Question

Non-relativistic protons accelerated by a potential difference `U` from a round beam with current `I`. Find the magnitude and direction of the Poynting vector the beam at a distance `r` from its axis.

Answer 1

Here `n e v = I//piR^(2)`
where `R =` radius of cross section of the conductor and `n =` change density (per unit volume)
Also `(1)/(2)mv^(2 = eU` or `v = sqrt((2eU)/(m))`
Thus, the moving protons have a charge per unit length
`= n epi R^(2) =I sqrt((m)/(2eU))`.
This gives rise to an electric field at a distance `r` given by
`E =(1)/(epsilon_(0)) sqrt((m)/(2eU))//2pi r`
The magnetic field is `H = (I)/(2pir)` (for `r gt R)`
Thus
`S = (I^(2))/(epsilon_(0)4pi^(2)r^(2)) sqrt((m)/(2eU))` radially outward from the axis This is the Poynting vector.