Moving Charges and Magnetism Problems - Numerical 13

An electron is accelerated from rest through a potential difference of 3 KV. It enters into the region of a uniform, perpendicular magnetic field of 0.2T. What is the radius of the path of electron inside the field?

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Suggested solution:

Suggested solution:

Let v be the speed acquired by the electron when accelerated by a potential difference of V.

Then,

(1/2) mv² = eV

or v = 2eV/m

The radius R of the circular path inside the magnetic field is

mv² / R = BeV

Given: B = 0.2 T, V = 3000 V, e = 1.6 x 10¹⁹ c, m = 9.1 x 10³¹ kg

= R = (1/0.2) 2 x 9.1 x 10^-³¹ x 3 x 10³ /1.6 x 10¹⁹ (m)

= 2.925 x 10^-²m = 2.295cm