Physics question from my test, having a hard time solving it (Lorentz Force)

[SevenSeeds5]

Homework Statement

http://www.solutioninn.com/physics/...ith-charge-2.15uc-and-mass-3.20-x-10-11-kg-is

I am just going to link this because I don't know how to draw a nice diagram like that and put it here. But essentially that is the question. More specifically I have problems with part d)

Homework Equations

F qvBsin(theta)
W= Fd = EkE (I used this to solve the question but I got it wrong.)

The Attempt at a Solution

Some answers suggest to use trigonometry to find the second deflection. However, how can we know that the angle of the second deflection is same as the angle of the arc?
I figure we need to do something with kinematics but I can't seem to figure it out.

 

[berkeman]

Homework Statement

http://www.solutioninn.com/physics/...ith-charge-2.15uc-and-mass-3.20-x-10-11-kg-is

I am just going to link this because I don't know how to draw a nice diagram like that and put it here. But essentially that is the question. More specifically I have problems with part d)

Homework Equations

F qvBsin(theta)
W= Fd = EkE (I used this to solve the question but I got it wrong.)

The Attempt at a Solution

Some answers suggest to use trigonometry to find the second deflection. However, how can we know that the angle of the second deflection is same as the angle of the arc?
I figure we need to do something with kinematics but I can't seem to figure it out.

Can you please show us your work on this, so we can look for errors? Thanks. :-)

 

[SevenSeeds5]

F = qvBsin(theta)
F = (2.15 x 10^-6)(1.45 x 10⁵)(0.42T)
F = 0.13N [Towards the centre]

W = F*d = E_net
(0.13N)(x) = 1/2mv²
(0.13N)(x) = 1/2(3.2 x 10^-11)(1.45 x 10⁵)²
x = 2.59m

I am 99% sure that I cannot use force since the force on the particle disappears after it leaves the field.
I reckon that I need to do something with kinematics (find time it took for a particle to travel on y- direction) use that and a projectile motion equation?

 

[rude man]

F = qvBsin(theta)
F = (2.15 x 10^-6)(1.45 x 10⁵)(0.42T)
F = 0.13N [Towards the centre]

W = F*d = E_net
(0.13N)(x) = 1/2mv²
(0.13N)(x) = 1/2(3.2 x 10^-11)(1.45 x 10⁵)²
x = 2.59m

I am 99% sure that I cannot use force since the force on the particle disappears after it leaves the field.
I reckon that I need to do something with kinematics (find time it took for a particle to travel on y- direction) use that and a projectile motion equation?

Computing work W makes no sense.
Once outside the B field the force is zero, so the particle will continue in a straight line to the wall, or so Mr. Newton posited. Since you computed Δx1 alrerady, calculating Δx is a a piece of cake.

 

[SevenSeeds5]

I realized this morning that the path of a particle after it leaves a field is actually a tangent of the arc, which means the angle is 90 degrees! I think I know how to solve this question now thanks :)

 

[rude man]

[QUOTE="SevenSeeds5, post: 4983557, member: 520137
" ... the angle is 90 degrees! [/QUOTE]
?
Does any angle LOOK like 90 deg. on your figure?

 

[SevenSeeds5]

I mean, that's just a diagram... but isn't the pathway of the particle after it has left the field perpendicular to the radius of the arc?
If this is not true, I don't know how to solve this.

 

[rude man]

If by "radius" you mean the line from the center of curvature to where the B field ends, that is correct.

The path of the particle from that point on until it hits the wall is a straight line tangent to the curve the particle described within the B field at the point where the particle exited the field.