At time $\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$, electrons in a vertical copper wire (Figure 23.122) are accelerated downward for a very short time $\mathit{\Delta }\mathit{t}$(by a power supply that is not shown). A proton, initially at rest, is located a very large distance L from the wire.

Explain in detail what happens to the proton (neglect gravity), and at what time.

The electron is accelerated initial at $t=0for\Delta t$.

The distance of proton from the wire is $L$.

The expression to calculate the magnitude of the radiative electric field is given as follows.

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{e}\mathbf{a}}{{\mathbf{c}}^{\mathbf{2}}\mathbf{r}}$ …(i)

Here, e is the charge, a is the acceleration of electron, r is the distance and c is the speed of the light.

Initially electron is at the rest but at $t=0$, the electron is accelerated for the short duration and at this time only static electric field is act upon it.

Consider the arrangement shown below.

The radiative electric field is directed into the tangential acceleration direction, after time $t$, the radiated electric field is act in downward direction and the magnitude of it can be calculated by using the equation (i). the wave is propagated into right direction due to which direction of the magnetic field in into the page and charge particle experienced the force which is in the left direction.

The time$t$ after which radiated electric field act on electron is given by,

$t=\frac{r}{c}$

Hence the direction of the force on charge particle is left and time is $t=\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$.