ is in the correct direction of wave propagation.
Also, check perpendicularity: $\vec{E}_0 \cdot \hat{B} = (2\hat{i} – 1.5\hat{j}) \cdot (-\hat{k}) = 0$ and $(3\hat{i} + 4\hat{j}) \cdot (-\hat{k}) = 0$, confirming $\vec{E}$ and $\vec{k}$ are perpendicular to $\vec{B}$.
Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $\hat{B}$.