A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.25 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Be =

Question

Alberto Leach

Physics

20 August, 04:36

A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.25 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Be =

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Bradford · Answer 1

Answer: Be = 1.363 * 10 ^-4 T

Explanation:

From r = mv/qB

For proton with mass Mp = 1.67 * 10^-27 and charge + e and field Bp = 0.25T

r = mpv / + e * Bp where v is the velocity

For electron with mass Me = 9.11*10^-31, charge - e and field Be, the radius is

r = mev / - e * Be

Note that radius and velocities are the same

r = mev / + e * Bp = mev / - e * Be

Be = meBp/Mp

= 9.11*10^-31 * 0.25 / 1.67 * 10^-27

= 1.363*10^-4 T