The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude?
of 6 × 10−5 T. A proton is
moving horizontally toward the west in this
field with a speed of 1.7 × 106 m/s.
The charge on the proton is 1.60218 ×
10−19 C and its mass is 1.67262 × 10−27 kg.
What is the magnitude of the magnetic
force the field exerts on this charge?
Answer in units of N
What is the radius of the circular arc followed
by this proton?
Answer in units of m
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Answers & Comments
Verified answer
(1) F = qvB
(2) r = mv^2 / F
Your numbers are in SI units, so it's easy to plug and chug