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: Electromagnetic Wave Relation — in a vacuum, $c = f\lambda$ and $E_0 = c B_0$, with equal average energy densities in the electric and magnetic fields.

(a) Wavelength:

(b) Magnetic field amplitude:

(c) Average energy densities:

$u_E = \frac{1}{2}\varepsilon_0 E_0^2 \times \frac{1}{2} = \frac{1}{4}\varepsilon_0 E_0^2, \quad u_B = \frac{1}{2\mu_0} B_0^2 \times \frac{1}{2} = \frac{1}{4\mu_0} B_0^2$ …

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.

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:

• $f = 2.0 \times 10^{10}\ \text{Hz}$
• $c = 3 \times 10^{8}\ \text{m s}^{-1}$

So:

That’s 1.5 cm — a microwave wavelength.

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:

Given $E_0 = 48\ \text{V m}^{-1}$:

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:

$\langle u_E \rangle = \frac12 \varepsilon_0 \langle E^2 \rangle = \frac12 \varepsilon_0 \cdot \frac{E_0^2}{2} = \frac14 \varepsilon_0 E_0^2$ …

Method: Standard Wave Relations for EM Waves

This problem uses the fundamental wave equation and the intrinsic relation between E and B in free space, plus the energy density equality property of EM waves.

(a) Wavelength of the wave

Step 1: Recall the wave equation

For any electromagnetic wave in vacuum:

Step 2: Substitute given values

Step 3: Compute

Answer: $\boxed{1.5 \times 10^{-2}\ \text{m}}$ (or 1.5 cm)

(b) Amplitude of the magnetic field

Step 1: Use the E–B amplitude relation in free space

Step 2: Rearrange and substitute

Step 3: Compute

Answer: $\boxed{1.6 \times 10^{-7}\ \text{T}}$

(c) Show average energy densities are equal

Step 1: Write the average energy density formulas

• Electric field:

• Magnetic field:

Step 2: Use $B_0 = E_0 / c$ and $c = 1/\sqrt{\varepsilon_0 \mu_0}$

Substitute into $\langle u_B \rangle$: …

Common Mistakes & How to Avoid Them

Mistake 1: Using wrong formula for wavelength

The error: Students often confuse $c = f\lambda$ with $v = f\lambda$ and forget that for EM waves in vacuum, $v = c$.

How to avoid: Always write the relation explicitly:

Then rearrange:

Key check: The answer should be in metres — if you get a tiny number like $1.5\ \text{cm}$, that's correct for such a high frequency.

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

The error: Students write $E_0 = B_0$ or $E_0 = cB_0$ incorrectly (swapping numerator/denominator).

How to avoid: Memorise the exact relation:

So:

Quick sanity check: $B_0$ is always much smaller than $E_0$ (by factor $c$), so $10^{-7}\ \text{T}$ is reasonable.

Mistake 3: Using wrong formula for energy density

The error: Students use $u_E = \frac{1}{2}\varepsilon_0 E^2$ but forget the average value, or use peak value $E_0$ instead of RMS value.

How to avoid: For sinusoidal variation:

• Instantaneous: $u_E = \frac{1}{2}\varepsilon_0 E^2$
• Average over one cycle: $\langle u_E \rangle = \frac{1}{4}\varepsilon_0 E_0^2$

Similarly for magnetic field:

• Instantaneous: $u_B = \frac{B^2}{2\mu_0}$
• Average: $\langle u_B \rangle = \frac{B_0^2}{4\mu_0}$

Mistake 4: Not proving equality — just stating it

The error: Students write "they are equal" without showing the algebra.

How to avoid: Show the derivation step-by-step:

1. Write $\langle u_E \rangle = \frac{1}{4}\varepsilon_0 E_0^2$
2. Write $\langle u_B \rangle = \frac{B_0^2}{4\mu_0}$
3. Substitute $B_0 = E_0/c$ and $c = 1/\sqrt{\mu_0\varepsilon_0}$: $\langle u_B \rangle = \frac{(E_0/c)^2}{4\mu_0} = \frac{E_0^2}{4\mu_0 c^2} = \frac{E_0^2}{4\mu_0 \cdot \frac{1}{\mu_0\varepsilon_0}} = \frac{1}{4}\varepsilon_0 E_0^2 = \langle u_E \rangle$ …