GATE 2020 EC Q.45: solution and answer

Official answer: 1Source: Official answer key, GATE 2020 EC (IIT Delhi)

1. Write the phase of a uniform plane wave as $\omega t-\mathbf{\beta }\cdot \mathbf{r}$, where $\mathbf{\beta }$ is the propagation (wave) vector and $\mathbf{r}=x{\hat{a}}_x+y{\hat{a}}_y+z{\hat{a}}_z$.

2. Compare this with the given phase: $\omega t+3x-y-z=\omega t-(-3x+y+z)$, so $\mathbf{\beta }=-3{\hat{a}}_x+{\hat{a}}_y+{\hat{a}}_z$ (in $\text{rad/m}$). The wave travels along $\frac{-3{\hat{a}}_x+{\hat{a}}_y+{\hat{a}}_z}{\sqrt{11}}$.

3. In vacuum there are no sources, and Gauss's law for magnetism gives $\nabla \cdot \mathbf{H}=0$. For a plane wave $\mathbf{H}={\mathbf{H}}_0\cos (\omega t-\mathbf{\beta }\cdot \mathbf{r})$ with a constant amplitude vector ${\mathbf{H}}_0$, this becomes $\nabla \cdot \mathbf{H}=(\mathbf{\beta }\cdot {\mathbf{H}}_0)\sin (\omega t-\mathbf{\beta }\cdot \mathbf{r})=0$ for every $\mathbf{r}$ and $t$. So $\mathbf{\beta }\cdot {\mathbf{H}}_0=0$, which means the field is transverse to the direction of propagation.

4. Apply the transversality condition with ${\mathbf{H}}_0={\hat{a}}_x+2{\hat{a}}_y+b{\hat{a}}_z$: $$({\hat{a}}_x+2{\hat{a}}_y+b{\hat{a}}_z)\cdot (-3{\hat{a}}_x+{\hat{a}}_y+{\hat{a}}_z)=-3+2+b=0$$

5. Solve: $b=1$.