Question

# Determine the magnitude of the force on an electron traveling 8.31E+5 m/s horizontally to the east...

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

B: 3.49

G: 1.06

The magnetic force on the electron is,

F = Bvq

= (0.8 T)(8.31x105 m/s)(1.6x10-19 C)

= 1.06x10-13 N

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 G : 1.06*10^-13

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