Wire I has a current of 12 A pointing to the right. Wire II is placed 1.5 m belo

Question

Wire I has a current of 12 A pointing to the right. Wire II is placed 1.5 m below Wire I with a current of 18 A pointing to the left. An electron travels along the midpoint between Wire I and Wire II with a velocity of 3.5E6 m/s in the same direction as the current in Wire II. a. Calculate the magnetic force acting on the electron and its direction b. Calculate the acceleration of the electron. c. What magnitude of electric field would be needed so that the electron would be undeflected and in what direction?

Explanation / Answer

Accelerating Voltage= V1=1.60 kV = 1600 V

Charge on alpha= q=3.2*10^-19 C

Mass = m =6.64*10^-27 kg

Kinetic energy= electric potential energy

(1/2)mv^2 = qV

v = sq rt [2qV/m]

Thealpha particleswill emerge undeflected from between the plates if magnetic force=elctric force

qvB =qE

E = Vo/d

B = E /v

B =(Vo/d) sq rt [m/2qV]

B =0.04968 T

B =4.968*10^-2 T
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As electric field points upward , electric force is upwards. For alpha to remain undeflected,direction of magnetic field should be out of the page

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