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A proton starts with a velocity of 5.7 x105 m/s and is accelerated through a potential...

A proton starts with a velocity of 5.7 x105 m/s and is accelerated through a potential difference of 2.5 kV. It then enters a magnetic field of 1.3 T. What is the radius of curvature of the path the proton will take? (ANS: 7.2mm)

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Answer #1

The charge of a proton is

Therefore, its change in kinetic energy is

Initial kinetic energy of the proton

Final Kinetic Energy of the proton is

Final velocity of the proton is

To calculate the radius of curvature, we note that for a circular path of the particle, the centrifugal force should be equal to the magnetic force.

Rearranging this, we get

Since

Substituting these into our equation, we get

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