Question

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65 T.
a) What is the potential difference in volts required in the first part of the experiment to accelerate electrons to a speed of 6.1 x 10^7 m/s?

Answer 1

Answer:

V = 10581.59 V

Explanation:

To solve this problem, we have to take into the relation between the kinetic energy of an electron and the energy generated by the potential difference of an electric field. This relation is given by

$E_{k} = E_{V}$

$\frac{1}{2}mv^{2} = eV$

where m is the mass of the electron, v is the velocity of the electron, e is the electric charge and V is the potential difference that accelerated the electron.

By taking V from the last expression we have:

$V = \frac{mv^{2}}{2e} = \frac{(9.1*10^{-31} )(6.1*10^{7})^{2}}{2*(-1.6*10^{-19})}} = 10581.59 V$

where we have taken that

me = 9.1*10^(-31) Kg

e = -1.6*10^(-19) C

v = 6.1*10^(7)

I hope this is useful for you

Regards