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The magnetic force (Lorentz force) experienced by the proton in the magnetic field is given by
since
, because the velocity v and the force F in this problem are perpendicular, and so also the angle 
between the velocity and the magnetic field B should be 
.
Let's find the magnitude of the magnetic field; this is given by
To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force
Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T
since
Let's find the magnitude of the magnetic field; this is given by
To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force
Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T
Explanation:
It is given that,
Radius of circular particle accelerator, r = 1 m
The distance covered by the particle is equal to the circumference of the circular path, d = 2πr
d = 2π × 1 m
(a) The speed of satellite is given by total distance divided by total time taken as :
Let t is the period of the particle.
d = distance covered
s = speed of particle
It is given that the charged particle is moving nearly with the speed of light
(b) On the circular path, the centripetal acceleration is given by :
Hence, this is the required solution.