Physics help...

Asked by stuff12 (220

) | asked March 16th, 2008 | 4 responses | “Great Question” (0

an electron moving with a speed of 2×10^6 m\s perpendicular to a uniform magnetic field of 10^-3 T will execute one revolution of a circular path in a time most nearly
(a) 1s (b) 10^-6s© 10^-8s (d) 10^-10s (e) 10^-15s

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No one is your class will talk to you about this? What part of that question don’t you understand?

Zaku (2266

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Is it©?

Lightlyseared (2946

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asked 4 weeks ago by this time… but i know the answer!
and why!

F= vq x B
F sub B = F sub C (a centripital force is being supplied by the magnetic force)
so vq x B = mv^2/r
r = mv/qB

2 pi r / v = path length/ velocity = period (this part is the only part i’m not sure about)
then T=2pi m/qB (where pi = pi, m= mass of an electron, q= charge on an electron, B=strength of magnetic field.)

I did this last week in AP physics, so i’m guessing you’re a HS student like me, in an AP class.

itsnotmyfault1 (49

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Self-censoring my own tease… never mind.

Zaku (2266

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