In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s ^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$
In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s ^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$
Magnetic field strength, $B=6.5 \,\,G=6.5 \times 10^{-4} \,T$
Speed of the electron, $V=4.8 \times 10^{6}\, m / s$
Charge on the electron, $e=1.6 \times 10^{-19} \,C$
Mass of the electron, $m_{e}=9.1 \times 10^{-31} \,kg$
Angle between the shot electron and magnetic field, $\theta=90^{\circ}$
Magnetic force exerted on the electron in the magnetic field is given as:
$F=e v B \sin \theta$
This force provides centripetal force to the moving electron. Hence, the electron starts moving
in a circular path of radius $r$ Hence, centripetal force exerted on the electron, $F_{e}=\frac{m v^{2}}{r}$
In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e., $F_{c}=F$
$\frac{m v^{2}}{r}=e v B \sin \theta$
$r=\frac{m v}{B e \sin \theta}$
$=\frac{9.1 \times 10^{-31} \times 4.8 \times 10^{6}}{6.5 \times 10^{-4} \times 1.6 \times 10^{-19} \times \sin 90^{\circ}}$
$=4.2 \times 10^{-2} \,m =4.2 \,cm$
Hence, the radius of the circular orbit of the electron is $4.2\, cm$
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