An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kv . then it passes through a region of transverse magnetic field, where it moves in a circular arc with a radius of 0.170 m .
what is the magnitude of the field?

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

where :

m is the mass of electron

v is the velocity

V is the potential difference

   eq 1

Radius of electron moving in magnetic field is given by:

      eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

B=9.1397*10^-4 Tesla

option number c is the right answer.