Answer to Question #89699 in Electricity and Magnetism for Khadeeja

Question

An electron emitted from a hot filament is accelerated through potential difference of 2.88kV and enters a region of uniform magnetic field of 0.1T at an angle of 30°C with the field. The trajectory of the electron is,
(1) circle of radius 1.6 millimetre
(2)circle of radius 0.9 millimetre
(3) helix of radius 0.9 millimetre
(4) helix of radius 1.6 millimetre
(5) none of the above

Answer 1

Solution. The motion of an electron can be viewed in two directions along the field and perpendicularly decomposing the velocity into two components. When moving along a field, no force acts on the electron and it moves uniformly. In the perpendicular direction to the field, the electron is affected by the Lorentz force, which causes the electron to move in a circle.

Having added two movements the trajectory of movement will be a spiral helix. Find the radius of the helix.

$$qvBsin30^0=\frac {m(vsin30^0)^2} {r}$$

where q=1.9*10^-19C is electron charge; v is electron velocity; B=0.1T is magnetic field; m=9.1*10^-31kg is electron mass.

Therefore

$$r=\frac {mvsin30^0} {qB}$$

Find the electron velocity using the equation

$$qV=\frac {mv^2} {2}$$

where V=2.88*10^3V is voltage. Hence

$$v=\sqrt {\frac {2qV} {m}}$$

Substituting the obtained value of speed in the radius equations we get.

$$r=\frac {\sqrt {mV} sin30^0} {B\sqrt {q}}=\frac {\sqrt {9.1*10^{-31}\times2.88\times10^3} sin30^0} {0.1\times \sqrt {1.6\times10^{-19}}}$$

$$r=9\times10^{-4}=0.9mm$$

Answer. (3) helix of radius 0.9 millimetre