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

D

Dx

Guest
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:
 
Last edited by a moderator:

HallsofIvy

Science Advisor
41,626
821
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)
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top