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
0 0
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Answer #1

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

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Answer #2

B: 3.49

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Answer #3

G: 1.06

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Answer #4

The magnetic force on the electron is,

               F = Bvq

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

                  = 1.06x10-13 N

Answer: option (G)

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Answer #5

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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Answer #7

G : 1.06*10^-13

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