A plane electromagnetic wave travels in free space along the x-direction. The electric field compone

Q: Why do we use $c$ and not some other velocity?

The problem explicitly states that the electromagnetic wave is traveling in 'free space' (a vacuum). In a vacuum, all electromagnetic waves travel at the speed of light, $c = 3 \times 10^8 \text{ m/s}$.

Q: How do I easily remember the cross product rule for directions?

Use the right-hand rule! Point the fingers of your right hand in the direction of the electric field ($\vec{E}$), then curl them towards the magnetic field ($\vec{B}$). Your thumb will point in the direction of wave propagation ($\vec{v}$).

Q: Can the magnetic field be in the x-direction?

No. Electromagnetic waves are transverse waves. This means both the electric and magnetic fields must oscillate perpendicular to the direction of propagation. Since the wave is moving in the x-direction, neither field can be along the x-axis.

Q: What if the wave was traveling in a medium like glass or water?

In a medium, the wave travels slower than $c$. You would use the velocity in that medium, $v = c/n$ (where $n$ is the refractive index), and the relationship becomes $B = E/v$.

Q: Why is the magnitude of the magnetic field so incredibly small compared to the electric field?

This is a fundamental property of electromagnetic waves. Because $B = E/c$ and $c$ is a massive number ($3 \times 10^8$), the numerical value of $B$ in Teslas will always be much smaller than $E$ in Volts/meter. However, they carry equal amounts of energy in the wave.

Question

A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is $E = 6	ext{ Vm}^{-1}$ along y-direction. Its corresponding magnetic field component, $B$ would be

Options:

(A) $2 	imes 10^{-8}	ext{ T}$ along z - direction

(B) $6 	imes 10^{-8}	ext{ T}$ along x - direction

(C) $6 	imes 10^{-8}	ext{ T}$ along z - direction

(D) $2 	imes 10^{-8}	ext{ T}$ along y - direction

Sigma Percentile · Accepted Answer

### Visualizing the Electromagnetic Wave

Imagine an electromagnetic wave traveling through the vast emptiness of free space. The problem tells us that this wave is moving along the positive x-direction. We can represent this velocity vector as $\vec{v} = c\hat{i}$.

At a specific point in space and time, the electric field component $\vec{E}$ is pointing along the y-direction with a magnitude of $6 	ext{ V/m}$. So, we can write this as $\vec{E} = 6\hat{j} 	ext{ V/m}$. Our goal is to find the corresponding magnetic field component, $\vec{B}$, both in magnitude and direction.

### Calculating the Magnitude

In an electromagnetic wave propagating through a vacuum, the magnitudes of the electric field $E$ and the magnetic field $B$ are intimately connected by the speed of light $c$. The relationship is beautifully simple:

$$ \frac{E}{B} = c $$

We can rearrange this to solve for the magnetic field magnitude:

$$ B = \frac{E}{c} $$

Now, let's substitute the values we know. The electric field $E$ is $6 	ext{ V/m}$, and the speed of light $c$ is a universal constant, $3 	imes 10^8 	ext{ m/s}$.

$$ B = \frac{6}{3 	imes 10^8} $$

Dividing $6$ by $3$ gives us $2$, leaving us with:

$$ B = 2 	imes 10^{-8} 	ext{ T} $$

### Decoding the Direction

Finding the magnitude was just half the battle. Now, we need to determine the direction of the magnetic field. This is where the geometry of electromagnetic waves comes into play. The electric field $\vec{E}$, the magnetic field $\vec{B}$, and the direction of propagation $\vec{v}$ are all mutually perpendicular.

More specifically, the direction of wave propagation is always given by the cross product of the electric and magnetic field unit vectors:

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

We know the wave is traveling in the x-direction ($\hat{i}$) and the electric field is in the y-direction ($\hat{j}$). Let's plug these into our cross product equation:

$$ \hat{i} = \hat{j} 	imes \hat{B} $$

Now, we must ask ourselves: what unit vector, when crossed with $\hat{j}$ from the right, yields $\hat{i}$? If we recall our standard right-hand rule for cross products, we know that $\hat{j} 	imes \hat{k} = \hat{i}$. Therefore, the magnetic field must be pointing in the positive z-direction:

$$ \hat{B} = \hat{k} $$

### The Final Picture

Combining our magnitude and direction, we get the complete magnetic field vector:

$$ \vec{B} = 2 	imes 10^{-8} \hat{k} 	ext{ T} $$

This means the magnetic field has a magnitude of **$2 	imes 10^{-8} 	ext{ T}$** and is directed along the **z-direction**. This perfectly matches option (a).

Select Answer:

Visualized Solution

Visualizing the Vectors

Magnitude Formula

Substituting Values

Calculating Magnitude

Direction Rule

Applying the Cross Product

Deducing Magnetic Field Direction

Final Answer

The Way Forward

The Sigma Insight: Characteristics of Electromagnetic Waves

Visualizing the Electromagnetic Wave

Calculating the Magnitude

Decoding the Direction

The Final Picture

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