Question

An electron has a kinetic energy of 4.2 x 10-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 3.3 x 10-5 T. Determine the radius of the path.

The electron in the problem has
Charge of the electron (q) = 1.6*10-19C
Mass of the electron (me) = 9.11*10-31kg

The equation I was told to use is,
r= √(2meKE) / qB

And the answer I got was 3.8 x 10 ^ -12



But that is not the correct answer. Please help!! THANK YOU!

Answer 1

I think I may have used to wrong equation. I tried...

r= √(me*v) / qB

and I found v = 9.6 *10 ^ 6

so r= 1.288


But I am being told that is wrong too...

Answer 2

The force on the electron has to equal the centripetal force.

qBv=mv^2/r

right? solve for r.

Answer 3

It would be helpful to you to stop looking for formulas, and start to try to analyze the basic relationships of force, energy, and motion, and derive your own solutions. There are only about a dozen important relations in all of physics, and the rest come from those.

Answer 4

I didn't mean to resend that message again...

But anyway, based on what you said, the equation I should use is...

(qBv)* (mv^2) = r??

Thank you for your help...

Answer 5

qBv=mV^2/r

r= mv^2/qbv=mv/qB

You need urgently to fix your deficit in math. The language of physics is mathematics, and you will never succeed until you can speak the language.

Answer 6

<b> check my work</b> I did it off the top of my head.

Answer 7

If your interested, the correct answer is 1.65517.

Thank you for your help.