Answer :
Answer:
Explanation:
Given a proton, a proton is a positively charge particle,
q=1.609×10^-19C
The mass of proton is given as
M=1.67×10^-27kg
Velocity of the proton is
V=531km/s
Since 1km=1000m
Then, V=531,000m/s
From the Cardinal point eastward is negative direction of x
The proton is then directed eastward
Then, v=-531,000 i m/s
Magnetic field is given as
B=21.5mT
B=21.5×10^-3T
Magnetic field is directed upward i.e positive direction of y axis
Then,
B=21.5 j mT
This shows that the angle between the magnetic field and the velocity is θ=90°. They are perpendicular to each other.
Radius=?
The magnetic force is given as
F=q V×B
Note that this is a cross product and it is defined as
A×B=|A||B|Sinθ
F=qVB•Sin θ
Since the Velocity has a magnitude of
V=531,000m/s
Also B=21.5×10^-3T
Therefore,
F=qVB•Sin θ
F=1.609×10^-19×531,000×21.5×10^-3Sin90
F=1.84×10^-15N
Then, the Force to keep the body in circular motion is centripetal force
Which is given as
Fc=Mv²/R
Fc=1.67×10^-27×531,000²/R
Fc=4.71×10^-16/R
Here, the magnetic force supplies the centripetal force.
This shows that Fc=F
Fc=F
4.71×10^-16 / R = 1.84×10^-15
Cross multiply
4.71×10^-16= 1.84×10^-15 R
R=4.71^-16/1.84×10^-15
R=0.25597m
R=0.256m
Or
R=255.98mm
The radius of the circular path is approximately 0.256m