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Homework Statement
A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.460 T magnetic field?
Note: I hope this question is meant in advanced physics >.<
If not, I apologize. Please direct me towards the proper place.
Homework Equations
r = (mv)/(qB)
E = qV
K = ½mv^2
The Attempt at a Solution
I let
E=K and so
½mv^2 = qV
I solved for v which gave me
v = √((2*q*V)/m)
I subbed this v into
r = (mv)/(qB)
I know from the information provided that
B = 0.460 T
m = 6.68*10^-27 kg
V = 4.00*10^3
and I know q = 2*e
I also found v so there aren't any other unknowns, I subbed in...
I got 0.0199 m for r but my answer is said to be wrong. Not sure where I made a mistake. Any help is appreciated! Thank you