Irodov 3.359: One Force, Two Fields – Amit & Deepali Sinha Foundation

Problem Irodov 3.359

An inertial frame $K$ contains a uniform magnetic field of induction $\vect B$ . A frame $K'$ moves relative to $K$ with a velocity $\vect v$ , small compared with the speed of light and perpendicular to $\vect B$ . Find the electric field strength $\vect E'$ in $K'$ . Hint: compare the forces on a test charge, as seen in the two frames, at an instant when the charge is at rest in $K'$ .

The frame K' (dashed axes) moving with velocity v along the shared x axis of K, through the uniform field B of K, perpendicular to v (drawn out of the page). — The frame $K'$ (dashed axes) moving with velocity $\vect v$ along the shared $x$ axis of $K$ , through the uniform field $\vect B$ of $K$ , perpendicular to $\vect v$ (drawn out of the page).

Given

uniform $\vect B$ in $K$
no electric field in $K$ (assumed)
$\vect v \perp \vect B$
$v \ll c$

Find

\vect E'

K'

Solution

Step 1The force seen in

K

Take a test charge $q$ that is at rest in $K'$ at the instant considered, so in $K$ it moves with the velocity $\vect v$ . Frame $K$ is taken to have no electric field (assumed), so the force there is the Lorentz force alone:

\vect F = q\,\vect v\times\vect B \tag{1}

Step 2The force seen in

K'

In $K'$ the charge is at rest, so no magnetic force can act on it whatever the magnetic field of $K'$ is. Any force it feels there comes from an electric field $\vect E'$ :

\vect F' = q\,\vect E' \tag{2}

In K: the charge moves with v through B (out of the page), and the magnetic force F acts on it. — In $K$ : the charge moves with $\vect v$ through $\vect B$ (out of the page), and the magnetic force $\vect F$ acts on it.

In K': the same charge at rest; the same force F' is now exerted by an electric field E'. — In $K'$ : the same charge at rest; the same force $\vect F'$ is now exerted by an electric field $\vect E'$ .

Step 3The force is the same in both frames

The charge’s acceleration is the same in $K$ and $K'$ , and so is its mass, so the two forces are one and the same. Setting (2) equal to (1) and dividing by $q$ :

\vect E' = \vect v\times\vect B

(3)

Its magnitude is $vB$ , since $\vect v \perp \vect B$ , and it is perpendicular to both $\vect v$ and $\vect B$ ; for the figure’s senses it points down the page.

Check

The result must hold for any charge, not only one at rest in $K'$ . A charge moving with $\vect u'$ in $K'$ moves with $\vect u = \vect u' + \vect v$ in $K$ ; with $\vect B' = \vect B$ at these speeds, the force in $K'$ is

q(\vect E' + \vect u'\times\vect B) = q(\vect v + \vect u')\times\vect B = q\,\vect u\times\vect B,

exactly the force in $K$ . Units: $vB$ is $(\text{m/s})\cdot\text{T} = \text{V/m}$ .

Remark

Whether a field is called electric or magnetic depends on the frame. A metal rod moving through $\vect B$ develops a motional emf: in the laboratory it is the magnetic force on its electrons, while in the rod’s own frame, where the electrons are at rest, it is the electric field (3). Problem 3.360 uses exactly this field to find the charges on a moving plate.