Assume the region to the right of a certain plane contains a uniform magnetic fi

Question

Assume the region to the right of a certain plane contains a uniform magnetic field of magnitude b = 1.37 mT and the field is zero in the region to the left of the plane as shown in the figure below. An electron, originally traveling perpendicular to the boundary plane, passes into the region of the field Determine the time interval required for the electron to leave the "field-filled" region, noting that the electron's path is a semicircle. 1.30e-8 Assuming the maximum depth of penetration into the field is 2.72 cm, find the kinetic energy of the electron. 8.0 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

Explanation / Answer

if speed is v, charge magnitude is e, magnetic field is B, and radius of the semi-circular path is R,

then during the circular motion, centripetal force balances the magnetic force

hence mass*speed^2/radius=charge*speed*magnetic field

==>m*v^2/r=e*v*B

==>v=e*B*r/m

using the given values, v=1.6*10^(-19)*1.37*10^(-3)*2.72*0.01/(9.1*10^(-31))=6.5519*10^6 m/s

part a:

time interval required=pi*r/v=pi*2.72*0.01/(6.5519*10^(6))=1.3042*10^(-8) seconds

part b:

then kinetic energy=0.5*m*v^2

=0.5*9.1*10^(-31)*(6.5519*10^6)^2

=1.9532*10^(-17) J

as 1 eV=1.6*10^(-19) J,

kinetic energy=(1.9532*10^(-17))/(1.6*10^(-19))=122.07 eV

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