Motion in B Field: Deduce Conditions & Find Angle for x>b

[bon]

Homework Statement

In a particular experiment, a particle of mass m and charge +q moves with speed v along the x-axis towards increasing x. Between x = 0 and x = b there is a region of uniform magnetic flied B in the y-direction. Deduce the conditions under which the particle will reach the region x>b. In the even that it does reach this region, find an expression for the angle to the x-axis at which it will enter it.

Homework Equations

The Attempt at a Solution

I\'m struggling to understand why the particle wouldn't reach the region x>b! Surely it has an x-component of velocity v which is unchanged because the B field produces a perp force (i.e. in the z-direction) so the x-position would grow as x=vt..?
What am i missing? THanks!

 

[Ray Vickson]

Homework Statement

In a particular experiment, a particle of mass m and charge +q moves with speed v along the x-axis towards increasing x. Between x = 0 and x = b there is a region of uniform magnetic flied B in the y-direction. Deduce the conditions under which the particle will reach the region x>b. In the even that it does reach this region, find an expression for the angle to the x-axis at which it will enter it.

Homework Equations

The Attempt at a Solution

I\'m struggling to understand why the particle wouldn't reach the region x>b! Surely it has an x-component of velocity v which is unchanged because the B field produces a perp force (i.e. in the z-direction) so the x-position would grow as x=vt..?
What am i missing? THanks!

When the particle's velocity points along the x-axis the force points in the +z direction. But, what about when the velocity no longer points along the x-axis? You need to
write out the equations of motion in detail.

RGV

 

[bon]

Aha. ok i see.

I work it out to be v > (qB/m)^2 b

is this right?

Thanks!

 

[berkeman]

Aha. ok i see.

I work it out to be v > (qB/m)^2 b

is this right?

Thanks!

Can you show your work for how you arrived at that answer?

 

[bon]

ell from the eom you get

x\'\' = -qB/m z\'

and z\'\' = qB/m x\'

and then solving for x you get x\' = v- (qB/m)^2 x

and so we need x\'(b) to be greater than 0, hence the condition..

I\'m now trying to work out the angle though and am having problems..

 

[bon]

For the angle i get tan theta = (qBb/m) / [v - b(qB/m)^2 ]

This seems too complicated! Have i gone wrong?

 

[Ray Vickson]

When I (or, rather, Maple) solve the DEs m*dvx/dt = -qB*vz and m*dvz/dt = qB*vx with vx(0)=v and vz(0)=0 I get a solution very different from yours. I then get a crossing condition different from yours, and a tangent of the crossing angle more complicated than yours.

Hint: try a solution of the form sin(w*t) and/or cos(w*t) for an appropriate w.

RGV

 

[berkeman]

ell from the eom you get

x\'\' = -qB/m z\'

and z\'\' = qB/m x\'

and then solving for x you get x\' = v- (qB/m)^2 x

and so we need x\'(b) to be greater than 0, hence the condition..

I\'m now trying to work out the angle though and am having problems..

For the angle i get tan theta = (qBb/m) / [v - b(qB/m)^2 ]

This seems too complicated! Have i gone wrong?

Another way to approach the problem is to realize that the charged particle will trace a circular path due to the perpendicular magnetic field, and calculate the Lamor radius of that motion...

http://en.wikipedia.org/wiki/Larmor_radius

.

 

[bon]

Another way to approach the problem is to realize that the charged particle will trace a circular path due to the perpendicular magnetic field, and calculate the Lamor radius of that motion...

http://en.wikipedia.org/wiki/Larmor_radius

.

Ok thanks I've now seen where I went wrong.

I get x = v/w sin wt where w = qB/m

So the condition is v > qbB/m

is this right by any chance?

Thanks!

 

[bon]

I get the angle to be given by:

tan theta = cot[ arcsin(wb/v) ]

 

[berkeman]

Ok thanks I've now seen where I went wrong.

I get x = v/w sin wt where w = qB/m

So the condition is v > qbB/m

is this right by any chance?

Thanks!

Drawing the circular orbit shows that r = b for the Lamor radius to keep the particle from going past b. But does the Lamor radius go up or down with velocity?

 

[bon]

But if x = v/w sin wt then x max = v/w

so it reaches it x max > b i.e. v > bw surely?

Also is my angle right?

thanks.

 

[berkeman]

But if x = v/w sin wt then x max = v/w

so it reaches it x max > b i.e. v > bw surely?

Also is my angle right?

thanks.

Sorry, I'm not tracking your sin(wt) path -- I was just using the equation for the Lamor radius from the page I linked for you.

 

[bon]

hmm surely the same applies?

the radius gets bigger with v, so my equation is right?

Is the angle right also?

 

[berkeman]

hmm surely the same applies?

the radius gets bigger with v, so my equation is right?

Is the angle right also?

Yes, the radius gets bigger with v. But in your answer above, you wrote "v > qbB/m" for containment...

I didn't check the angle -- if you could post a sketch, that would make it easier to check.

And please don't call me Shirley.

 

[bon]

No - I said v > ... for the particle to reach x>b, which is what the question asks! :S

And I didn\'t call you Shirley.

 

[berkeman]

No - I said v > ... for the particle to reach x>b, which is what the question asks! :S

And I didn\'t call you Shirley.

Oh nuts, I misread the original question! So your solution is correct. Sorry I haven't checked the angle, but it's probably right since you understand what the orbit looks like.

 

[berkeman]

I get the angle to be given by:

tan theta = cot[ arcsin(wb/v) ]

I've had time to look at the entry angle for cases when the orbit radius r > b. I get a simpler answer than you did, and different. For one thing, the mass m is not in your equation, but it affects the radius of the orbit. And w is not one of the fundamental things you are given in the problem, so your answer should probably be expressed in terms of m, B, q and v.

I used a sketch of the orbit and the x-y axes, and used the triangle that is formed by the radius of the orbit and the dimension b inside the orbit circle...