An electron moves between two parallel plates, as shown in (Figure 1) . A 480-N/

Question

An electron moves between two parallel plates, as shown in (Figure 1) . A 480-N/C E field points from the upper plate toward the lower plate. A 0.12-Tmagnetic field points into the plane of the picture.

Determine the magnitude of the electric force that the electric field exerts on an electron between the plates.

Determine the direction of the electric force that the electric field exerts on an electron between the plates.

At what speed must the electron move so that the magnetic field exerts an opposing force that balances the electric force? (If narrow slits are placed at the entrance and exit of the plates, the device allows charged particles of only one speed to pass through both slits. Such a device is called a velocity selector.)

In which direction must the electron move so that the magnetic field exerts an opposing force that balances the electric force?

Explanation / Answer

A) Electric Force = F = -qE = 1.6*10^-19*480 = 7.68*10^-17 N

B) As the electric field is downward and charge is negative, so the direction of electric force will be upward.

C) Now magnetic force = Electric force

qvB = qE =======> v= E/B

v = 480/0.12 = 4000 m/s

D) The electron should move to the right.

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