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

The magnitude of the magnetic field is 0.83 T

Step-by-step 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.