Question

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

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

Answer 2

B: 3.49

Answer 3

G: 1.06

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)

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

Answer 6

$\left. F = qvB = (1.6 \times 10^{-19} \, C)\left(8.31 \times 10^5 \, \frac{m}{s}\right) \left(0.800 \, T \right ) = 1.06 \times 10^{-13} \, N \rightarrow \boxed{G}$

Answer 7

G : 1.06*10^-13