An ion source is producing 6Li ions which have a charge e an

An ion source is producing 6Li ions, which have a charge +e and mass 9.99 times 10^-27 kg. The ions are accelerated by a potential difference of 10.0 kV and pas horizontally into a region in which there is uniform vertical magnetic field of magnitude B = 1.01 T. Calculate the strength of the smallest electric field to be set up over the same region, that will allow the 6Li ions to pass through undeflected.

Solution

KE gain = q deltaV

m v^2 /2 = q deltaV

(9.99 x 10^-27 ) v^2 /2   = (1.6 x 10^-19) ( 10 x 10^3)

v = 565968.48 m/s

electric force = q E

magnetic force = q v B

hence Fnet = qE - q vB = 0

v B = E

(565968.48) (1.01) = E

E = 571628.16 N/C