An electron travels with speed 0.50 x 107 m/s between the two parallel charged p

Question

An electron travels with speed 0.50 x 107 m/s between the two parallel charged plates shown in the figure(Figure 1). The plates are separated by 1.0 cm and are charged by a 200 V battery Part A What magnetic field strength will allow the electron to pass between the plates without being deflected? Express your answer to two significant figures and include the appropriate units BValue mlT Submit Request Answer Part B What is the direction of the magnetic field? Figure 1 of 1 upward O downward into the page O out of the page Submit Request Ans 1.0 cm Provide Feedback Next >

Explanation / Answer

Given

electron speed v = 0.50*10^7 m/s

separation of plates is d = 0.01 m ,

V = 200 V

E = V/d ==> E = 200/0.01 = 20000 V/m

the charge (electron will not deflect if the equivalent magnetic force act on the electron

so the electric force acting on the eleactron is  

Part A

F_e = E*q

F_m = B*q*V sin theta

equating these

E*q = B*q*V sin theta

here theta is 90 degrees that means the magnetic field should be perpendicular to the motion of the charge

B = E/ V  

B = 20000 /(0.50*10^7) T

B = 0.004 T

B = 4 mT

PArt B

the electron is of -ve charge will deflect towards the +ve plate so with out having deviation by the electron the magnetic force should be towards +ve plate

from Right hand rule the manetic field should be into the page

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