How Many Turns Does a Proton Make in a Solenoid?

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The discussion focuses on calculating the number of turns a proton makes while traveling through a solenoid with specific parameters. Key points include the need to determine the magnetic field inside the solenoid using the formula B = μ₀in, where n is the turns per length. The proton's motion is analyzed as circular due to the magnetic field, requiring the calculation of the radius and period of motion. The angle of the proton's velocity vector affects the components of motion, with only the perpendicular component contributing to the circular path. Participants emphasize the importance of correctly applying physics equations to solve for the total turns and time taken.
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Homework Statement


Hello,

The question of the problem is: Solenoid, calculate laps of a proton?

Someone could help me solve this exercise, but I try not to do this achievement . Would appreciate your help . This is the information that I provided to do the exercise:

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A proton with velocity v inside a solenoid of 10,000 laps , Long L and 1 A current (see figure) is triggered. The proton velocity vector has an angle of 40 ° with respect to the magnetic field direction . Calculate the turns while supplementing the proton travels a distance L , the time to complete each lap and total time T to travel L.

The values ​​of the magnitude of the velocity ( in m / s ) and the length of the coil ( L in m) chosen according to the first letter of your name using the following code :
A- 10 B- 20 C- 30 D -40 E -50 F- 60 G -70 H-80 ...
N -140 O- ​​150 P -160 Q -170 R -180 S -190 U - 200 U -210 V -220 W -230 ...

If your name is Domingo Pérez then v = 40 m / s and L = 160 m . If your name will begin with the same letter then use the number of the next letter to the family name. So if your name is Candido Colón use v = 30 m / s and L = 40 m .

Homework Equations


The Attempt at a Solution


I try so many times to solve the problems, but for me is impossible to solve, i really appreciate all the help.
 
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Ok, well the first thing to do will be to find the magnetic field inside the solenoid:
##B = \mu_{0}in##, where ##n = \frac{turns}{length}##

Then since the proton moves across field lines, it'll go into circular motion, so think of these equations:
##F_{r} = ma_{r} = m\frac{v^{2}}{r}##

Remember that only the vertical component of the velocity will cause the circular motion.

So from there you can find the period of the motion, pitch of the helix, etc.
 
Data:

Turns: 10,000
Lenght: 180m
Electic current: 1A
Velocity vector: 40°
Magnitude of the velocity: 100 m/s

This is what it got at the moment:

n=\frac{10,000}{180}
n=55.6m

B=(4πx10^{-7}T*m/A)(1A)(55.6m)
B=6.99x10^{-5}

Now i need to Calculate the turns the proton complete while it travels a distance L, the time to complete each turn T and the total time to travel L.

I don't have any clue, someone can give a help to approach the problem?

I aprreciate your help, thank you guys.
 
Well, try starting with the period, T. Period is defined as: ##T = \frac{2\pi r}{v}##, which is just the circumference of the circular path divided by the speed, so can you solve for T from here?
 
Well, i don't have r. But i saw a video in which they calculate R in the following way:

R=(lenght/2∏)
R=(180m/2∏)
R= 28.65

T= \frac{(2∏)(28.65)}{100}
T=1.80

Turns: 10,000
Lenght: 180m
Electic current: 1A
Velocity vector: 40°
Magnitude of the velocity: 100 m/s

Im not sure if the calculates are correct :(
 
Well, r is the radius of the circular path (which will be a helix), so you don't want to use the 180m -- that's the length of the solenoid, not the circumference of anything. Look at the equation for circular motion I posted above -- you can solve for the radius from that.

Also, since the particle is launched at an angle, it won't be the entire magnitude that goes into creating the circular path -- it will only be the component that's perpendicular to the field (since magnetic fields only act on things moving perpendicularly to them). So that's where the angle comes into play. The component that's instead parallel to the field is responsible for moving the particle down the solenoid, and pulling the circular path out into a helix.
 
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