Assume the region to the right of certain plane contains a uniform magnetic fiel
ID: 1416424 • Letter: A
Question
Assume the region to the right of certain plane contains a uniform magnetic field of magnitude B = 1 mT and the field is zero in the region to the left of the plane as shown in the figure. An electron, originally travelling perpendicular to the boundary plane, passes into the region of the field. Determine the time interval required for electron to leave the "field-filled" region, noting that the electron's path is a semicircle. Assume the maximum depth of penetration into the field is 2 cm, find the kinetic energy of the electron As the electron enters the field, in what direction (up, down, into the page, out of page) will be deflected?Explanation / Answer
consider the horizontal direction to be +ve x axis, vertically upward direction on the paper to be +ve y axis and direction out of the page to be +ve z axis.
then the magnetic field is along -ve z axis.
original direction of velocity of the electron is +ve x axis.
force on the electron=q*(cross product of v and B)
as v is along +ve x axis and B is along -ve z axis, the force will be along -ve y axis (as charge of electron is -ve and cross product of v and B is +ve y axis)
hence the electron will start moving in a circlular path with magnetic force acting towards the centre of the circle and with a constant magnitude =q*v*B
the electron will eventually reach the boundary after completing a semi-circle and then move out of the magnetic field.
as we know, centripetal force=m*v^2/r
where m=mass of electron=9.1*10^(-31) kg.
equating the two forces:
m*v^2/r=q*v*B
==>r=m*v/(q*B)
hence radius of motion is m*v/(q*B).
total path travelled in a semi circle=pi*radius=pi*m*v/(q*B)
time taken=path travelled/speed=pi*m*v/(q*B*v)=pi*m/(q*B)
using the values of all the constants and using B=0.001 T
we get time taken=1.78678*10^(-8) seconds
part b:
maximum deppth=radius of the path=m*v/(q*B)
==>m*v/(q*B)=0.02 m
==>v=0.02*q*B/(m)=3.51648*10^6 m/s
then kinetic energy of the electron=0.5*m*v^2=0.5*9.1*10^(-31)*(3.51648*10^6)^2=5.6263*10^(-18) J
part c:
as we found out in part a, electron will be deflected in downward direction.