How Does Magnetic Field Influence the Path of a Doubly Charged Helium Atom?

[superjen]

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 0.450 T magnetic field?


the equation i was using was
r = mV/|q|B

m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19

I don't think this is right. any help pr tips?
Thanks :)

 

[LowlyPion]

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 0.450 T magnetic field?

the equation i was using was
r = mV/|q|B

m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19

I don't think this is right. any help pr tips?
Thanks :)

First figure the kinetic energy, from the work done on the charge.

W = q*ΔV

From ½mv² you can derive a value for v .

Then you can use your second equation derived from

F = mv² /R = qv*B

R = mv/qB