Question

An electron inside of a television tube moves with a speed of 2.80×107 m/sm/s . It encounters a region with a uniform magnetic field oriented perpendicular to its trajectory. The electron begins to move along a circular arc of radius 0.190 mm . What is the magnitude of the magnetic field?

Answer 1

Answer:

The magnitude of the magnetic field is 0.83 T

Explanation:

Given :

Speed of electron inside television tube (v) = $2.8$×$10^7 m/s$

Radius of electron moving path ($r$) = $0.190$×$10^-3m$

According lorentz force in magnetic field is,

∴ $F = q (v$ ×$B)$

Where, $q$ = charge of electron, $v$ = speed of electron, $B$ = magnetic field and

$F$ = force on electron.

when electron is moving in circular orbit the force is given by,

∴ $F = mv^2/r$

so we equate above both equation.

∴ $mv^2/r = q(v$ × $B)$

∴ $B = mv/qr$

∴ $B = 9.1$×$10^-31$×$2.8$×$10^7$$/1.6$×$10^-19$×$0.190$×$10^-3$

∴ $B =$ 0.83 Tesla

Therefore, the magnitude of the magnetic field is 0.83 T.