Determine the magnitude of the force on an electron traveling
8.31E+5 m/s horizontally to the east in a vertically upward
magnetic field of strength 0.800 T.
(in N)
A: 2.79 |
F = q(vxB)
As v and B are perpendicular to each other hence F = 1.6E-19*8.31E5*0.8 = 1.06368 E -13
Hence option G is correct
The magnetic force on the electron is,
F = Bvq
= (0.8 T)(8.31x10^{5} m/s)(1.6x10-19 C)
= 1.06x10^{-13} N
Answer: option (G)
G) 1.06*10^-13 N
F = q*v*B*sin(90)
= 1.6*10^-19*8.31*10^5*0.8*1
= 1.06*10^-13 N
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