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GATE 2026 PH – Question 11

Electromagnetic theory · electromagnetic waves in free space, non-conducting and conducting media · 1 mark · Multiple choice

In free space, an electromagnetic wave is travelling whose wavevector is $\mathbf k=10(\hat x+\sqrt3\hat y)\ \mathrm{m^{-1}}$. The electric field component of this electromagnetic wave is given by $\mathbf E(\mathbf r,t)=\hat z\,600\cos(\mathbf k\cdot\mathbf r-\omega t)\ \mathrm{V\,m^{-1}}$. The speed of light in free space is $c=3.0\times10^8\ \mathrm{m\,s^{-1}}$. The corresponding magnetic field $\mathbf B(\mathbf r,t)$ is

  1. $2\times10^{-6}(\sqrt3\hat x-\hat y)\cos(\mathbf k\cdot\mathbf r-\omega t)\ \mathrm{V\,m^{-2}s}$
  2. $10^{-6}(\sqrt3\hat x-\hat y)\cos(\mathbf k\cdot\mathbf r-\omega t)\ \mathrm{V\,m^{-2}s}$
  3. $2\times10^{-5}(-\sqrt3\hat x+\hat y)\cos(\mathbf k\cdot\mathbf r-\omega t)\ \mathrm{V\,m^{-2}s}$
  4. $10^{-5}(\sqrt3\hat x-\hat y)\cos(\mathbf k\cdot\mathbf r-\omega t)\ \mathrm{V\,m^{-2}s}$

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Correct answer: (B) $10^{-6}(\sqrt3\hat x-\hat y)\cos(\mathbf k\cdot\mathbf r-\omega t)\ \mathrm{V\,m^{-2}s}$

Explanation

**Plane electromagnetic wave in free space.** The magnetic field is perpendicular to both the propagation direction and the electric field:
$$\mathbf B=\frac{\hat{\mathbf k}\times\mathbf E}{c},$$
with the same phase factor $\cos(\mathbf k\cdot\mathbf r-\omega t)$.

**Unit vector along $\mathbf k$.** $\mathbf k=10(\hat x+\sqrt3\hat y)$ has magnitude $10\sqrt{1+3}=20$ m⁻¹, so
$$\hat{\mathbf k}=\frac{\hat x+\sqrt3\hat y}{2}.$$

**Cross product** with $\mathbf E=600\,\hat z\cos(\dots)$, using $\hat x\times\hat z=-\hat y$ and $\hat y\times\hat z=\hat x$:
$$\hat{\mathbf k}\times\hat z=\frac{-\hat y+\sqrt3\hat x}{2}=\frac{\sqrt3\hat x-\hat y}{2}.$$

**Magnitude:**
$$\mathbf B=\frac{600}{3\times10^8}\cdot\frac{\sqrt3\hat x-\hat y}{2}\cos(\dots)=10^{-6}\left(\sqrt3\hat x-\hat y\right)\cos(\mathbf k\cdot\mathbf r-\omega t)\quad(\text{option B}).$$