Question

Consider a particle of charge Q, moving perpendicular to a magnetic field of flux density B with a velocity v. 

A) Electrons are accelerated through a potential difference of 2000 V. Calculate their kinetic energy in J.

B) Calculate their speed.

C)If they enter a magnetic field of flux density 1.5 mT, what will be the radius of the circle that they will move in

Answer 1

(A)

$$KE=q\Delta V=1.6\cdot10^{-19}\ C\cdot2000\ V=3.2\cdot10^{-16}\ J.$$

(B)

$$KE=\dfrac{1}{2}m_ev^2,$$ $$v=\sqrt{\dfrac{2KE}{m_e}}=\sqrt{\dfrac{2\cdot3.2\cdot10^{-16}\ J}{9.1\cdot10^{-31}\ kg}}=2.65\cdot10^7\ \dfrac{m}{s}.$$

(C) Applying the Newton's Second Law of Motion, we get:

$$qvB=\dfrac{m_ev^2}{r},$$ $$r=\dfrac{m_ev}{qB},$$ $$r=\dfrac{9.1\cdot10^{-31}\ kg\cdot2.65\cdot10^7\ \dfrac{m}{s}}{1.6\cdot10^{-19}\ C\cdot1.5\cdot10^{-3}}=0.1\ m.$$