Electromagnetism of particle problem

[MathematicalPhysicist]

Homework Statement

We are given surface current density, K which is constant (K is negative) along the z-axis which is placed in x=a plane (the board which contains this density is neutral), in x=0 plane there's an infinite plane which is charged by a surface density $$\sigma$$ which is constant.
we have a particle with charge q and mass m which starts at rest at r=0.
Now,
1. given sigma find out what is the minimal K s.t the particles doesn't cross the boards.
2. given the K you found assume now K is twice the K in 1, and show that:
the minimal distance from the board with the current is
$$d_{min} = a- \frac{m\sigma}{\pi q (3\sigma^2 +\frac{4m\sigma}{\pi q a})}$$

Homework Equations

There's a file with a picture of the system attached to this post here:
http://img2.tapuz.co.il/forums/1_143953502.pdf
It's question number 1, you shouldn't mind that the text is in hebrew, I translated what it asks in English.

The Attempt at a Solution

For 1 I thought of looking at the force euqation:

mv^2/a = qE+ qv x B where E is the electric field from the charged board and B is from the above board (situated at x=a), and then just need to find out when there's only one solution to this quadratic equation with regards to v, the veclotiy of the particle.

The problem arises when I need to prove (2).
Don't know how to do it.

Thanks.

 

[hikaru1221]

I guess things look like my picture? (see the picture attached) If so, then let me show you a way to obtain the general solution.

It's easily shown that: $$\vec{B}=\frac{\mu _o|K|}{2}\hat{y}$$ and $$\vec{E}=\frac{\sigma}{2\epsilon _o}\hat{x}$$.

The Newton's 2nd law equation for the particle: $$m\frac{d\vec{v}}{dt}=e\vec{E}+q\vec{v}\times\vec{B}$$

Therefore:

(1): $$\frac{dv_x}{dt}=A+Bv_z$$

(2): $$\frac{dv_z}{dt}=-Bv_x$$

$$v_y=0$$ (the particle only moves inside the y=0 plane)

where: $$A=e\frac{\sigma}{2m\epsilon _o}$$ and $$B=\frac{\mu _o|K|}{2m}q$$

From (1) and (2), you can solve for a relation between v_x and v_z. Remember the initial condition: $$\vec{v}(\vec{r}=0)=0$$. From (2), you can solve for the relation between x and v_z. Then from the 2 relations, you should get an equation relating v_x and x. Notice that when the particle reaches to the closest distance to the board with current, v_x=0.

For part 1, we have x_max=a when v_x=0. Solve for K (or |K|). For part 2, we simply substitute K with 2 times value of K from part 1.

By the way, I don't see $$\epsilon _o$$ and $$\mu _o$$ anywhere in d_min needed to show. Maybe the question assumes using another unit system rather than SI system?

P.S: By "mv^2/a = qE+ qv x B", you assumed that at the time the particle reaches x=a, the particle's radius of curvature at that point is a, which hasn't been shown. It can be 2a, 3a, or 1000a; we don't know. You have to show it.

 

[MathematicalPhysicist]

We're using cgs system of units and c=1.
Thanks I think I understand what to do in 1.

as of 2, I am not sure I understand what to do (I get in cgs units and c=1), that for part 1:

K=m/(2qa), after solving the differential equation:

$$d^2v_x/dt^2 = -(\frac{2\pi K q}{m})^2 v_x$$
which with v(0)=0 we get a sine function, and for x=a v_x=0 we get what I wrote in the above.

I think that in part 2, I need to write down x_max=d (where d is the distance from the lower plane situated at x=0), and then again v_x=0, but I don't get the correct final solution.

 

[hikaru1221]

I've never worked with the CGS system before. Maybe that's how the term "pi" arises.
You don't need to solve that differential equation. Just divide (1) by (2), you will get:
$$\frac{dv_x}{dv_z}=\frac{A+Bv_z}{-Bv_x}$$
Then do the integration: $$-\frac{B}{2}v_x^2=Av_z+\frac{B}{2}v_z^2$$
Integrate (2): $$v_z=-Bx$$
From the 2 equations above: $$-\frac{B}{2}v_x^2=-ABx+\frac{B^3}{2}x^2$$
I don't get the answer either. However, I think it's wrong. Consider the extreme case where $$\sigma = 0$$. That means there is no electric field, no initial force to cause the particle to move, and the particle remains at rest at r=0. In that case, d_min=a. But from the answer:
$$d_{min}=a-\frac{m\sigma}{3\pi q\sigma ^2+4m\sigma /a}=a-\frac{m}{3\pi q\sigma +4m/a}$$
when $$\sigma = 0$$ , we get $$d_{min}=3a/4$$ which is wrong.

 

[MathematicalPhysicist]

Ok I understnad what you write, Now when I solve the equation I get K~sqrt(sigma), so indeed if sigma equals zero then also K=0 zero cause otherwise the particle will cross the board situated at x=a.

Again thanks for your help with this question.

 

[hikaru1221]

Now when I solve the equation I get K~sqrt(sigma), so indeed if sigma equals zero then also K=0 zero cause otherwise the particle will cross the board situated at x=a.
.

Logically this is not the reason. We haven't had anything to verify our solution yet, since the only hint (the answer) is different from ours. I only pointed out that the statement d_min =... in the question is logically wrong based on another reasoning, not on our result K~sqrt(sigma).

 

[MathematicalPhysicist]

But it doesn't make sense, even if there isn't an electric field, the particle would feel the force from the B-field, unless there is no also a magnetic field.

 

[hikaru1221]

Initially v=0, so if there is no electric field, even if there is B-field, the particle won't move at all.

 

[MathematicalPhysicist]

I see your point, thanks for clearing this out.