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stuff12's avatar

Physics help...

Asked by stuff12 (245points) March 16th, 2008 from iPhone

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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4 Answers

Zaku's avatar

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

Zaku (30375points)“Great Answer” (0points) Flag as…
itsnotmyfault1's avatar

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 (203points)“Great Answer” (1points) Flag as…
Zaku's avatar

Self-censoring my own tease… never mind.

Zaku (30375points)“Great Answer” (0points) Flag as…

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