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: Understanding the Concept:
An electromagnetic plane wave in free space has its electric field, magnetic field, and direction of travel mutually perpendicular. Once we know the wave's direction (given by the wavevector \(\vec{k}\)) and the electric field \(\vec{E}\), the magnetic field is fixed by the relation \(\vec{B} = \frac{1}{c}\hat{k}\times \vec{E}\), where \(\hat{k}\) is the unit vector along the direction of propagation.

Step 2: Key Formula or Approach:
1. Find the unit vector \(\hat{k} = \vec{k}/|\vec{k}|\).
2. Use \(\vec{B}(\vec{r},t) = \frac{1}{c}\,\hat{k}\times \vec{E}(\vec{r},t)\), since \(\vec{B}\) carries the same phase \(\cos(\vec{k}\cdot\vec{r}-\omega t)\) as \(\vec{E}\) for a wave travelling in a single direction.

Step 3: Detailed Explanation:
The wavevector is \(\vec{k} = 10(\hat{x}+\sqrt{3}\hat{y})\) m\(^{-1}\), so its magnitude is \(|\vec{k}| = 10\sqrt{1^2+(\sqrt3)^2} = 10\sqrt{4} = 20\) m\(^{-1}\).
The unit vector along propagation is \(\hat{k} = \frac{10(\hat{x}+\sqrt{3}\hat{y})}{20} = \frac{1}{2}(\hat{x}+\sqrt{3}\hat{y})\).
The electric field amplitude vector is \(600\hat{z}\) V.m\(^{-1}\). Cross it with \(\hat{k}\):
\[ \hat{k}\times(600\hat{z}) = \frac{600}{2}\left[(\hat{x}\times\hat{z}) + \sqrt{3}(\hat{y}\times\hat{z})\right] \]
Using \(\hat{x}\times\hat{z} = -\hat{y}\) and \(\hat{y}\times\hat{z} = \hat{x}\):
\[ \hat{k}\times(600\hat{z}) = 300\left[-\hat{y}+\sqrt{3}\hat{x}\right] = 300(\sqrt{3}\hat{x}-\hat{y}) \]
Divide by \(c = 3.0\times10^{8}\) m.s\(^{-1}\):
\[ \vec{B}_0 = \frac{300(\sqrt{3}\hat{x}-\hat{y})}{3.0\times10^{8}} = 10^{-6}(\sqrt{3}\hat{x}-\hat{y}) \]
Since \(\vec{B}\) tracks the same phase as \(\vec{E}\):
\[ \vec{B}(\vec{r},t) = 10^{-6}(\sqrt{3}\hat{x}-\hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t) \text{ V.m}^{-2}\text{.s} \]

Step 4: Why the other options are wrong.
Option (A) has the right direction \((\sqrt3\hat{x}-\hat{y})\) but a magnitude twice too large. Option (C) reverses the direction, using \(-\sqrt3\hat{x}+\hat{y}\) instead, and is off by a further factor of 10 in size, the kind of mistake you get from crossing the vectors in the wrong order and misplacing a power of ten. Option (D) has the correct direction but is 10 times too large.

Final Answer:
The magnetic field is \(10^{-6}(\sqrt{3}\hat{x}-\hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)\) V.m\(^{-2}\).s, which is option (B). \[ \boxed{\vec{B}(\vec{r},t) = 10^{-6}(\sqrt{3}\hat{x}-\hat{y})\cos(\vec{k}\cdot\vec{r}-\omega t)\ \text{V.m}^{-2}\text{.s}} \]