Question

(ex) The electric field is 1.20 times 10^6 N/C between plates 1 and 2, but is zero everywhere else in the figure. The magnetic field is 0.600 T everywhere in the figure. After passing straight through the region between the plates, aluminum ions enter the region with only a magnetic field, move in a circular path of radius 0.899 m there, and finally hit the photographic plate. a) The region between plates 1 and 2 is a "velocity selector": Only particles with a certain velocity pass through in a straight line. For an ion in the "velocity detector": i) Draw a free-body diagram. ii) Determine the velocity for a straight line path. (Symbols first, then numbers, as always!) iii) What happens if an ion moves faster than this? Slower? Explain. b) Using the data for the circular path, determine the mass of an aluminum ion. c) Determine mass number of this aluminum isotope. (The mass number is the mass in atomic mass units, or u, rounded to the nearest integer. 1 u = 1.66 times 10^-27 kg.)

Answer 1

a)the strength of the electric field $E=1.2\times 10^{6}N/C$

the strength of the magnetic field B=0.60T

i)

ii)when charged particle entered in the electric and magnetic fields which are perpendiular to one another, the force acting on the charged particle must be equal to get stright line motion i.e.

magnetic force=electric force

Bqv=Eq

the velocity of the charged particle $v=\frac{E}{B}=\frac{1.2\times 10^{6}}{0.6}=2\times 10^{6}m/s$

iii)if charged particle is faster than actual speed due to the magnetic field strength it deflected from its path to wards anti clockwise direction

if charged particle is slower than actual speed due to the electric field strength , it deflected from its path towards negative charged particle.

b)the charged paticle get circular path of radius r=0.899m due to magntic field.now the forces acting on it is

centrifugal force=magnetic force

$\frac{mv^{2}}{r}=Bqv$, the charge $q= 1.6\times 10^{-19}C$

then the mass of the charged particle $m=\frac{Bqr}{v}=\frac{0.6\times 1.6\times 10^{-19}\times 0.899}{2\times 10^{6}}$

$=0.4315\times 10^{-25}Kg=43.15\times 10^{-27}Kg$

c)if mass number=m then the mass of the charged particle=mass number x mass of each proton

then mass number=mass of charged particle/mass of proton

$=\frac{43.15\times 10^{-27}}{1.66\times 10^{-27}}=25.99U=26U$