Question

What is the kinetic energy of an electron moving at speed 0.1c in a magnetic field of 0.0006T perpendicular to it c is speed of light in vacuum

Answer 1

The kinetic energy of the electron is 2.53 KeV

Explanation:

Given:

Speed of the electron $v=0.1c=0.1\times 3\times 10^8=3\times 10^7$ m/s

Magnetic Field $B=0.0006$ T

To find out:

Kinetic Energy of the electron

Solution:

We know that

Mass of an electron $m=9\times 10^{-31}$ kg

Charge of en electron $q=1.6\times 10^{-19}$ C

In magnetic field since the movement of the electron is perpendicular Therefore, the force acting on it will not cause any change in kinetic energy

Thus, kinetic energy is

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

$\implies K=\frac{1}{2}\times 9\times 10^{-31}\times (3\times 10^7)^2$ Joule

$\implies K=40.5\times 10^{-17}$ Joule

$\implies K=4.05\times 10^{-16}$ Joule

$\implies K=\frac{4.05\times 10^{-16}}{1.6\times 10^{-19}}$ eV     (∵ $1eV=1.6\times 10^{-19}\text{ J}$

$\implies K=2.53\times 10^3$ eV

$\implies K=2.53$ KeV

Hope this answer is helpful.

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