A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-directio

Q: Why do we use the speed of light $c$ here?

Because the problem states the wave is traveling in 'free space' (vacuum), where all electromagnetic waves travel at the speed of light $c = 3 \times 10^8$ m/s.

Q: What happens to the frequency of 50 MHz?

The frequency is extra information in this specific problem! The relationship between the amplitudes of $\mathbf{E}$ and $\mathbf{B}$ depends only on the speed of the wave, not its frequency.

Q: How did we know $\hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}$?

This comes from the standard right-hand rule for the Cartesian coordinate system. If you curl your fingers from the y-axis ($\hat{\mathbf{j}}$) to the z-axis ($\hat{\mathbf{k}}$), your thumb points along the x-axis ($\hat{\mathbf{i}}$).

Q: Could the magnetic field be in the negative z-direction?

No. If $\mathbf{B}$ was in the $-\hat{\mathbf{k}}$ direction, the cross product $\hat{\mathbf{E}} \times \hat{\mathbf{B}}$ would be $\hat{\mathbf{j}} \times (-\hat{\mathbf{k}}) = -\hat{\mathbf{i}}$, which means the wave would travel in the negative x-direction. But the problem specifies it travels in the positive x-direction.

Q: Are $\mathbf{E}$ and $\mathbf{B}$ always in phase?

Yes, in a plane electromagnetic wave traveling in free space, the electric and magnetic fields oscillate perfectly in phase with each other.

Question

A plane electromagnetic wave of frequency $50	ext{ MHz}$ travels in free space along the positive $x$-direction. At a particular point in space and time, $\mathbf{E} = 6.3\hat{\mathbf{j}}	ext{ V/m}$. The corresponding magnetic field $\mathbf{B}$, at that point will be

Options:

(A) $18.9 	imes 10^8 \hat{\mathbf{k}}	ext{ T}$

(B) $6.3 	imes 10^{-8} \hat{\mathbf{k}}	ext{ T}$

(C) $18.9 	imes 10^{-8} \hat{\mathbf{k}}	ext{ T}$

(D) $2.1 	imes 10^{-8} \hat{\mathbf{k}}	ext{ T}$

Sigma Percentile · Accepted Answer

### The Symphony of Electromagnetic Waves

Imagine you are standing in a vast, empty 3D space. Suddenly, an invisible wave of energy surges past you. This is an electromagnetic wave, a perfect, self-sustaining dance between electric and magnetic fields.

The question tells us that this wave is traveling through free space along the positive x-direction. At a specific moment, the electric field $\mathbf{E}$ is pointing straight up along the y-axis, with a value of $6.3 \hat{\mathbf{j}}$ V/m.

Our mission? To find its partner in this dance—the magnetic field $\mathbf{B}$.

### The Master Equation: Magnitude

In the vacuum of free space, electromagnetic waves travel at the ultimate speed limit of the universe: the speed of light, $c = 3 	imes 10^8$ m/s.

There is a beautiful, fundamental relationship that binds the electric and magnetic fields together. The ratio of their magnitudes is exactly equal to the speed of the wave.

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

This means we can easily find the magnitude of the magnetic field by rearranging the formula:

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

### Executing the Calculation

Now, let's bring our numbers into the equation. We know the magnitude of the electric field is $6.3$ V/m.

Substituting this into our rearranged formula, we get:

$$|\mathbf{B}| = \frac{6.3}{3 	imes 10^8}$$

Dividing $6.3$ by $3$ gives us $2.1$. And the $10^8$ in the denominator comes up as $10^{-8}$.

$$|\mathbf{B}| = 2.1 	imes 10^{-8} 	ext{ T}$$

We now have the magnitude, but a vector is not complete without its direction.

### The Right-Hand Rule: Finding the Direction

Electromagnetic waves are transverse waves. This means the electric field, the magnetic field, and the direction of propagation are all mutually perpendicular to each other.

Mathematically, the direction of propagation is given by the cross product of the electric and magnetic field vectors:

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

We know the wave is traveling in the positive x-direction, so $\hat{\mathbf{v}} = \hat{\mathbf{i}}$. We also know the electric field is oscillating in the y-direction, so $\hat{\mathbf{E}} = \hat{\mathbf{j}}$.

Plugging these into our cross product relationship:

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

Now, ask yourself: what unit vector, when crossed with $\hat{\mathbf{j}}$, gives $\hat{\mathbf{i}}$? According to the standard right-hand rule for Cartesian coordinates, $\hat{\mathbf{j}} 	imes \hat{\mathbf{k}} = \hat{\mathbf{i}}$.

Therefore, the magnetic field must be pointing in the positive z-direction, which is $\hat{\mathbf{k}}$.

### The Final Answer

We have successfully found both the magnitude and the direction of the magnetic field. Combining them, we get our final vector:

$$\mathbf{B} = 2.1 	imes 10^{-8} \hat{\mathbf{k}} 	ext{ T}$$

This perfectly matches option (d).

**A Thought to Ponder:** What if this wave wasn't in a vacuum, but traveling through water or glass? The speed of the wave would decrease to $v = \frac{c}{n}$, where $n$ is the refractive index. Consequently, the magnetic field would become stronger for the same electric field! Always keep an eye on the medium.

Animated Solution for Physics - Electromagnetic Waves: A plane electromagnetic wave of frequency $50 MHz$ travels in free space along the positive $x$ -direction. At a particular point in space and time, $E = 6.3 \hat{j} V/m$ . The corresponding magnetic field $B$ , at that point will be

Select Answer:

Visualized Solution

$Visualizing the EM Wave$

$∣ B ∣ = \frac{∣ E ∣}{c}$

$∣ B ∣ = \frac{6.3}{3 \times 1 0 ^{8}}$

$∣ B ∣ = 2.1 \times 1 0^{- 8} T$

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

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

$B = 2.1 \times 1 0^{- 8} \hat{k} T$

$What if it was in a medium?$

The Sigma Insight: Characteristics of Electromagnetic Waves

The Symphony of Electromagnetic Waves

The Master Equation

Executing the Calculation

The Right-Hand Rule

The Final Answer

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Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣B∣=c∣E∣​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣B∣=3×1086.3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣B∣=2.1×10−8 T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E^×B^=v^

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } j^​×B^=i^

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B=2.1×10−8k^ T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it was in a medium?