Question

An electron with a velocity given by v⃗ =(1.6×10^{5} m/s
)x^+(6700 m/s )y^ moves through a region of space with a magnetic
field B⃗ ==(0.26 T )x^−(0.10 T )z^ and an electric field E⃗ =(220
N/C )x^.

Using cross products, find the magnitude of the net force acting on the electron. (Cross products are discussed in Appendix A.)

Express your answer using two significant figures.

Answer #1

Net force on electron due to magnetic and electric field will be:

F_net = Fe + Fm

Fe = Force due to electric field = q*E

Fm = Force due to magnetic field = q*(VxB)

q = charge on electron = -1.6*10^-19 C

E = Electric field = (220 N/C) x^ = 220 i

Using x^ = i, y^ = j, z^ = k

V = (1.6*10^5 i + 6700 j) m/s

B = (0.26 i - 0.10 k) T

So,

So,

F_net = q*E + q*(VXB) = q*[E + VxB]

F_net = -1.6*10^-19*[220 i - 670 i + 1.6*10^4 j - 1742 k]

F_net = 1.6*10^-19*450 i - 1.6*1.6*10^-15 j + 1.6*10^-19*1742 k

F_net = 7.2*10^-17 i - 2.56*10^-15 j + 2.787*10^-16 k

In two significant figures:

**F_net = 7.2*10^-17 i - 2.6*10^-15 j + 2.8*10^-16
k**

**F_net = (7.2*10^-17 N) x^ - (260*10^-17 N) y^ +
(28*10^-17 N) z^**

**Let me know if you've any query.**

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