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