A proton has a speed of 3 x 10^6 m/s

  • Thread starter Dx
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  • #1
Hello,
I need help deriving a formula to solve for this please.

A proton has a speed of 3 x 10^6 m/s in a direction perpendicular to a uniform magnetic field and the proton moves in a circle radius .2 m. what is the magnitude of the magnetic field?

If someone can show me the all formulas to use i can manipulate from here.
Thanks!
Dx :wink:
 
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  • #2
HallsofIvy
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A particle with mass m, charge q, moving at speed v in a constant magnetic field of magnitude B at right angles to the velocity vector of the particle, will move in a circle with radius mv/(qB)
 

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