A velocity selector consists of electric and magnetic fields that are perpendicu
ID: 1413646 • Letter: A
Question
A velocity selector consists of electric and magnetic fields that are perpendicular to one another. If in such an arrangement a B-field of 12.0 T is used, then what should be the value of the electric filed so that an electron with velocity of 0.2c passes the velocity selector undeflcctcd? Now we turn off the electric field. Assuming the electron is moving to the right and magnetic field is horizontal and points into the paper, what is the magnitude and direction of the force acting on the electron at that instance? Ignore the effect of gravity. What is the radius of curvature of the electron in .Explanation / Answer
a) If an electron travels undeflected through a velocity selector,then the forces acting on the electron balances each other. Neglecting gravity ,there are electric and magnetic forces acting perpendicular to the direction of the velocity of the electron
The magnetic force is
FB = q ( v x B )
= q v B ( since v and B are perpendicular )
The electric force is
FE = q E
At equilibrium FB = FE
q v B = q E
E = v B
E = 0.2 c x 12 T = 0.2 x 3 x 108 x 12
E = 7.2 x 108 N / C
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b) The magnetic force is
FB = q ( v x B )
Where v is the velocity of the electron and q is the charge of the electron
F = q v B
F = -1.6 x 10-19 x 0.2 c x 12
F = - 11.52 x 10-11 N
The force is along the vertically downward direction
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c) Since the force is perpendicular to the direction of motion , the electron will follow a circular path
The centripetal force is provided by the magnetic force
m v2 / r = q v B
Where r is the radius of the circular path
r = m v / q B
r = 9.11 x10-31 x 0.2 x 3 x 108 / (1.6 x 10-19 x 12)
r = 0.285 x 10-4 m