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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 magneticfield 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 ? |
Answer» Solution :`(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`. |
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