Question

Electrons execute 1.4*10^6 revolutions per sec in a magnetic field of flux density

Answer 1

The charge of the electrons is $8.02 \times 10^{-19} \ C$.

The given parameters:

• Number of revolution of the electron, N = 1.4 x 10⁶ rev per sec
• Flux density, B = 10⁻⁵ T
• Mass of electron, m = 9.11 x 10⁻³¹ kg

The angular speed of the electron is calculated as follows;

$\omega = 2\pi N\\\\\omega = 2\pi \ \frac{rad}{1 \ rev} \times 1.4 \times 10^6 \ rev/s\\\\\omega = 8.798 \times 10^6 \ rad/s$

The charge of the electrons is calculated as follows;

$F_c= m\omega ^2 r\\\\F_{B}= qvB = q\omega r B \\\\F_c = F_B\\\\m\omega ^2 r = q\omega r B \\\\q = \frac{m\omega ^2 r}{\omega r B} \\\\q = \frac{m \omega }{B} \\\\q = \frac{9.11 \times 10^{-31} \times 8.798 \times 10^6}{10^{-5}} \\\\q = 8.02 \times 10^{-19} \ C$

The charge of the electrons is $8.02 \times 10^{-19} \ C$.

The complete question here:

Electrons execute 1.4*10^6 revolutions per sec in a magnetic field of flux density 10⁻⁵ T. What will be the value of charge of the electron?

Learn more about magnetic force of electrons here: brainly.com/question/3990732