Electrons enter a region of perpendicular electric and magnetic fields. The regi

Question

Electrons enter a region of perpendicular electric and magnetic fields. The region occupied by the fields is 40.0 mm long in the x direction, where x is taken to be the original direction of motion. After passing through this region, the electrons then travel an additional 300 mm in the x direction before hitting a phosphorescent screen that lights up where the electrons strike it. The electrons pass undeflected through the region of perpendicular fields when E = 2.0 × 103 N/C and B = 1.7 × 105 T.(Figure 1)

Part A

Part complete

Determine the speed of the undeflected electrons.

1.2×108 ms

Part B

If the magnetic field is turned off but the electric field is left on, by what distance have the electrons been deflected when they reach the phosphorescent screen?

v =

1.2×108 ms

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Explanation / Answer

A) for no deflection, Fe = FB

q*E = q*v*B

==> v = E/B

= 2*10^3/(1.7*10^-5)

= 1.2*10^8 m/s

B) time taken to travel additional 300 mm,

t = d/v

= 300*10^-3/(1.8*10^8)

= 1.67*10^-9 s

acceleration of electron in y-direction, ay = q*E/m

= 1.6*10^-19*2*10^3/(9.1*10^-31)

= 3.52*10^14 m/s^2

deflection in y-direction, delta_y = (1/2)*ay*t^2

= (1/2)*3.52*10^14*(1.67*10^-9)^2

= 4.9*10^-4 m or 0.49 mm

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