Kinetic Energy and Magnetic Fields

[PhilCam]

1. Homework Statement [/b
An electron has a kinetic energy of 5.80 10-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 5.10 10-5 T. Determine the radius of the path?



I know

KE = 1/2mv^2

Using KE = 1/2mv^2 and saying KE = 5.8 x 10^-17, and m = 9.10938 x 10^-31 KG

I get that v= 11284559 m/s

Then using F = mv^2/r, I end up with an answer of 2.27 x 10^-12 m.

It tells me this answer is wrong.

Any help would be great.

 

[korican04]

1. Homework Statement [/b
An electron has a kinetic energy of 5.80 10-17 J. It moves on a circular path that is perpendicular to a uniform magnetic field of magnitude 5.10 10-5 T. Determine the radius of the path?



I know

KE = 1/2mv^2

Using KE = 1/2mv^2 and saying KE = 5.8 x 10^-17, and m = 9.10938 x 10^-31 KG

I get that v= 11284559 m/s

Then using F = mv^2/r, I end up with an answer of 2.27 x 10^-12 m.

It tells me this answer is wrong.

Any help would be great.

I obtained the same v as you, but what did you use for the force?
I used F=q(vxB), in this case v cross B is vB because they are perpendicular.
F= qvB = mv^2/r
r=mv/qB

 

[PhilCam]

Ahh thanks, I was using B for force, instead of multiplying by the charge of the electron.

Correct answer is 1.26 m.