Kinetic energy of each electron in a beam of television picture tube is `12.0 keV`. The electrons are emitted horizontally from geomagnetic south to geomagnetic north. The vertial componenet of earths magnetic field points down and has a magnitude `55.0 mu T`. `(a)` In which direction will the electrons deflect ? `(b)` How far will the beam deflect in moving `20.00 cm` through the television tube ?

Explanation

`(a)` Figure shows the electron traveling north. The magnetic field points inward. From right hand rule `vec(v)xxvec(B)` vector is toward west. Since electron is negative, the force on it is toward east.

`(b)` The electron speed can be determined from kinetic energy.
`K=(1)/(2)m upsilon^(2), " "upsilon=sqrt((2K)/(m))`
`=sqrt((2(12.0xx10^(3))(1.60xx10^(-19)))/(9.11xx10^(-31))`
`=6.49xx10^(7)m//s`
The magnetic force on the electron is `F_(B)e upsilon B sin theta` where `theta` is angle betweent he electron velocity an the magnetic field. In this case, velocity `upsilon` and magnetic field B are perpendicular, `theta =90^(@)`. The magnetic force provides centripital force for circular motion, therefore,
`e upsilonB=(m upsilon^(2))/(R)`
`R=(m upsilon)/(eB)=(9.11xx10^(-31)xx6.49xx10^(7))/(1.6xx10^(-19)xx(55xx10^(-6)))=6.72m`
Let the arc traced by the electron beam subtend an angle `theta` at the centre, l be the length of the ube, and d the deflection.
From figure, `l=Rsin theta`
`d=R-R cos theta ` or `R cos theta =R-d`
On squaring and adding these two equation, we have
`R^(2)=(R-d)^(2)+l^(2)` or `d^(2)-2Rd+l^(2)=0`
On solving the above quadratic equation, we get
`d=R+-sqrt(R^(2)-l^(2))`
The plus sign corresponds to angle `180^(@)-theta` and minus sign corresponds to out case.
For `llt lt R, " "(R^(2)-l^(2))^(1//2)~=R R-(1)/(2)(l^(2))/(R)`
Hence `d=(1)/(2)(l^(2))/(R)=(1)/(2)xx((0.200)^(2))/(6.72)=0.00298m`.