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
Answers
Answer.... ✍
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 10^7v=0.1c=0.1×3×10
8
=3×10
7
m/s
Magnetic Field B=0.0006B=0.0006 T
To find out:
Kinetic Energy of the electron
Solution:
We know that
Mass of an electron m=9\times 10^{-31}m=9×10
−31
kg
Charge of en electron q=1.6\times 10^{-19}q=1.6×10
−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^2K=
2
1
mv
2
\implies K=\frac{1}{2}\times 9\times 10^{-31}\times (3\times 10^7)^2⟹K=
2
1
×9×10
−31
×(3×10
7
)
2
Joule
\implies K=40.5\times 10^{-17}⟹K=40.5×10
−17
Joule
\implies K=4.05\times 10^{-16}⟹K=4.05×10
−16
Joule
\implies K=\frac{4.05\times 10^{-16}}{1.6\times 10^{-19}}⟹K=
1.6×10
−19
4.05×10
−16
eV (∵ 1eV=1.6\times 10^{-19}\text{ J}1eV=1.6×10
−19
J
\implies K=2.53\times 10^3⟹K=2.53×10
3
eV
\implies K=2.53⟹K=2.53 KeV
thnx.....
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