An electron moves along the z-axis with vz=2.7×10^7m/s. As it passes the origin,

Question

An electron moves along the z-axis with vz=2.7×10^7m/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 (1 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 , 2 cm ) Express your answers using two significant figures. Enter your answers numerically separated by commas.

* Part C (0 cm , 2 cm , 2 cm ) Express your answers using two significant figures. Enter your answers numerically separated by commas.

Please show steps and label all answers clearly, thank you.

Explanation / Answer

For a charge q moving with velocity v, the magnetic field at a distance r from it is given as:

B = (o/4) (q v x r / r3) where v x r is the cross product of the distance vector and the velocity vector.

Now, we know that the electron is moving with the velocity of V = 2.7 x 10^7 m/s along the z axis. We will now use the above to solve the given problems.

Part A.) For point P(1 , 0 , 0)

we have the angle made between the distance vector and velocity vector is 90 degrees, hence we get:

B = (o/4) (q v x r / r3) = (10^-7)(1.60217662 × 10-19 x 2.7 x 10^7 x 10^-2 / 10^-6)

or, B = 4.3259 x 10^-15 T

Also, we know that the direction of the magnetic field can be determined by curling our right hand along the direction of the current and the curl gives the direciton of field. Here, electron moves along z axis, hence current is along -z axis. On curling our fingers, we get that the direction of magnetic field at 1, 0, 0 is along -Y axis.

Part B.) For a point on the z axis the distance vector makes an angle of zero degrees with the velocity vector and hence the magnetic field induced would be zero,

Part C.) For point P(0, 2, 2)

Here again the perpendicular distance from the line of motion of the charge is 2 cm, hence the magnitude of the magnetic field would remain same as part a above.

That is, Magnetic field at 0,2,2 is B = 4.3259 x 10^-15 T

and direction is along +x axis.

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