Q.The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0 = 510\ \text{nT}$. What is the amplitude of the electric field part of the wave?

The key idea is that in vacuum, the electric and magnetic field amplitudes of an electromagnetic wave are related by the speed of light: $E_0 = c B_0$.

Given $B_0 = 510\ \text{nT} = 510 \times 10^{-9}\ \text{T}$ and $c = 3.0 \times 10^8\ \text{m/s}$:

1. Use the relation $E_0 = c B_0$.
2. Substitute: $E_0 = (3.0 \times 10^8) \times (510 \times 10^{-9})$. …

The key idea here is that in vacuum, electromagnetic waves travel at the speed of light $c$, and the electric and magnetic fields are not independent — they are linked by Maxwell’s equations. For a plane harmonic wave, the amplitudes satisfy $E_0 = c B_0$. This is a direct consequence of Faraday’s law and Ampère’s law in free space, where the rate of change of one field generates the other.

Why does this relation hold? Imagine a wave moving along the $x$-axis, with the electric field oscillating along $y$ and the magnetic field along $z$. Faraday’s law tells us that a changing magnetic field produces an electric field, and the magnitudes are tied by the wave speed. In vacuum, that speed is $c = 3.00 \times 10^8\ \text{m/s}$. So once you know $B_0$, multiplying by $c$ gives $E_0$ directly — no extra constants needed.

Let’s work it out step by step.

1. Write the given data. The amplitude of the magnetic field is $B_0 = 510\ \text{nT}$. Recall that $1\ \text{nT} = 10^{-9}\ \text{T}$, so

2. Recall the fundamental relation. For an electromagnetic wave in vacuum, the amplitudes of the electric and magnetic fields are related by

where $c = 3.00 \times 10^8\ \text{m/s}$ is the speed of light in vacuum.

This is not an approximation — it follows exactly from Maxwell’s equations for a plane wave in free space. The ratio $E/B$ equals $c$ at every point and at every instant for a wave in vacuum.

3. Substitute the values.

Multiply the numbers:

Multiply the powers of ten:

So

(Recall that $1\ \text{T} \cdot \text{m/s} = 1\ \text{V/m}$, so the units work out perfectly.) …

Method: Using the Speed of Light Relation for EM Waves

This problem uses the fundamental relation between electric and magnetic fields in an electromagnetic wave — the fields are linked by the speed of light.

Steps

1. Recall the key relation In vacuum, for any electromagnetic wave:

where

• $E_0$ = amplitude of electric field (V/m)
• $B_0$ = amplitude of magnetic field (T)
• $c$ = speed of light in vacuum = $3.0 \times 10^8\ \text{m/s}$

2. Convert units Given $B_0 = 510\ \text{nT}$

So:

3. Apply the formula

4. Calculate $E_0 = 3.0 \times 5.10 \times 10^{8-7} = 15.3 \times 10^{1}$ …

Here are the most common mistakes students make when solving this problem, along with how to avoid each one.

1. Forgetting the Speed of Light Constant

The Mistake:

Students often try to use $E_0 = c B_0$ but write $c$ as $3 \times 10^8$ without checking units, or they confuse $c$ with the speed of sound or another constant.

How to Avoid:

• Memorize $c = 3.0 \times 10^8\ \text{m/s}$ for vacuum.
• Always write the relation clearly:

This is derived from Maxwell’s equations — in vacuum, the electric and magnetic field amplitudes are linked by the speed of light.

2. Unit Conversion Errors (nT to T)

The Mistake:

Given $B_0 = 510\ \text{nT}$, students plug in $510$ directly without converting to tesla (T). This gives a wildly wrong answer.

How to Avoid:

• Always convert nano (n) to base SI units:

So:

3. Using the Wrong Formula (Mixing RMS and Amplitude)

The Mistake:

Some students use $E_{\text{rms}} = c B_{\text{rms}}$ but then treat $B_0$ as RMS, or vice versa. The problem explicitly asks for amplitude, so you must use the amplitude form.

How to Avoid:

• Read the question carefully: “amplitude of the magnetic field part” → $B_0$.
• The correct relation for amplitudes is:

(Not $E_{\text{rms}} = c B_{\text{rms}}$ unless the problem gives RMS values.)

4. Arithmetic Slip-Ups in Scientific Notation

The Mistake:

When multiplying $3 \times 10^8$ by $5.10 \times 10^{-7}$, students misplace the decimal or add exponents incorrectly.

How to Avoid:

• Do the multiplication step-by-step:

Multiply coefficients: $3.0 \times 5.10 = 15.3$

Add exponents: $10^{8 + (-7)} = 10^{1}$

So:

$E_0 = 15.3 \times 10^{1} = 153\ \text{V/m}$ …