1. Dec 7, 2004

### agirlsrepublic

a velocity selector consists of magnetic and electric fields. if B= 0.015 find the value of E such that a 750 eV electron is moving along the + x- axis and is undeflected.

this is what i did..

the magnetic force=electric force so qvB=qE or vB=E
K= .5 mv^2 and r= mv/qB so k =[.5r^2q^2B^2]/m and if you simplify you get:
r= [(2km)^.5]/qB

r= {[2*750*1.6*E-19*9.11E-31]^.5}/1.6E-19*.015= 1.38E-16

if v=rqm/m then v = [1.38E-6*1.6E-19*.015]/9.11E-31 = 3.64E3

since E= vB then E= 3.64E3*0.15= 5.45E1

but this is not the answer

thanks!

2. Dec 7, 2004

### Staff: Mentor

Right.
No. $K = 1/2 m v^2$, so $v = \sqrt{2K/m}$. Thus $E = B \sqrt{2K/m}$.

3. Dec 7, 2004

### agirlsrepublic

thanks..i appreciate it

4. Dec 7, 2004

### dextercioby

Well,Doc,for her sake,let's hope she got where your equation with E,B and v came from.If not,i'm willing to explain to her everything in detail.Yet,by the look of her last post,she aint't got a clue.

5. Dec 7, 2004

### agirlsrepublic

listen buddy.. if i know that E=vB then i know where he got the equation from

6. Dec 7, 2004

### dextercioby

Yo,pal, u definitely didn't show it.Else u wouldn't have gotten yourself in messy & useless calculations.My guess... :tongue2: