A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a pa

Q: Why didn't we use the frequency of 500 MHz?

In this specific problem, we only needed to find the instantaneous electric field vector corresponding to a given magnetic field vector. The relationship between their amplitudes ($E_0 = cB_0$) and directions ($\hat{\mathbf{v}} = \hat{\mathbf{E}} \times \hat{\mathbf{B}}$) is independent of the wave's frequency. The frequency was provided as a distractor.

Q: Can I use $\hat{\mathbf{B}} \times \hat{\mathbf{E}}$ for the direction of propagation?

No, the cross product is not commutative. $\hat{\mathbf{B}} \times \hat{\mathbf{E}}$ would give you the exact opposite direction of the wave's actual propagation. Always use $\hat{\mathbf{E}} \times \hat{\mathbf{B}}$.

Q: What if the wave was travelling in a medium other than vacuum?

If the wave was in a medium with refractive index $n$, the speed of light $c$ in the formula $E_0 = cB_0$ would be replaced by the velocity of the wave in that medium, $v = c/n$.

Q: How do I quickly remember the cross products of unit vectors?

Draw a circle with $\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}$ in clockwise order. Going clockwise (e.g., $\hat{\mathbf{x}} \times \hat{\mathbf{y}}$) gives the next vector positive ($\hat{\mathbf{z}}$). Going counter-clockwise (e.g., $\hat{\mathbf{x}} \times \hat{\mathbf{z}}$) gives the next vector negative ($-\hat{\mathbf{y}}$).

Q: Is the electric field always exactly in phase with the magnetic field?

Yes, for a plane electromagnetic wave travelling in a vacuum or a linear isotropic dielectric, the electric and magnetic fields oscillate perfectly in phase. When $\mathbf{B}$ reaches its maximum, $\mathbf{E}$ also reaches its maximum.

Question

A plane electromagnetic wave of frequency $500 	ext{ MHz}$ is travelling in vacuum along $y$-direction. At a particular point in space and time, $\mathbf{B} = 8.0 	imes 10^{-8} \hat{\mathbf{z}} 	ext{ T}$. The value of electric field at this point is (speed of light $= 3 	imes 10^8 	ext{ ms}^{-1}$; $\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}$ are unit vectors along $x, y$ and $z$-direction).

Options:

(A) $-24 \hat{\mathbf{x}} 	ext{ V/m}$

(B) $2.6 \hat{\mathbf{x}} 	ext{ V/m}$

(C) $24 \hat{\mathbf{x}} 	ext{ V/m}$

(D) $-2.6 \hat{\mathbf{x}} 	ext{ V/m}$

Sigma Percentile · Accepted Answer

The behavior of electromagnetic waves is one of the most elegant concepts in physics. In a vacuum, these waves consist of oscillating electric and magnetic fields that dance together in perfect harmony.

Imagine you are observing this wave as it travels through space. The problem tells us that the wave is propagating along the $y$-direction.

This gives us our first crucial piece of information: the velocity vector $\hat{\mathbf{v}}$ is simply $\hat{\mathbf{y}}$.

### The Magnitude: A Simple Scaling

An electromagnetic wave is not just a random collection of fields; the electric and magnetic components are intimately linked. Their peak magnitudes are connected by a fundamental constant of the universe: the speed of light, $c$.

The relationship is beautifully simple:

$E_0 = c B_0$

We are given the peak magnetic field $B_0 = 8.0 	imes 10^{-8} 	ext{ T}$ and we know $c = 3 	imes 10^8 	ext{ m/s}$.

Let's substitute these values into our master equation to find the magnitude of the electric field.

$E_0 = (3 	imes 10^8) 	imes (8.0 	imes 10^{-8})$

Notice how the powers of ten perfectly cancel each other out. The $10^8$ and $10^{-8}$ multiply to exactly $1$.

This leaves us with a straightforward multiplication: $3 	imes 8.0$.

$E_0 = 24 	ext{ V/m}$

We now have the magnitude, but an electric field is a vector. We must determine its direction to fully solve the problem.

### The Direction: The Right-Hand Rule

In any plane electromagnetic wave, the electric field $\mathbf{E}$, the magnetic field $\mathbf{B}$, and the direction of propagation $\hat{\mathbf{v}}$ are mutually perpendicular.

They follow a strict geometric rule governed by the cross product. The direction of propagation is always given by crossing the electric field into the magnetic field.

$\hat{\mathbf{v}} = \hat{\mathbf{E}} 	imes \hat{\mathbf{B}}$

We already know that the wave travels along the $y$-axis, so $\hat{\mathbf{v}} = \hat{\mathbf{y}}$. The problem states the magnetic field is along the $z$-axis, so $\hat{\mathbf{B}} = \hat{\mathbf{z}}$.

Let's plug these unit vectors into our cross product relationship.

$\hat{\mathbf{y}} = \hat{\mathbf{E}} 	imes \hat{\mathbf{z}}$

Now, we must use our knowledge of the cyclic properties of unit vectors. We need a vector that, when crossed with $\hat{\mathbf{z}}$, results in positive $\hat{\mathbf{y}}$.

Recall that crossing $\hat{\mathbf{x}}$ with $\hat{\mathbf{z}}$ gives $-\hat{\mathbf{y}}$. Therefore, to get a positive $\hat{\mathbf{y}}$, we must cross $-\hat{\mathbf{x}}$ with $\hat{\mathbf{z}}$.

$(-\hat{\mathbf{x}}) 	imes \hat{\mathbf{z}} = \hat{\mathbf{y}}$

This definitively proves that the electric field must oscillate along the negative $x$-axis. So, $\hat{\mathbf{E}} = -\hat{\mathbf{x}}$.

### Final Calculation and The Trap

We have successfully found both the magnitude and the direction of the electric field. Now, we simply combine them into a single vector expression.

$\mathbf{E} = E_0 \hat{\mathbf{E}}$

Substituting our calculated values, we get the final answer.

**$\mathbf{E} = -24 \hat{\mathbf{x}} 	ext{ V/m}$**

Before we conclude, let's address the elephant in the room: the frequency of $500 	ext{ MHz}$. Why was it given?

This is a classic trap set by examiners. They often provide superfluous information to test your confidence in the core principles.

Because we only needed to relate the instantaneous amplitudes and directions of the fields, the frequency of the wave was entirely irrelevant to our calculation. Always trust your fundamental equations!

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Visualized Solution

$Coordinate System$

$Magnitude Relation$

$Substitution$

$Calculation$

$Direction Relation$

$Vector Substitution$

$Cross Product Logic$

$Final Electric Field$

$The Distractor$

The Sigma Insight: Characteristics of Electromagnetic Waves

The Magnitude

The Direction

Final Calculation and The Trap

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Coordinate System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Magnitude Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vector Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cross Product Logic

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Electric Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Distractor