What is the radius of curvature of an electron beam in a uniform magnetic field?

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SUMMARY

The radius of curvature of an electron beam in a uniform magnetic field can be calculated using the equation \( r = \frac{Mv}{qB} \). In this scenario, an electron beam with an energy of 2.0 MeV enters a magnetic field of 0.05 weber/m². To find the velocity of the electrons, the relativistic energy formula \( E = \gamma mc^2 \) must be applied, where \( \gamma = (1 - \beta^2)^{-1/2} \) and \( \beta = v/c \). The mass-energy equivalence for an electron is 0.511 MeV, which is essential for accurate calculations.

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Physics students, particle physicists, and engineers working with electron beams and magnetic fields will benefit from this discussion.

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MeV related to Velocity ??

Homework Statement


An electron beam from a 2.0 MeV Van De Graff enters at right angles to a uniform magnetic field of .05 weber/m^2. What is the radius of curvature of the path of the elctrons?

Homework Equations


A couple of them. First off what the hell is a weber... after taking 3 years of physics in college this is my first encounter with such a unit. But more importantly... how do i convert MeV into Velocity?


The Attempt at a Solution



I know

qv x B = (Mv^2)/r
v=E/B

I tried converting MeV into joules and then plugging it into the Kinetic energy equation .. no go.. i started getting velocities much bigger than the speed of light. Any help is appreciated, as this also is used in the next problem.
 
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The weber is a unit of magnetic flux, equal to one telsa-meter^2. See
http://en.wikipedia.org/wiki/Tesla_(unit)

You need to use a relativistic formula to get the velocity, E=\gamma mc^2, with \gamma=(1-\beta^2)^{-1/2} and \beta=v/c. For the electron, mc^2=0.511\,\rm MeV. (This is how particle physicists always quote particle masses, by the rest energy mc^2; it's much more convenient than converting everything to MKS units.)
 

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