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# 工數向量及Laplace問題

請問有沒有大大會這一題,麻煩請解答一下,非常感謝

When an electron with charge e and mass m moves with velocity V in a magnetic field B it experiences a force proportional to V×B.

If this is the only external force, then the equation of motion for the electron is

m(dV/dt)= eV×B

Suppose that B is a constant and the axes are chosen so that B = Bˆk(^k為z方向單位向量), electron passes through the origin at t = 0 with V(0) = u0(下底為0)ˆi (^i為x方向單位向量)+ w0(下底為0)ˆk.

(a) Write the equations and initial conditions in Cartesian coordinates that govern the electron's velocity components, V1, V2 and V3.

(b) Solve the equations using the Laplace Transform. It will be helpful to define ω= eB/m.

(c) From the velocity found in part b find R(t) and show that the electron traverses a helical path of circular cross section and constant pitch.

(d) Find the electron's distance from the origin at time t = 10π/ω

### 2 Answers

- mathmanliuLv 71 decade agoFavorite Answer
粗體字表向量 (R為位置函數)

設ω= eB/m, V=(u, v, w), R=(x,y,z), B=(0, 0, B), V(0)=(u0, 0, w0)

VB=(Bv, -Bu, 0)

a) m(dV/dt)= eVB 即 (u'(t), v'(t), w'(t))= (ωv, -ωu, 0)或

u'(t)= ωv

v'(t)= -ωu

w'(t)= 0

u(0)=u0, v(0)=0, w(0)=w0

b) Laplace

sU-u0=ωV

sV-0= -ωU

sW-w0=0

=>U(s)= u0 s/(s^2+ω^2) => u(t)= u0*cosωt

V(s)= -ωu0/(s^2+ω^2) => v(t)= - u0* sinωt

W(s)= w0/ s => w(t)=w0

c) 求R(t)=(x(t), y(t), z(t))

x'(t)=u(t)=u0*cosωt => x(t)= (u0/ω)sinωt

y'(t)=v(t)= -u0*sinωt=> y(t)= (u0/ω)cosωt

z'(t)=w(t)= w0 => z(t)=w0*t

=>x^2+y^2=(u0/ω)^2 (投影至xy平面為圓, 週期 ωT=2π=>T=2π/ω)

又z=w0*t 隨時間增加而增加,每單位時間上升ω0高度

故質點路徑為螺線, 半徑 u0/ω, 2π/ω單位時間繞一圈,

每圈上升高度z= w0*2π/ω(螺距pitch固定)

d)求曲線長S

S=∫√[(dx/dt)^2+(dy/dt)^2+(dz/dt)^2] dt , t=0~ 10π/ω

=∫√(u^2+v^2+w^2) dt , t=0~ 10π/ω

=∫√[(u0)^2+(w0)^2] dt , t=0~10π/ω

= √[(u0)^2+(w0)^2] * 10π/ω

希望沒打錯字!

Source(s): me