Answer:
a) 0.114 meters is the radius of the circular orbit for an electron with given speed.
b) 14.6 meters is the radius of the circular orbit for a proton with given speed.
Explanation:
Force on the charge particle :
[tex] F = qvB[/tex]
Since, the particle is revolving around in a circular path ;
[tex]F'=\frac{mv^2}{r}[/tex]
particle in constantly moving in circular path under action of magnetic field:
F = F'
[tex]qvB=\frac{mv^2}{r}[/tex]
[tex]r=\frac{mv}{qB}[/tex]
Where :
r = radius of the circular path
m = mass of the particle
v = velocity of the particle
B = magnetic field
q = magnitude of  charge on the particle
a) Â Velocity of the electron = Â [tex]v=1.0\times 10^6 m/s[/tex]
Mass of electron = m = [tex]9.11\times 10^{-31} kg[/tex]
Charge on an electron , q= [tex]1.602\times 10^{-19} C[/tex]
B = [tex]5\times 10^{-5} T[/tex]
Radius of the circular orbit for an electron with given speed =[tex]r_e[/tex]
[tex]r_e=\frac{9.11\times 10^{-31} kg\times 1.0\times 10^6 m/s}{1.602\times 10^{-19} C\times 5\times 10^{-5} T}[/tex]
[tex]r_e = 0.114 m[/tex]
0.114 meters is the radius of the circular orbit for an electron with given speed.
b) Â Velocity of the proton = Â [tex]v=7.0\times 10^4 m/s[/tex]
Mass of electron = m = [tex]1.67\times 10^{-27} kg[/tex]
Charge on an electron , q= [tex]1.602\times 10^{-19} C[/tex]
B = [tex]5\times 10^{-5} T[/tex]
Radius of the circular orbit for a proton with given speed =[tex]r_p[/tex]
[tex]r_e=\frac{1.67\times 10^{-27} kg\times 7.0\times 10^4 m/s}{1.602\times 10^{-19} C\times 5\times 10^{-5} T}[/tex]
[tex]r_p = 14.6 m[/tex]
14.6 meters is the radius of the circular orbit for a proton with given speed.
