Question

Problem 8: 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.

What is the potential difference in volts required in the first part of the experiment to accelerate electrons to a speed of 6 1 × 107 m/s?

Answer 1

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=$v=6.1\times 10^7m/s$

Charge on electron, $q=e=1.6\times 10^{-19} C$

Mass of electron,$m_e=9.1\times 10^{-31} kg$

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

$V=\frac{v^2m_e}{2e}$

Where V=Potential difference

$m_e=$Mass of electron

v=Velocity of electron

Using the formula

$V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}$

$V=10581.59 V$

Hence, the potential difference=10581.59 V