- Repeat The Analysis In Example 5 2 Except Replacing The E Field With A Gravitational Field Acceleration G Does The D 1 (24.75 KiB) Viewed 39 times
THAT I CAN READ. I REALLY WANT TO UNDERSTAND WHAT I'M DOING IN THIS
PROBLEM AND HOW TO ANSWER THESE QUESTIONS
- Repeat The Analysis In Example 5 2 Except Replacing The E Field With A Gravitational Field Acceleration G Does The D 2 (174.04 KiB) Viewed 39 times
Example 5.2 Cycloid Motion A more exotic trajectory occurs if we include a uniform electric field, at right angles to the magnetic one. Suppose, for instance, that B points in the x-direction, and E in the z-direction, as shown in Fig. 5.7. A particle at rest is released from the origin; what path will it follow? Solution: Let's think it through qualitatively, first. Initially, the particle is at rest, so the magnetic force is zero, and the electric field accelerates the charge in the z-direction. As it picks up speed, a magnetic force develops which, according to Eq. 5.1, pulls the charge around
E bana 0 y B Figure 5.7 to the right. The faster it goes, the stronger Fmag becomes; eventually, it curves the particle back around towards the y axis. At this point the charge is moving against the electrical force. so it begins to slow down--the magnetic force then decreases, and the electrical force takes over, bringing the charge to rest at point a, in Fig. 5.7. There the entire process commences anew, carrying the particle over to point b, and so on. Now let's do it quantitatively. There being no force in the x-direction, the position of the particle at any time t can be described by the vector (0, y(t), z(t)); the velocity is therefore v= (0, y, z), where dots indicate time derivatives. Thus Î ŷ V x B= 0 Ý ż B 0 0 = Bżỹ - Byî, and hence, applying Newton's second law, F= Q(E+ v x B) = Q(E 2 + Bž – Byî) = ma = m(jỹ + ža). Or, treating the ĝ and ĉ components separately, QB2 = my, QE – QBy = mz. For convenience, let QВ WE (5.41 m (This is the cyclotron frequency, at which the particle would revolve in the absence of any electric field.) Then the equations of motion take the form E j wż, w 5 = už, = (- ). (5.51 B y(t) z(t) Their general solution” is Cicos wt + C2 sin wt + (E/B)t + C3, C2 cos w1 - Ci sin wt + C4. 2 As coupled differential equations, they are easily solved by differentiating the first and using the second to eliminate 2. } (5.6) =
= 1 But the particle started from rest (y(0) = 2(0) = 0), at the origin (y(0) = z(0) = 0); these four conditions determine the constants C1, C2, C3, and C4: E E y(t) = (wt - sin wt), z(t) = (1 --- Coswt). (5.7) ωB ωB In this form the answer is not terribly enlightening, but if we let E R= (5.8) WB' ωB and eliminate the sines and cosines by exploiting the trigonometric identity sin? wt +cos2 1, we find that (y – Rw1)+ (z – R)2 = R2. (5.9) This is the formula for a circle, of radius R, whose center (0, Rwt, R) travels in the y-direction at a constant speed, E v=WR= = (5.10) B The particle moves as though it were a spot on the rim of a wheel, rolling down the y axis at speed v. The curve generated in this way is called a cycloid. Notice that the overall motion is not in the direction of E, as you might suppose, but perpendicular to it.