Saved Bookmarks
| 1. |
In an inertial reference frame Kthere is only a unifromelectricfield E = 8 kV//min strength. Find themodulus and direction (a) of the vector E' (b) of the vectorB'in the interialrefrernceframe K'movingwith aconstantvelocityvrelative to the frameK at an angle alpha = 45^(@)to the vectorE. The velocityof the frameK' is equlatoabeta = 0.60fractionof the velocityof light. |
|
Answer» Solution :Choose `VEC(E)` in thedirection of the z-axis, `vec(E) = (0,0,E)`. The frame `K'` is movingwith velocity `vec(v) = (v sin ALPHA, 0, v cosalpha)`, in the`x-z` plane . Thenin the frame`K'` `vec(E')_(|\|) = vec(E)_(|\|) vec(B')_(|\|) = 0` `vec(E')_(_|_) = (vec(E)_(_|_))/(sqrt(1 - v^(2)//c^(2))) vec(B)'_(_|_) = (-vec(v) xx vec(E)//c^(2))/(sqrt(1 - v^(2)//c^(2)))` The vectoralong `vec(v)` is `vec(e) = (sin alpha, 0, cos alpha)` ANDTHE perpendicular vectorin the `x-z` planeis, `vec(f) = (-cos alpha, 0, sin alpha)`, (a) Thususing `vec(E) = E cos alpha vec(e) + E sin alpha vec(f)`. `E'_(|\|) = E cos alpha` and `E'_(|\|) = (E sin alpha)/(sqrt(1 - v^(2)//c^(2))` So `E' = E sqrt((1 - BETA^(2) cos^(2) alpha)/(1 - beta^(2)))` and`tan alpha' = (tan alpha)/(sqrt(1 - v^(2)//c^(2)))` (b) `B'_(|\|) = 0, vec(B')_(_|_) = (vec(v) xx vec(E)//c^(2))/(sqrt(1 -v^(2)//c^(2)))` `B' = (beta E sin alpha)/(c sqrt(1 - beta^(2)))` |
|