Question

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?

Answer 1

B=9.1397*10^-4 Tesla

Step-by-step explanation:

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

K.E=P.E

`(1)/(2)*m*v^(2)  =e*V`

where :

m is the mass of electron

v is the velocity

V is the potential difference

`v=\sqrt{(2*e*V)/(m) }` eq 1

Radius of electron moving in magnetic field is given by:

`R=(m*v)/(q*B)` 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

`R=\frac{m*\sqrt{(2*e*V)/(m) } }{e B}`

`B=\sqrt{(2*m*V)/(e*R^(2) ) }`

`B=\sqrt{(2*(9.31*10^(-31))*(2.12*10^(3))  )/((1.60*10^(-19))*(0.170)^(2)  ) }`

B=9.1397*10^-4 Tesla