Question

In free space, an electromagnetic wave is travelling whose wavevector is \(\vec{k} = 10(\hat{x} + \sqrt{3}\hat{y})\) m\(^{-1}\). The electric field component of this electromagnetic wave is given by \(\vec{E}(\vec{r},t) = \hat{z}\, 600\cos(\vec{k}\cdot\vec{r} - \omega t)\) V.m\(^{-1}\). The speed of light in free space is \(c = 3.0 \times 10^{8}\) m.s\(^{-1}\). The corresponding magnetic field \(\vec{B}(\vec{r},t)\) is

• A. \(\vec{B}(\vec{r},t) = 2\times10^{-6}(\sqrt{3}\hat{x} - \hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)\) V.m\(^{-2}\).s
• B. \(\vec{B}(\vec{r},t) = 10^{-6}(\sqrt{3}\hat{x} - \hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)\) V.m\(^{-2}\).s
• C. \(\vec{B}(\vec{r},t) = 2\times10^{-5}(-\sqrt{3}\hat{x} + \hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)\) V.m\(^{-2}\).s
• D. \(\vec{B}(\vec{r},t) = 10^{-5}(\sqrt{3}\hat{x} - \hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)\) V.m\(^{-2}\).s

Hint:

Use \(\vec{B} = \frac{1}{c}\hat{k}\times\vec{E}\) with \(\hat{k}\) the unit vector along \(\vec{k} = 10(\hat{x}+\sqrt3\hat{y})\), whose magnitude is 20 m\(^{-1}\).

Answer 1

The Correct Option is B

Step 1: Set up Faraday's law for a plane wave.
Instead of quoting the ready-made formula for $\vec{B}$, get it straight from Maxwell's equation $\nabla\times\vec{E} = -\partial\vec{B}/\partial t$. Write the phase as $\varphi = \vec{k}\cdot\vec{r}-\omega t$, so $\vec{E} = \vec{E}_0\cos\varphi$ with the constant vector $\vec{E}_0 = 600\hat{z}$.

Step 2: Take the curl of $\vec{E}$.
For a constant vector $\vec{E}_0$, $\nabla\times(\vec{E}_0\cos\varphi) = (\nabla\cos\varphi)\times\vec{E}_0 = -\sin\varphi\,(\vec{k}\times\vec{E}_0)$, because $\nabla\varphi = \vec{k}$. Faraday's law then gives $\partial\vec{B}/\partial t = \sin\varphi\,(\vec{k}\times\vec{E}_0)$.

Step 3: Integrate over time.
Try $\vec{B} = \vec{B}_0\cos\varphi$. Then $\partial\vec{B}/\partial t = \vec{B}_0\,\omega\sin\varphi$, since $\partial\varphi/\partial t = -\omega$ and the derivative of cosine brings a minus sign that cancels it. Matching with Step 2:
\[ \vec{B}_0\,\omega = \vec{k}\times\vec{E}_0 \implies \vec{B}_0 = \frac{\vec{k}\times\vec{E}_0}{\omega} \]

Step 4: Put in the numbers.
$\vec{k}\times\vec{E}_0 = 10(\hat{x}+\sqrt3\hat{y})\times 600\hat{z} = 6000\left[(\hat{x}\times\hat{z}) + \sqrt3(\hat{y}\times\hat{z})\right] = 6000(\sqrt3\hat{x}-\hat{y})$, using $\hat{x}\times\hat{z}=-\hat{y}$, $\hat{y}\times\hat{z}=\hat{x}$.
The angular frequency comes from the dispersion relation $\omega = c|\vec{k}|$. Here $|\vec{k}| = 10\sqrt{1+3} = 20$ m$^{-1}$, so $\omega = (3.0\times10^8)(20) = 6.0\times10^9$ rad.s$^{-1}$.
\[ \vec{B}_0 = \frac{6000(\sqrt3\hat{x}-\hat{y})}{6.0\times10^9} = 10^{-6}(\sqrt3\hat{x}-\hat{y}) \]

Step 5: Check against the options.
This gives $\vec{B}(\vec{r},t) = 10^{-6}(\sqrt3\hat{x}-\hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)$ V.m$^{-2}$.s, exactly option (B). The other three options scale the magnitude wrongly or flip the direction, all traceable to slipping on the factor $|\vec{k}|/\omega = 1/c$.

Final Answer:
Starting from Faraday's law directly gives the same field as the ready-made formula. \[ \boxed{\vec{B}(\vec{r},t) = 10^{-6}(\sqrt3\hat{x}-\hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)\ \text{V.m}^{-2}\text{.s}} \]