An electron is moving along x – axis in x – y plane with velocity of 102 m/s. A magnetic field of magnitude 10 T is present in the region along z – axis. If the magnitude of force on the particle is 1.6 m×10-16 N, find m.
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Answer:
Explanation:
use
f=qvbsintheta
so here theta =90
v=102m/s
b=10t
q=1.6*10^-19c you know about it
and then proceed for your answer through formula all things we know
which i mention above
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