Thursday

April 17, 2014

April 17, 2014

Posted by **Determine the radius** on Friday, April 3, 2009 at 9:32am.

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!

- physics -
**Determine the radius**, Friday, April 3, 2009 at 9:42amI 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...

- physics -
**bobpursley**, Friday, April 3, 2009 at 9:49amThe force on the electron has to equal the centripetal force.

qBv=mv^2/r

right? solve for r.

- physics -
**bobpursley**, Friday, April 3, 2009 at 9:50amIt 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.

- physics -
**Determine the radius**, Friday, April 3, 2009 at 10:27amI 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...

- physics -
**Determine the radius**, Friday, April 3, 2009 at 10:29amI didnt 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...

- physics -
**bobpursley**, Friday, April 3, 2009 at 10:44amqBv=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.

- physics -
**bobpursley**, Friday, April 3, 2009 at 10:45am**check my work**I did it off the top of my head.

- physics -
**Determine the radius**, Friday, April 3, 2009 at 2:34pmIf your interested, the correct answer is 1.65517.

Thank you for your help.

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