Angle of deviation from a magnetic field

In summary: E/c} = \frac{2rqB}{E} where the last equality comes from the previous equation p = 2rqB. In summary, the proton beam experiences a magnetic force while traveling through a cylinder, causing it to move in a circular orbit with a radius determined by its mass, velocity, and the magnetic field. Upon leaving the field, the beam deviates at an angle θ, which can be calculated using the equation tanθ = (2rqB)/E. The length of the arc and radius of the circle can also be used to find the value of θ.
  • #1
tanaygupta2000
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Homework Statement
Protons of kinetic energy 10^12 eV are injected into a uniform magnetic field of strength
10 Tesla. The magnetic field exists only inside a cylindrical region of diameter 50 cm
and is parallel to the axis of the cylinder. At the point of injection, the proton beam is
directed towards the axis of the cylinder and is perpendicular to it. By the time the beam
exits the magnetic field, it changes its direction by
[Hints: (i) The protons in the beam are ultrarelativistic, i.e., v = c.
(ii) Use the relativistic expressions for energy and momentum
(iii) F = dp/dt]

(a) 2.2e-3 radian
(b) 2.4e-4 radian
(c) 1.0e-2 radian
(d) 1.5e-3 radian
(e) 9.0e-4 radian
Relevant Equations
Relativistic energy equation, E^2 = p^2c^2 + mo^2c^4
Radius of revolution in a perpendicular magnetic field, r = mv/qB
The beam of protons are directed towards the axis of the cylinder, perpendicular to the direction of the field.
While traveling through the cross-section of the cylinder, the proton beam experiences a magnetic force, which tends to move the beam in a circular orbit of the radius given by:

r = mv/qB = (E/c)/qB (Since in E^2 = p^2c^2 + mo^2c^4, given v = c, so rest massmo = 0)

IMG_20200720_055858.jpg


After leaving the field region, the beam will move tangentially from the point of leaving, as I've made in the diagram, causing a deviation θ from the initial incident direction. The question requires to find the value of θ.
Now I think that mathematically, the angle θ subtanded by the tangent to the arc in the diagram with the initial direction will be the same as the angle θ subtanded by the same arc at the center.
And knowing the value of radius of the arc, r = mv/qB = (E/c)/qB (>>> 25cm), I'm having trouble finding the relation between the length of this arc and the radius of the circle in which this arc is present to get the required value of θ.
Please help !
 
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  • #2
Say proton gets momentum from Lorentz force p=Ft. t is time the proton is in cylinder area so t=l/v where l is path length and v is speed of proton. Since deflection is rather small ##l=2\pi r## where r is radius of cylinder
The small deviation angle is ##\theta = \frac{p}{P(v)}##
 
  • #3
anuttarasammyak said:
Say proton gets momentum from Lorentz force p=Ft. t is time the proton is in cylinder area so t=l/v where l is path length and v is speed of proton. Since deflection is rather small l=2πr where r is radius of cylinder
The small deviation angle is
Should that l=2πr be l=2r? Or did I miss something?
 
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  • #4
You are right. I made a mistake.
 
  • #5
anuttarasammyak said:
Say proton gets momentum from Lorentz force p=Ft. t is time the proton is in cylinder area so t=l/v where l is path length and v is speed of proton. Since deflection is rather small ##l=2\pi r## where r is radius of cylinder
The small deviation angle is ##\theta = \frac{p}{P(v)}##
Taking p = Ft = (qvB)*(2r/v) = 2rqB
and p(v) = E/c
I am getting θ = 2crqB/E = 0.0015 radian

which is Option - (d)
But how should I derive this - "##\theta = \frac{p}{P(v)}##"
Thanks !
 
  • #6
Speed or magnitude of momentum does not change during the process. Direction changes with
[tex]tan\theta = \frac{p}{P}[/tex]
 
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Related to Angle of deviation from a magnetic field

What is the angle of deviation from a magnetic field?

The angle of deviation from a magnetic field is the angle between the direction of a charged particle's motion and the direction of the magnetic field it is moving through.

How is the angle of deviation calculated?

The angle of deviation is calculated using the equation θ = qvB/m, where θ is the angle of deviation, q is the charge of the particle, v is the velocity of the particle, B is the strength of the magnetic field, and m is the mass of the particle.

What factors affect the angle of deviation?

The angle of deviation is affected by the strength of the magnetic field, the charge and mass of the particle, and the velocity of the particle. Additionally, the angle of deviation can also be affected by the shape and orientation of the magnetic field.

How does the angle of deviation impact a charged particle's trajectory?

The angle of deviation determines the curvature of a charged particle's trajectory in a magnetic field. A larger angle of deviation results in a more curved path, while a smaller angle of deviation results in a less curved path.

Can the angle of deviation be controlled?

Yes, the angle of deviation can be controlled by adjusting the strength and orientation of the magnetic field, as well as the initial velocity and charge of the particle. This is important in many applications, such as particle accelerators and magnetic confinement fusion reactors.

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