Question

An ion source is producing 6Li ions, which have charge +e and mass 9.99 × 10-27 kg. The ions are accelerated by a potential difference of 13 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.0 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.

Answer 1

Answer:

6.45 x 10^5 N/C

Explanation:

q = + e = 1.6 x 10^-19 C

m = 9.99 x 10^-27 kg

V = 13 kV = 13000 V

B = 1 T

Let v be the speed and E be the strength of electric field.

1/2 mv^2 = eV

v^2 = 2 e v / m

v^2 = (2 x 1.6 x 10^-19 x 13000) / (9.99 x 10^-27)

v = 6.45 x 10^5 m/s

As the charge particle is undeflected, the force due to magnetic field is counter balanced by the force due to electric field.

q E = q v B

E = v B = 6.45 x 10^5 x 1 = 6.45 x 10^5 N/C