# B-field and Magnetic forces, speed of a particle

Let's say I have a particle moving perpendicular to a uniform magnetic field of magnitude $x \ T$, and it moves in circle with a fixed radius. How do I find the speed of this electron? Initially I though to fusing the Biot-Savart law, but then the mass of the particle wouldn't affect it, so I figured that's not the right approach. Can anyone point me out to the right direction?

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Hint: Biot-Savart Law tells you how a moving charge creates a magnetic field. You need to find an equation that given a magnetic field, creates a force on a moving charged particle. The magnetic field in your problem is created by some other moving charges 'off stage'. – DJBunk Apr 1 '13 at 13:35
@DjBunk so my approach using the Biot-Savart law is correct? – Shelby. S Apr 1 '13 at 13:41
No, as DJ said : you need not calculate $\vec B$ but you have to visualise the motion of particle in effect of $\vec B$.You just deal like particle is experiencing a force $\vec F=q(v\times \vec B)$ and proceed. – ABC Apr 1 '13 at 13:44
Thank you, I'll see what I can figure out. – Shelby. S Apr 1 '13 at 13:45

Force $\vec F=q(\vec v\times \vec B)$ acts on particle . As $\vec F \perp \vec v$ the paricle moves in a circular path with constant speed.
This Magnetic foce will provide the particle necessary centripetal force for moving in circular motion. So, $$F=qvB=mv^2/r \ ;\text{ taking magnitudes.}$$