Irodov 3.340: Electric Field Matching One Tesla – Amit & Deepali Sinha Foundation

Problem Irodov 3.340

How strong must an electric field in vacuum be for its energy per unit volume to equal that of a magnetic field of induction $B = 1.0$ T, also in vacuum?

Given

$B = 1.0$ T
vacuum

Find

the field strength

E

with

w_e = w_m

Solution

Step 1The two energy densities

In vacuum the energy per unit volume of an electric field $E$ and of a magnetic field $B$ are

\begin{aligned} w_e &= \frac{\varepsilon_0E^2}{2}\\ w_m &= \frac{B^2}{2\mu_0} \end{aligned} \tag{1}

The electric density grows as $E^2$ from zero, so it equals the fixed magnetic density at exactly one positive $E$ .

The electric energy density w_e against E (gold) and the energy density B^2/2μ_0 of the given magnetic field, w_m (dashed level). They meet at E = cB. — The electric energy density $w_e$ against $E$ (gold) and the energy density $B^2/2\mu_0$ of the given magnetic field, $w_m$ (dashed level). They meet at $E = cB$ .

Step 2Equal densities

Setting $w_e = w_m$ in (1) and cancelling the factor $1/2$ :

\varepsilon_0E^2 = \frac{B^2}{\mu_0}

Taking the positive root, and using $1/\sqrt{\varepsilon_0\mu_0} = c$ , numerically

\frac{B}{\sqrt{\varepsilon_0\mu_0}} = cB = (3.00\times10^{8})(1.0)

E = \frac{B}{\sqrt{\varepsilon_0\mu_0}} \approx 3.0\times10^{8}\text{ V/m}

(2)

Check

Both densities at these values come out the same:

\begin{aligned} \frac{B^2}{2\mu_0} &= \frac{(1.0)^2}{2\,(1.257\times10^{-6})} = 4.0\times10^{5}\text{ J/m}^3\\ \frac{\varepsilon_0E^2}{2} &= \frac{(8.85\times10^{-12})(3.0\times10^{8})^2}{2} = 4.0\times10^{5}\text{ J/m}^3 \end{aligned}

Units: $B/\sqrt{\varepsilon_0\mu_0}$ is tesla times metres per second, and $\text{T m/s} = (\text{V s/m}^2)(\text{m/s}) = \text{V/m}$ .

Remark

An ordinary laboratory magnet of $1$ T stores as much energy per unit volume as an electric field of $300$ MV/m, a hundred times the breakdown field of air (about $3$ MV/m). That is why large field energies are stored and handled magnetically, in inductors, transformers and motors, while capacitors hold far less per unit volume. In a light wave the two densities are equal at every point, because there $E = cB$ exactly.