Saved Bookmarks
| 1. |
A particle with specific charge q//mmoves in the region of space where there are unifrommutuallyperpendicularelectric and magneticfieldswith strengthE and inductionB (fig). At the momentt = 0 the particlewas locatedat the pointO and had zero velocity. Forthe non-relativisticcase find: (a) the law of motionx(f) and y(t) of theparticle, the shape of the trajectory, (b) the lenghtof the segment of the trajectory betweentwo nearestpoints at whichthevelocityof the particleturns into zero, (c) the meanvalue of the particle's velocityvectorprojectionson the x axis(the drift velocity). |
|
Answer» Solution :The equaciton of motion is, `m(d^(2) vec(R))/(dt^(2)) = q(vec(E) + vec(v) xx vec(B))` Now, `vec(v) xx vec(B) = |(HATI,hatj,hatk),(DOT(x),dot(y),dot(z)),(0,0,B)| = vec(i) B doty - vec(j) B dotx` So, the equaction becomes, `(dv_(x))/(dt) = (q B v_(y))/(m), (dv_(y))/(dt) = (qE)/(m) - (qB)/(m) v_(x)`, and `(dv_(z))/(dt) = 0` Here, `v_(x) = dotx, v_(y) = doty, v_(z) = dotz`. The last equactions is easy to intergate, `v_(z)` = CONSTANT = 0, since `v_(z)` is zeroinitially. Thusintergating again, `z` = constant = 0, and motionis condifned to the `x-y` plane. We now multiply the secoundequaction by `i` and add to the first equatiion. `xi = v_(x) + iv_(y)` we get the equation, `(d xi)/(dt) =i omega xi omega = (qB)/(m)` This equation afterbeing mutiplied by `e^(i omega t)` can be rewrittenas, `(d)/(dt) (xi e^(iomega t)) = i omega e^(i omega t) (E)/(B)` and intergrated at once to give, `xi = (E)/(B) + C e^(-i omega t -i alpha)`, where `C` and `alpha` two real constants. Takingreal andimaginary parts. `v_(x) = (E)/(B) + C cso (omega t + alpha)` and`v_(y) = -C sin (omega t + alpha)` Since `v_(y) = 0`, when `t = 0`, we can takek `alpha = 0`, then`v_(x) = 0` at `t = 0` gives, `C = - (E)/(B)` we get, `v_(x) = (E)/(B) (1 - cos omega t)` and`v_(y) = (E)/(B) sin omega t`. Intergating again and using `x = y = 0`, at `t = 0`, we get `x(t) = (E)/(B) (t -(sin omega t)/(omega)), y(t) = (E)/(omega B) (1 - cos omegat)`. This is the equcation of a cycloid. (b) The velocityis zero, when `omega t = 2n pi`. We see that `v^(2) = v_(x)^(2) + v_(y)^(2) = ((E)/(B))^(2) (2 - 2 cos omega t)` pr, `v = (dx)/(dt) = (2E)/(B) |"sin" (omega t)/(2)|` The quanttiy inside the modulus is positive for `0 lt omega t lt 2pi`. Thuswe candropthe modulusand write for the distancetraversedbetween two successikve zeroesof velocity. `S = (4E)/(omega B) (1 - "cos" (omega)/(2))` Putting`omega t = 2pi`, we get `S = (8E)/(omega B) = (8 mE)/(q B^(2))` (c) The drift velocity is the x-directionand has the magnitude, `lt v_(s)gt = lt(E)/(B) (1 - cos omega t) gt = (E)/(B)`. |
|