An electron moves at speed 7.0 × 10^ 6 m/s toward the right between two parallel

Question

An electron moves at speed 7.0×10^6 m/s toward the right between two parallel plates, such as shown in (Figure 1) . A 0.15-T magnetic field points into the paper parallel to the plate surfaces.

A) Determine the magnitude of the magnetic force that the magnetic field exerts on the electron. Express your answer to two significant figures and include the appropriate units

B) Determine the direction of the magnetic force that the magnetic field exerts on the electron. Upward or Downward?

C) What should be the magnitude of an E field caused by oppositely charged plates to produce an electric force that just balances the magnetic force? Express your answer to two significant figures and include the appropriate units.

D) What should be the direction of an E field caused by oppositely charged plates to produce an electric force that just balances the magnetic force? Upward or Downward?

Explanation / Answer

magnetic force exerted by the magnetic field = Fb = q*v*B*sintheta

theta = 90

q = .6*10^-19

B = 0.15 T

Fb = 1.6*10^-19*7*10^6*0.15*sin90

Fb = 1.68*10^-13 N <<<-------answer

+++++++++++

B)

the magnetic force is in down ward direction

+++++++++++++

c)

E = Fb/q = 105*10^4 N/C

+++++++

d)

the direction E is in down ward direction

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