An electron moves along the z -axis with v z =6.0×107m/s. As it passes the origi

Question

An electron moves along the z-axis with vz=6.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (x, y, z) positions?

Part A

(2 cm , 0 cm, 0 cm)

Express your answers using two significant figures.Enter your answers numerically separated by commas.

Part B

(0 cm, 0 cm, 1 cm )

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Part C

(0 cm, 2 cm , 1 cm )

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Explanation / Answer

Vz = 6*10^7 m/s

B = (uo/4pi)(qvsin(theta)/r^2)

(a) r = (2 cm , 0 ,0)

B = (10^-7)(1.6*10^-19*6*10^7*sin(90)/0.02^2)

B = 2.4*10^-15 T

By definition the current is going opposite the direction of the electron. Use the right hand rule to see the direction of the magnetic field is in the negative y-axis.

B = -2.4*10^-15 j T

(b) r = (0,0 , 1cm)

angle between velocity and distance is theta = 0

B =0

(c) r = (0,2 cm , 1 cm)

r^2 = (2^2 +1^2) = 5 cm^2

tan(theta) = 2/1

theta = 63.4 dergees

B = (10^-7)(1.6*10^-19*6*10^7*sin(63.4)/0.05)

B = 1.7*10^-15 T

By definition the current is going opposite the direction of the electron. Use the right hand rule to see the direction of the magnetic field is in the positive x-axis.

B = 1.7*10^-15 i T

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