Question

The picture tube in an old black-and-white television uses magnetic deflection coils rather than electric deflection plates. Suppose an electron beam is accelerated through a 51.0-kV potential difference and then through a region of uniform magnetic field 1.00 cm wide. The screen is located 10.9 cm from the center of the coils and is 50.6 cm wide. When the field is turned off, the electron beam hits the center of the screen. Ignoring relativistic corrections, what field magnitude is necessary to deflect the beam to the side of the screen? (In mT)
_____mT

Answer 1

R = radius of path

fom fig

tan $\Theta$ = 25.3/10.9  

$\Theta$ = 66.7 degree

radius would be

sin $\Theta$ = 1/R

sin66.7 = 1/R

R = 1/sin66.7 = 1.088 cm

R = 1.088 x10^-2 m

as an electron beam is accelerated through a 51.0-kV potential difference

from energy conservation

KEi + PEi = KEf + PEf

0 + eV = 1/2 mv² + 0

1/2 mv² = e V

v = $\sqrt{2eV/m}$

This speed remains constant until they hit the screen, at w = 25.3 cm above its center.

we know that

B = mv/qR

B = m/eR [ $\sqrt{2eV/m}$ ]

B = 1/R [ $\sqrt{2mV/e}$ ]

B = 1/(1.088 x10^-2) [ $\sqrt{2*9.1*10^-^3^1 * 51000 / (1.6 *10^-^1^9)}$

B = 69.9 mT

answer