Field Relation E0 = cB0

PHYOrigin & Nature

In a plane EM wave the electric and magnetic fields are in phase and locked together by $E_0=cB_0$ . This single relation lets you convert between field amplitudes (or between rms values) and is one of the most reliable one-mark/two-mark numericals.

Because $c$ is large, $B_0$ is always many orders of magnitude smaller than $E_0$ .

Peak field relation

E_0 = c\,B_0

E_0

= peak electric field (V m

^{-1}

B_0

= peak magnetic field (T),

c=3\times10^{8}\ \text{m s}^{-1}

RMS field relation

E_{rms} = c\,B_{rms}

Same ratio holds for rms values;

E_{rms}=E_0/\sqrt{2}

and

B_{rms}=B_0/\sqrt{2}

Solving for B

B_0 = \dfrac{E_0}{c}

Use this form when the electric amplitude is given and the magnetic amplitude is required.

$\vec E$ and $\vec B$ oscillate in phase: both peak at the same instant and both vanish together.
The ratio $E/B=c$ holds at every instant, and equally for amplitudes and rms values, since both are scaled by the same $1/\sqrt{2}$ .
$\vec E$ , $\vec B$ and the direction of propagation are mutually perpendicular, with $\hat E\times\hat B=\hat k$ (direction of travel).
Magnetic field magnitudes are tiny ( $\sim10^{-7}$ to $10^{-8}$ T for everyday fields) precisely because of division by $c\approx3\times10^{8}$ .
In a medium of refractive index $n$ , the wave speed is $v=c/n$ and the local relation becomes $E_0=v\,B_0$ , not $cB_0$ .
Units must be SI: $E$ in V m $^{-1}$ , $B$ in tesla; then $E/B$ comes out in m s $^{-1}$ as a speed.

Where the marks go

Using $B_0=E_0\times c$ instead of $B_0=E_0/c$ — divide by $c$ to get the small magnetic value.
Mixing peak and rms: don't equate $E_0$ with $B_{rms}$ ; keep both peak or both rms.
Applying $E_0=cB_0$ inside a medium where the correct speed is $v=c/n$ .
Carrying $B$ in gauss or $E$ in unusual units instead of tesla and V m $^{-1}$ .

How the board asks it

Numericalthe peak field relation $E_0=cB_0$
The amplitude of the electric field of a plane electromagnetic wave in vacuum is $E_0 = 120\ \text{V m}^{-1}$ . Calculate the amplitude of the magnetic field $B_0$ of the wave.
Numericalthe rms field relation $E_{rms}=cB_{rms}$
The rms value of the magnetic field in a plane electromagnetic wave travelling in free space is $B_{rms} = 2.0 \times 10^{-8}\ \text{T}$ . Calculate the rms value of the electric field of the wave.
Numericalthe local relation $E_0=vB_0$ with $v=c/n$
A plane electromagnetic wave travels through a medium of refractive index $n = 1.5$ . If the peak electric field is $E_0 = 90\ \text{V m}^{-1}$ , calculate the peak magnetic field $B_0$ in the medium.
Define / statethe amplitude relation and mutual perpendicularity of $\vec E$ , $\vec B$ and the direction of propagation
State the relation between the amplitudes of the electric and magnetic fields in a plane electromagnetic wave travelling in vacuum, and state the mutual orientation of $\vec E$ , $\vec B$ and the direction of propagation.
Give reasonsthe magnetic field being tiny because of division by $c$
In an electromagnetic wave the magnitude of the magnetic field is extremely small compared with that of the electric field. Give a reason for this.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.