In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 imes 10^{10}\ ext{Hz}$ and amplitude $48\ ext{V m}^{-1}$.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that th

Question

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 	imes 10^{10}\ 	ext{Hz}$ and amplitude $48\ 	ext{V m}^{-1}$.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the $\mathbf{E}$ field equals the average energy density of the $\mathbf{B}$ field. [$c = 3 	imes 10^{8}\ 	ext{m s}^{-1}$.]

Accepted Answer

> [!TLDR]
> For a plane EM wave, the wavelength is found from $c = f\lambda$, the magnetic amplitude from $E_0 = c B_0$, and the equality of average energy densities follows from $u_E = \frac12\varepsilon_0 E^2$ and $u_B = \frac{B^2}{2\mu_0}$ together with $c = 1/\sqrt{\mu_0\varepsilon_0}$.

This is a classic problem that tests your understanding of the fundamental relationships in an electromagnetic wave. In free space, the electric and magnetic fields are not independent — they are linked by the speed of light, and their energy densities are always equal on average. Let’s see why.

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### 1. Wavelength from frequency

For any wave, the speed, frequency, and wavelength are related by $v = f\lambda$. For an electromagnetic wave in vacuum, $v = c$.

Given:
- $f = 2.0 	imes 10^{10}\ 	ext{Hz}$
- $c = 3 	imes 10^{8}\ 	ext{m s}^{-1}$

So:
$$
\lambda = \frac{c}{f} = \frac{3 	imes 10^{8}}{2.0 	imes 10^{10}} = 1.5 	imes 10^{-2}\ 	ext{m}
$$

That’s 1.5 cm — a microwave wavelength.

> [!TIP]
> Notice the frequency is $2 	imes 10^{10}$ Hz, which is 20 GHz — right in the microwave band. The wavelength of 1.5 cm confirms this.

---

### 2. Magnetic field amplitude from electric field amplitude

In a plane EM wave, the instantaneous magnitudes are related by $E = cB$. This holds for the amplitudes too:
$$
E_0 = c B_0
$$

Given $E_0 = 48\ 	ext{V m}^{-1}$:
$$
B_0 = \frac{E_0}{c} = \frac{48}{3 	imes 10^{8}} = 1.6 	imes 10^{-7}\ 	ext{T}
$$

> [!WARNING]
> A common mistake is to forget that $B_0$ is in tesla, not gauss. $1.6 	imes 10^{-7}$ T is $1.6$ milligauss — a very small field, which is typical for EM waves.

---

### 3. Showing that average energy densities are equal

The instantaneous energy densities are:
- Electric: $u_E = \frac12 \varepsilon_0 E^2$
- Magnetic: $u_B = \frac{B^2}{2\mu_0}$

For a sinusoidal wave, $E = E_0 \sin(kx - \omega t)$ and $B = B_0 \sin(kx - \omega t)$. The time average of $\sin^2$ over one cycle is $1/2$.

So:
$$
\langle u_E angle = \f

Q.In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10}\ \text{Hz}$ and amplitude $48\ \text{V m}^{-1}$ .

Concept understanding — Electromagnetic Wave Relation

Electromagnetic Wave Relation: From Intuition to Precision

What is frequency? What is wavelength?

Why does this matter for exams?

A concrete example

Why this formula?

Electromagnetic Wave Relation: Why $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

1. The Starting Point: Maxwell's Equations in Vacuum

2. Deriving the Wave Equation for $\mathbf{E}$

3. Identifying the Wave Speed

4. The Magnetic Field Follows Suit

5. The Crucial Relationship Between $\mathbf{E}$ and $\mathbf{B}$

1. Wavelength from frequency

2. Magnetic field amplitude from electric field amplitude

3. Showing that average energy densities are equal

Method: Standard Wave Relations for EM Waves

(a) Wavelength of the wave

(b) Amplitude of the magnetic field

(c) Show average energy densities are equal

Common Mistakes & How to Avoid Them

Mistake 1: Using wrong formula for wavelength

Mistake 2: Forgetting the factor of $c$ in $E_0$ and $B_0$ relation

Mistake 3: Using wrong formula for energy density

Mistake 4: Not proving equality — just stating it

Concept and Intuition

Step-by-Step Solution

Q.In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0×1010 Hz2.0 \times 10^{10}\ \text{Hz}2.0×1010 Hz and amplitude 48 V m−148\ \text{V m}^{-1}48 V m−1.

1. Wavelength from frequency

2. Magnetic field amplitude from electric field amplitude

3. Showing that average energy densities are equal

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Q.In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10}\ \text{Hz}$ and amplitude $48\ \text{V m}^{-1}$ .

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