physic prelab question THEORY When an electron is accelerated by a potential dif
ID: 1496156 • Letter: P
Question
physic prelab question
THEORY When an electron is accelerated by a potential difference (voltage) V, the resulting change in potential energy is converted to kinetic energy, so 2eV where m is the electron mass and e is the magnitude of its charge. An electron in a magnetic field B is subjected to a force whose magnitude is 2eV This force is perpendicular to the electron's velocity and the magnetic field, so it causes the electron to move in a circle. The centripetal force required for a circular orbit of radius r is Equating the two expressions for F gives eBr Combining with Equation 1 gives the charge-to-mass ratio 2eV eBr e 2V m Br Thus a measurement of the voltage, V, the magnetic field B, and the radius r of the circular electron path gives the charge-to-mass ratio As in lab 7, the magnetic field is created using a coil. At a distance z from the center of a coil the field is B(z)HoNIR where ,-4TX10" Tesla-m/Amp, R is the radius of the coil, N is the number of turns of wire, and l is the current. In this case we have two coils, both a distance D/2 from the electron beam in the vacuum tube. The field is thus R24fD For true Helmhotz coils, the separation D is the same as the coil radius R. The field is then 8 NI 125 R uoNIR 3/2Explanation / Answer
1)
e/m=(1.6*10^-19)/(9.1*10^-31)
=1.76*10^11 C/kg
2)
radius, R=10.65cm
separation, D=5cm
no of turns, N=128
current, i=3.9 A
magnetic field, B=uo*N*i*R^2/(R^2+(D/2)^2)^3/2
=4pi*10^-7*128*3.9*(10.65*10^-2)^2/((10.65*10^-2)^2+(0.05/2)^2)^(3/2)
=5.43*10^-3 T
=5.43 mT
3)
voltage, v=89 v
orbit radius, r=0.75cm
e/m=2*V/(B^2*r^2)
=(2*89)/((5.43*10^-3)^2*(10.65*10^-2)^2)
=5.32*10^8 C/kg
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