Tricky Magnetic Field Question (Proton Enters Magnetic Field )

In summary, the problem involves a proton entering a magnetic field at a right angle and experiencing circular motion. The initial strategy of using the relation between centripetal force and magnetic force led to a radius of 5.18 meters, which did not make sense. The correct approach is to use the angle formed by the radius vector in the diagram and the horizontal, which is the same angle as the particle's path turning from the vertical. The effect of gravity is negligible due to the uniform magnetic field and the particle's initial kinetic energy in the vertical direction. The change in energy is 0, which can be proven by calculating the energy at two different points and solving for the final velocity, which is the same as the initial velocity.
  • #1
Plutonium88
174
0
Tricky Magnetic Field Question... (Proton Enters Magnetic Field...)

Here are 2 pictures.

1. This is the written question: (It's a long one.)
CHARGE1.jpg


2. Here is the given diagram for the question..
c_HARGEDIAGRAM.jpg


Possible Equations to use:
fm = qvbsinAThoughts / attempts:

Initially i thought because the charge entered the field perpendicular to both the magnetic force.. that it would be a relation between Centrepetal force and magetnic force..

FM = FC

I solved for this and found the radius to be 5.18meters... which amkes no sense, so obviusly this technique can't be used..

Another strategy I'm thinking at the moment, is perhaps the angle at which the particle deviates could be related to the Radius and small (d) distance triangle... to solve using pythagereom thereom...However I'm not sure how i could solve this angle, unless i relate the Magnetic Force to something.. Perhaps to Fnet = Ma, FM = Fneti'm also having a problem considering the accelleration, and i would like to know, Gravity is involved in the acceleration of this object at all, cause it does not mention that gravity is negligible,

Honestly, I just need a couple tips to be pointed in the right/ how to consider things... and last but not least...I am really thinking, that the direction of the Force (FM) is in the direction where the charge is accelerating constantly(to the left), and that the direction it travels 'upward' from the diagram, is the direction in which the object travels in uniform motion, with no acceleration.. I believe this because the magnetic field vector is in the direction in which deflection occurs... (left) and thus must only be able to affect that 'horizontal' components.. leaving the vertical component unaffected...

Another reason i believe there is no accelleartion 'vertically' (from diagram) is because the charges initially velocity, has only one component.. It reminds me of When you reach the Maximum H in a projectile... Where your component vertically is 0, and your horizontal component is (vx*T) in relation to this..

I can see that after entering the field, the Charge gets two velocity components...

So yea.. please give me some reccomendations ppzl <3So yea any pointers... I don't want to throw any solutions up yet.. I just want some advice before i put something half decent up and worth some ones time to look at.I really want to treat it like a projectile... in some way shape or form *******
 
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  • #2


Plutonium88 said:
Thoughts / attempts:

Initially i thought because the charge entered the field perpendicular to both the magnetic force.. that it would be a relation between Centrepetal force and magetnic force..

FM = FC

I solved for this and found the radius to be 5.18meters... which amkes no sense, so obviusly this technique can't be used..
Why do you think that this radius makes no sense? It's a radius of curvature of the path; it doesn't have to correspond to any particular dimension of the apparatus.
Another strategy I'm thinking at the moment, is perhaps the angle at which the particle deviates could be related to the Radius and small (d) distance triangle... to solve using pythagereom thereom...
Look at the diagram provided and where the shown radius vector is "rooted". The angle that this vector makes with the horizontal will be the same angle that the particle's path turns from the vertical (you should convince yourself that this is so).

At the outset you also might want to convince yourself that the effect of gravity will be negligible. What's the initial kinetic energy of the particle in the vertical direction, and what's the change in energy that can occur?
 
  • #3


gneill said:
Why do you think that this radius makes no sense? It's a radius of curvature of the path; it doesn't have to correspond to any particular dimension of the apparatus.
Because.. I assumed that the radius would kind of moderately proportional to the diagram, which i should never think, but honestly i showed another student, and he's like that makes no sense... So i kind took his opinion.. which i shouldn't have...
gneill said:
Look at the diagram provided and where the shown radius vector is "rooted". The angle that this vector makes with the horizontal will be the same angle that the particle's path turns from the vertical (you should convince yourself that this is so).

Okay this is clear, thank you.
gneill said:
At the outset you also might want to convince yourself that the effect of gravity will be negligible. What's the initial kinetic energy of the particle in the vertical direction, and what's the change in energy that can occur?

Okay I'm going to start looking more closely at the energies, do you mind if i post back what i have in terms of a solution later tonight/tomorrow?
 
  • #4


Plutonium88 said:
Okay I'm going to start looking more closely at the energies, do you mind if i post back what i have in terms of a solution later tonight/tomorrow?

