Proton enters magnetic field in spiral trajectory

[songoku]

Homework Statement

A proton moves in a spiral path when enters a magnetic field of 5 T. If the pitch is 5 cm, find the angle between v and B.

Homework Equations

F = B q v sin θ
F = m v² / r

The Attempt at a Solution

I am guessing that I need to break down the velocity into two components but still don't know how to use it. What is pitch actually?

Thanks

 

[Basic_Physics]

That is it went 5 cm up or down after one revolution.

 

[apelling]

The pitch of a spiral is the distance moved after one complete revolution.
The diagram in section 2.5 of this wiki article is fairly clear:-
http://en.wikipedia.org/wiki/Screw_thread

The two velocity components would be
vsinθ perpendicular to B
vcosθ parallel to B

vsinθ should be used in both the equations you give not just the first one.
The two forces can then be equated.
The pitch of the spiral is equal to vcosθ x (time period for one revolution, T) ie distance=speed x time
But T also is equal to 2πr/vsinθ.

Using the above ideas I found an expression for θ that still had one unknown in it. Is there any other information given in the question?

 

[songoku]

That is it went 5 cm up or down after one revolution.

The pitch of a spiral is the distance moved after one complete revolution.
The diagram in section 2.5 of this wiki article is fairly clear:-
http://en.wikipedia.org/wiki/Screw_thread

The two velocity components would be
vsinθ perpendicular to B
vcosθ parallel to B

vsinθ should be used in both the equations you give not just the first one.
The two forces can then be equated.
The pitch of the spiral is equal to vcosθ x (time period for one revolution, T) ie distance=speed x time
But T also is equal to 2πr/vsinθ.

Using the above ideas I found an expression for θ that still had one unknown in it. Is there any other information given in the question?

That's all the information given. I get your idea. We have to know either the speed or radius to be able to solve for angle.

Thanks a lot for your help :)

 

[Nickie]

for your question. The pitch in this context refers to the distance between each revolution of the spiral trajectory. To find the angle between the velocity and the magnetic field, we can use the equation F = Bqv sin θ, where F is the force on the proton, B is the magnetic field strength, q is the charge of the proton, v is the velocity, and θ is the angle between v and B. We can also use the equation F = mv^2/r, where m is the mass of the proton and r is the radius of the spiral path.

First, we need to find the radius of the spiral path. We can use the formula r = mv/qB, where m is the mass of the proton, v is the velocity, q is the charge of the proton, and B is the magnetic field strength. Plugging in the given values, we get r = (1.67 x 10^-27 kg)(v)/(1.6 x 10^-19 C)(5 T) = 10^-8v m.

Next, we can calculate the speed of the proton using the equation F = mv^2/r. Since the proton is moving in a circular path, the force is provided by the magnetic field, so we can say F = Bqv. Setting these two equations equal to each other and solving for v, we get v = Br/q. Plugging in the given values, we get v = (5 T)(10^-8v m) / (1.6 x 10^-19 C) = 3.125 x 10^11 m/s.

Now, we can use trigonometry to find the angle between v and B. Drawing a diagram, we can see that the angle between v and B is the same as the angle between the radius of the spiral path and the tangent to the path at that point. This angle can be found using the tangent function, tan θ = opposite/adjacent. In this case, the opposite side is the pitch (5 cm) and the adjacent side is the radius of the spiral path (10^-8v m). So, tan θ = (5 cm) / (10^-8v m). Solving for θ, we get θ = arctan (5 cm / 10^-8v m) = 2.86 x 10^-9 radians.

In conclusion, the angle between the velocity and the