Explanation:
Given that,
Potential difference, V = 412 V
Magnitude of magnetic field, B = 188 mT
(a) The potential energy of electron is balanced by its kinetic energy as :
[tex]eV=\dfrac{1}{2}mv^2[/tex]
v is speed of the electron
[tex]v=\sqrt{\dfrac{2eV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 412}{9.1\times 10^{-31}}} \\\\v=1.2\times 10^7\ m/s[/tex]
(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :
[tex]r=\dfrac{mv}{eB}\\\\r=\dfrac{9.1\times 10^{-31}\times 1.2\times 10^7}{1.6\times 10^{-19}\times 188\times 10^{-3}}\\\\r=3.63\times 10^{-4}\ m[/tex]
Hence, this is the required solution.