No problem. Take your time :smile:
 
  • #5


since this is a uniform magnetic field, and the charge enters at a right angle,

the charge experiences circular motion. There is a constant force perpindicular to its motion.

therefore..
A)
Fm = Fc

R = mv/qb

R= 5.14m

b) now i was thinking in terms of that angle, if my original radius is correct, i should be able to make a trianglee... with the Radius R, and the distance (d) of the magnetic field

and solve..

SinA=d/R
A=2.79 degrees

now.. i don't really want to move on, yet because I'm not positive if this is an okay strategy to solve for this angle.

and in terms of energy, i did a calculation...

Et1= 1/2MVo^2 (initial point) potential energy lvl zero..

The change in energy i believe would be, no change.. There are no non conservative forces acting upon the charge...

and in circular motion work done on an object is = 0 because the force is perpindicular to the velocity..

so therefore the change in energy is 0...?

and this can be proven i noticed when i calculated

Et2 = 1/2Mvf^2 + mgd

If i make energy totals the same and solve for Vf

Et1 = Et1
1/2mvo^2 = mgd + 1/2mvf^2
Vf = sqrtall(mv^2 -2gd)
Vf = 145000
which is its original velocity

can you tell me how I'm doing :O
 
  • #6


Plutonium88 said:
since this is a uniform magnetic field, and the charge enters at a right angle,

the charge experiences circular motion. There is a constant force perpindicular to its motion.

therefore..
A)
Fm = Fc

R = mv/qb

R= 5.14m

b) now i was thinking in terms of that angle, if my original radius is correct, i should be able to make a trianglee... with the Radius R, and the distance (d) of the magnetic field

and solve..

SinA=d/R
A=2.79 degrees

now.. i don't really want to move on, yet because I'm not positive if this is an okay strategy to solve for this angle.
No, it works fine. That's a good value for the angle.
and in terms of energy, i did a calculation...

Et1= 1/2MVo^2 (initial point) potential energy lvl zero..

The change in energy i believe would be, no change.. There are no non conservative forces acting upon the charge...

and in circular motion work done on an object is = 0 because the force is perpindicular to the velocity..

so therefore the change in energy is 0...?

and this can be proven i noticed when i calculated

Et2 = 1/2Mvf^2 + mgd

If i make energy totals the same and solve for Vf

Et1 = Et1
1/2mvo^2 = mgd + 1/2mvf^2
Vf = sqrtall(mv^2 -2gd)
Vf = 145000
which is its original velocity

can you tell me how I'm doing :O
Doing okay so far.

For the check to make sure that gravity is an insignificant factor, you might have just compared the initial kinetic energy with the maximum change in potential energy across the extent of the apparatus.
 
  • #7


ughhhhhhhh. alright... umm I'm going to have to think about this one some more then...

but i have to solve from a time t1, and the x1 is the angle of deflection... i thought that kinematics would be required in that case ( b, and c)
 
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  • #8


I don't understand why you're using formulas for motion under gravitational acceleration when you've decided that gravity can be ignored.

The particle's path curves while it is within the magnetic field, after which it travels in a straight line. After the "bend" in the trajectory due to the magnetic field, the problem boils down to geometry.
 
  • #9


okay so..using the angle i had originally 2.79

i can use that angle to solve for x1
becaues the angle is the point which the proton deviates.

tanA = d/x1
x1=d/tanA

now in order to get t1 i need the hypoteneuse where it travels..

cosa = d/z
z= d/cosA

z = vot
t= z/vo
 

1. What happens to a proton when it enters a magnetic field?

When a proton enters a magnetic field, it experiences a force due to the interaction between its charge and the magnetic field. The direction of the force is perpendicular to both the direction of motion of the proton and the direction of the magnetic field. This causes the proton to move in a curved path, known as a circular motion.

2. How does the strength of the magnetic field affect the path of the proton?

The strength of the magnetic field affects the path of the proton by determining the amount of force it experiences. A stronger magnetic field will cause the proton to curve at a sharper angle, while a weaker magnetic field will result in a wider curve.

3. Why does the proton move in a circular path?

The proton moves in a circular path due to the Lorentz force, which is the force exerted on a charged particle in a magnetic field. This force is always perpendicular to the direction of motion and causes the proton to continuously change direction, resulting in a circular path.

4. How does the velocity of the proton affect its path in a magnetic field?

The velocity of the proton also plays a role in determining its path in a magnetic field. The faster the proton is moving, the stronger the force it experiences and the tighter the curvature of its path. If the proton is moving too slowly, it may not have enough force to overcome the inertia of its original path and will continue moving in a straight line.

5. Can the direction of the magnetic field affect the path of the proton?

Yes, the direction of the magnetic field can affect the path of the proton. If the magnetic field is reversed, the direction of the force on the proton will also be reversed, causing it to curve in the opposite direction. Additionally, the angle between the direction of the magnetic field and the direction of motion of the proton can also affect its path, resulting in a spiral motion rather than a perfect circle.

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