Irodov 3.370: The Frame Where Only B Survives – Amit & Deepali Sinha Foundation

Problem Irodov 3.370

In an inertial frame $K$ there are a uniform electric field of strength $E = 40$ kV/m and a uniform magnetic field of induction $B = 0.20$ mT, at right angles to each other. In the inertial frame $K'$ in which only one of the two fields, electric or magnetic, is observed, find that field: the strength $E'$ or the induction $B'$ . Hint: use the two quantities that are the same in every inertial frame, $\vect E\cdot\vect B$ and $E^2 - c^2B^2$ .

The uniform fields of the frame K at right angles, drawn with E up the page and B out of it (senses chosen). — The uniform fields of the frame $K$ at right angles, drawn with $\vect E$ up the page and $\vect B$ out of it (senses chosen).

Given

$E = 40$ kV/m
$B = 0.20$ mT
$\vect E\perp\vect B$ , both uniform, in $K$

Find

E'

B'

in the frame

K'

with only one field

Solution

Step 1The two invariants in K

The fields are perpendicular, so the first invariant is zero. For the second, compare $E$ with $cB$ , which for $c = 3.00\times10^{8}$ m/s is $60$ kV/m:

\vect E\cdot\vect B = 0, \qquad E^2 - c^2B^2 = (40^2 - 60^2)\ (\text{kV/m})^2 < 0 \tag{1}

Step 2Which field survives

In $K'$ there is only $\vect B'$ , so the second invariant there is $-c^2B'^2$ . Setting it equal to its value (1) in $K$ :

-c^2B'^2 = E^2 - c^2B^2 \quad\Longrightarrow\quad B'^2 = B^2 - \frac{E^2}{c^2}

Numerically, with $E/(cB) = 40/60$ :

B' = 0.20\sqrt{1 - (40/60)^2}\ \text{mT}

B' = B\sqrt{1 - \Bigl(\frac{E}{cB}\Bigr)^2} \approx 0.15\text{ mT}

(2)

The first invariant is satisfied too: with $\vect E' = 0$ , $\vect E'\cdot\vect B' = 0$ , as in $K$ .

Step 3Such a frame exists

The invariants say what the field must be if the frame exists. A frame moving across both fields removes the electric field when $\vect E + \vect v\times\vect B = 0$ , which holds for $\vect v$ along $\vect E\times\vect B$ with the speed

v = \frac{E}{B} = \frac{40\times10^{3}}{0.20\times10^{-3}}\text{ m/s} = 2.0\times10^{8}\text{ m/s} = \frac{2}{3}\,c \tag{3}

This is below $c$ , exactly because $E < cB$ , so the frame is a real one.

The frame K' in which only a magnetic field is seen moves across both fields, along E· B, with the speed E/B. — The frame $K'$ in which only a magnetic field is seen moves across both fields, along $\vect E\times\vect B$ , with the speed $E/B$ .

Check

Direct transformation into the frame of (3): there $\vect v\times\vect E$ is along $\vect B$ with magnitude $vE = v^2B$ , so

B' = \frac{B - v^2B/c^2}{\sqrt{1-v^2/c^2}} = B\sqrt{1 - \frac{v^2}{c^2}} = 0.20\sqrt{1 - 4/9}\ \text{mT}

the same $0.149$ mT, since $v/c = E/(cB)$ . Units: $E/c$ is $(\text{V/m})/(\text{m/s}) = \text{T}$ , so $B^2 - E^2/c^2$ is consistent.

Remark

Had $E$ exceeded $cB$ , the same argument would give a frame with only an electric field, $E' = \sqrt{E^2 - c^2B^2}$ , moving along $\vect E\times\vect B$ at $c^2B/E$ . For $E = cB$ exactly no such frame exists: perpendicular fields of that ratio, as in a light wave, stay perpendicular and of that ratio in every frame, so neither can be removed. The crossed fields of a velocity filter are of this kind: the particles that pass undeflected move at $E/B$ , the speed of the frame in which they are at rest and, with only a magnetic field there, feel no force.