jkwiens.com: Answer: Particle in electric and magnetic field

Answer: Particle in electric and magnetic field

Saturday, 1 of December , 2007 at 11:53 am

Question: Write down the equation of motion for a charged particle in uniform, parallel electric and magnetic fields, both in the z-direction, and solve it, given that the particle starts from the origin with the velocity . A screen is placed at x=a where $a \ll \frac{mv}{qB}$ . Show that the locus of points of arrival of particles with given m and q, but different speeds , is approximately a parabola.

Answer: The equation of motion is given by the Lorentz Force equation:

$m\ddot{r} = q \vec{E} + q \dot{r} \times \vec{B}$

However, since $\vec{E} = E \hat{k}$ and $\vec{B} = B \hat{k}$ for this problem, the equation of motion becomes

$m\ddot{r} = q E \hat{k} + q \dot{r} \times B \hat{k}$

which reduces to

$m \ddot{x} = q\dot{y}B, m\ddot{y} = - q\dot{x}B, m \ddot{z} = qE$

First, lets solve $\ddot{z} = \frac{qE}{m}$ . If we integrate $\ddot{z}$ , we get

$\dot{z}(t) = \int \frac{qE}{m}dt = \frac{qE}{m}t + C$

since the initial velocity at t=0 is , this means $\dot{z}(0) = 0$ .

Therefore, $\dot{z} = \frac{qE}{m}t$ .
If we integrate again, we get:

$z(t) = \int \frac{qE}{m}tdt = \frac{qE}{2m}t^{2} +C$

since the initial position at t=0 is , this means .

Therefore, $z = \frac{qE}{2m}t^{2}$ .

Next, we will reduce the other two equations

$\ddot{x} = \frac{qB}{m}\dot{y}$

$\dot{x} = \int \frac{qB}{m}\dot{y}dt = \frac{qB}{m}y + C$

since $\dot{x}(0) = v_0$ , therefore $\dot{x} = \frac{qB}{m}y + v_0$ .

We can similar reduce $\ddot{y}$

$\ddot{y} = -\frac{qB}{m}\dot{x}$

$\dot{y} = -\int\frac{qB}{m}\dot{x}dt = -\frac{qB}{m}x +c$

since $\dot{y}(0) = 0$ , therefore $\dot{y} = -\frac{qB}{m}x$ .

If we put these equations into the original equations of motion, we can decouple the x and y terms.

$\ddot{x} = \frac{qB}{m}(-\frac{qB}{m}x) = - (\frac{qB}{m})^{2}x$

which has a general solution of

$x = Asin(\frac{qB}{m}t) + Bcos(\frac{qB}{m}t)$

However, since , therefore $x = Asin(\frac{qB}{m}t)$ .
In order to calculate A, we need to take the derivative.

$\dot{x} = \frac{AqB}{m}cos(\frac{qB}{m}t)$

since $\dot{x}(0) =v_0$ , the following reduction can be made $\frac{AqB}{m} = v_0 \Rightarrow A = \frac{m v_0}{qB}$ .

Therefore, $x = \frac{m v_0}{qB}sin(\frac{qB}{m}t)$ and $\dot{x} =v_0 cos(\frac{qB}{m}t)$ .

Also, since $\dot{x} = \frac{qB}{m}y +v_0$ , we can solve for y

$v_0 cos(\frac{qB}{m}t) = \frac{qB}{m}y + v_0$

$y = \frac{v_0 m}{qB}(cos(\frac{qB}{m}t) - 1)$

This means the equation of motion is:

$x = \frac{m v_0}{qB}sin(\frac{qB}{m}t)$

$y = \frac{v_0 m}{qB}(cos(\frac{qB}{m}t) - 1)$

$z = \frac{qE}{2m}t^{2}$

Next we will analysis what happens when x=a

$x = a = \frac{m v_0}{qB}sin(\frac{qB}{m}t)$

$t = \frac{m}{qB}sin^{-1}(\frac{aqB}{m v_0})$

Next take the taylor expansion

$t = \frac{m}{qB} $\displaystyle\sum_{n=1}^{\infty}\frac{(2n)!}{4^{n}(n!)^{2}(2n+1)}(\frac{aqB}{mv_0})^{2n+1}$$

$t = \frac{m}{qB} ((\frac{aqB}{m v_0} + \frac{1}{66}(\frac{aqB}{m v_0})^{3}+\cdots))$

Since $a \ll \frac{mv}{qB}$ , this means $\frac{aqB}{m v_0} \ll 1$ . This means that a good approximation of t would be the first term:

$t = \frac{m}{qB} (\frac{aqB}{m v_0}) = \frac{a}{v_0}$

Next. lets find $y(\frac{a}{v_0})$

$y = \frac{v_0 m}{qB}(cos(\frac{qBA}{m v_0}) - 1)$

Again, we will take the taylor expansion

$y = \frac{v_0 m}{qB}($\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n+1)!}(\frac{qBa}{m v_0})^{2n+1}$ - 1)$

$y = \frac{v_0 m}{qB}((1 - \frac{1}{6}(\frac{qBa}{m v_0})^{2}+\cdots) - 1)$

Since $a \ll \frac{mv}{qB}$ , this means $\frac{aqB}{m v_0} \ll 1$ . This means that a good approximation of y would be the first two term:

$y = \frac{v_0 m}{qB}(1 - \frac{1}{6}(\frac{qBa}{m v_0})^{2} - 1)$

$y = -\frac{1}{6}\frac{qBa^{2}}{m}(\frac{1}{v_0})$

Next, lets find $z(\frac{a}{v_0})$

$z = \frac{q E}{2m}(\frac{a}{v_0})^{2} = \frac{q E a^{2}}{2m}(\frac{1}{v_0})^{2}$

If we merge the results from $y(\frac{a}{v_0})$ and $z(\frac{a}{v_0})$ , we get the following relation:

$z = \frac{q E a^{2}}{2m}(\frac{1}{v_0})^{2} = \frac{q E a^{2}}{2m}(-6\frac{m}{qBa^{2}}y)^{2}$

$z = 18 \frac{mE}{qa^{2}B^{2}}y^{2}$

This relation shows that the locus of points on the screen with given m and q, but different speeds , is approximately a parabola.

Category: Answers, Electricity and Magnetism

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