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

true is the correct answer

hope i

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west