Answer:
Radius, r = 0.051 meters
Explanation:
It is given that,
Potential difference, [tex]V=4.55\ kV=4.55\times 10^3\ V[/tex]
Magnetic field, [tex]B=4.4\ mT=4.4\times 10^{-3}\ T[/tex]
The magnetic field is perpendicular to the velocity of the electron. We need to calculate the radius of the path this electron will follow in the magnetic field.
The magnetic force is balanced by the centripetal force as:
[tex]\dfrac{mv^2}{r}=qvB[/tex]
[tex]r=\dfrac{mv}{qB}[/tex]..........(1)
Using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=eV[/tex]
[tex]v=\sqrt{\dfrac{2eV}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 4.55\times 10^3}{9.1\times 10^{-31}}}[/tex]
[tex]v=4\times 10^7\ m/s[/tex]
Equation (1) becomes :
[tex]r=\dfrac{9.1\times 10^{-31}\times 4\times 10^7}{1.6\times 10^{-19}\times 4.4\times 10^{-3}}[/tex]
r = 0.051 m
So, the radius of the path this electron will follow in the magnetic field is 0.051 meters. Hence, this is the required solution.
