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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. |
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Answer» Solution :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. |
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