Protons move in a 10.2-cm diameter circle in the presence of a uniform 0.625-T m
ID: 1490505 • Letter: P
Question
Protons move in a 10.2-cm diameter circle in the presence of a uniform 0.625-T magnetic field
a) What is the angle between the direction of motion of the protons and the magnetic field?
b) What is the speed of the protons? (I got 1747)
c) What is the kinetic energy of one of these protons?
d) What is the magnitude of the magnetic force experienced by each proton?
f) An electric field is now generated in the same region as the magnetic field. What are the magnitude and direction of the electric field if the protons move in a line?
Explanation / Answer
given that
the charge of protron Q=1.6*10^-19 C
diameter d=10.2 cm
radius r=5.1cm =5.1*10^-2 m
magnetic field B=0.625 T
this problem belong motion in a magnetic field
force applied on a moving charge in magnetic field
the magnitude of the force experiened by a charged particle in magnetic field
F=qVB sin(theta) ........................(1)
but centripetal force F=mv^2/r ...................(2)
basing on above two equations we get[ mv=Bqr]
there fore
(a) ans
the angle b/w the direction of motion of proton and magnetic field is 90
(b) ans
the speeds of protons
v=Bqr/m
=0.625*1.6*10^-19*5.1*10^-2 /1.67*10^-27
=5.1*10^-21/1.67*10^-27
=3.054*10^6 m/s
(c) ans
the kinetic energy of the each proton=B^2 q^2r^2/2m
=(0.625)^2 (1.6*10^-19)^2 (5.1*10^-2)^2/2*1.67*10^-27
=26.01*10^-42/3.34*10^-27
=7.79*10^-15 J
(d) ans
the magnetic force Fmax=Bqv
=0.625*1.6*10^-19*3.054*10^6
=3.054*10^-13 N
(f) ans
the electric field of proton E=F/q
E=3.054*10^-13/1.6*10^-19
=1.90875*10^6 N/c
the direction of the electric field is counter - clock wise
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