Question

6. A proton at the origin is briefly accelerated with ãl (a, 0,0) The observation location P is a distance d from the origin, at an angle θ = 30° to the +x axis. (a) Determine the vector ãi and express your result as a vector triple (..., your result as a vector triple. your result as a vector triple. your result as a vector triple. (Assume that the displacement (b) Determine the radiative electric field at P and express (c) Determine the radiative magnetic field P and express d) Determine the Coulomb electric field at P and express of the proton is very small compared to d.)

Answer 1

A^vector = (a_0, 0, 0) (given)Vertical-bar a_perpendicular^vector Vertical-bar = a_0 sin theta In component form we have a_perpendicular^vector = Vertical-bar a_perpendicular Vertical-bar (sin theta - cos theta, 0) = (a_0 sin^2 theta, - a_0 sin theta cos theta,0) equation (i) b) Radiative electric field at P is E^vector(P)_rad = 9/4pi epsilon_0 d/(d^vector middot u^vector)^3 (d^vector times (u^vector times u^vector)) where u^vector = Cd^^- v^vector = C(cos theta i^^+ sin theta j^^) - a-0 t i^6 Now, d^vector times (u^vector times a^vector) = u^vector (d^vector middot a^vector) - a^vector(d^vector middot u^vector) = u^vector (a_0d cos theta) - a^vector (cd - a_0 td cos theta) \r\n d^vector times (u^vector times a^vector) = (a_0 d cos theta)[(c cos theta - a_0 t) i^^+ C sin theta j^^] - d(c - a-0 t cos theta)i^^= (a_0 cd cos^2 theta - a^2 td cos theta - a-0 cd + a-0^2 dt cos theta)i^^+ a_0 cd cos theta sin theta j^^= -a_0 cd sin^2 theta i^^+ a_0 cd sin theta cos theta j^^