Answer: a) r = 281m, b) r = 5.13×10^-7m
Explanation: From the question, we realised that the particle enters the earth magnetic field with it velocity perpendicular to the magnetic field thus making the particle have a circular motion.
The force exerted on a charge in a magnetic field perpendicular to the velocity of the particle is responsible for the centripetal force required to give the object it circular motion.
Magnetic force = centripetal force.
qvB = mv²/r
By dividing "v" on both sides, we have that
qB = mv/r
Above is the formulae that defines the circular motion of a particle in the earth's magnetic field
Where q = magnitude of electronic charge.
B = strength of magnetic field = 1.62×10^-7 T
v = speed of particle = 8.03×10^6 m/s
A) If the particle where to be an electron, q ( magnitude of electron) =1.609×10^-19c.
m = mass of electronic charge = 9.11×10^-31 kg.
By substituting these parameters into the formulae we have that
1.609×10^-19× 1.62×10^-7 = 9.11×10^-31 × 8.03×10^6/ r
By cross multiplying, we have that
1.609×10^-19× 1.62×10^-7 × r = 9.11×10^-31 × 8.03×10^6
r = 9.11×10^-31 × 8.03×10^6 / 1.609×10^-19× 1.62×10^-7
r = 7.32*10^(-24)/ 2.61×10^-26
r = 2.81×10²
r = 281m
B)
If the particle is proton, q = magnitude of a proton charge = 1.609×10^-19c,
m = mass of proton = 1.673×10^-27 kg
By substituting these parameters into the formulae we have that
1.609×10^-19× 1.62×10^-7 = 1.673×10^-27×8.03×10^6/ r
By cross multiplying, we have that
1.609×10^-19× 1.62×10^-7 × r = 1.673×10^-27× 8.03×10^6
r = 1.673×10^-27 × 8.03×10^6 / 1.609×10^-19× 1.62×10^-7
r = 1.34*10^(-32)/ 2.61×10^-26
r = 0.513×10^-6 m
r = 5.13×10^-7m
