Answer :
Given:
The potential difference V=3000 V
The radius of curvature of the curved path of the electron, r=26.0 cm=0.26 m
The velocity of the electron, v=32.5 Mm/s=32.5×10⁶ m/s
To find:
The magnitude of the magnetic field.
Explanation:
The magnetic force acting on the electron provides it the neccessary centripetal force to trace a curved path.
Thus the centripetal force acting on the electron is equal to the magnetic force acting on it.
Therefore,
[tex]\frac{mv^2}{r}=\text{evB}[/tex]Where m is the mass of the electron, e is the charge of the electron, and B is the magnitude of the magnetic field.
On rearranging the above equation,
[tex]B=\frac{mv}{er}[/tex]On substituting the known values,
[tex]\begin{gathered} B=\frac{9.1\times10^{-31}\times32.5\times10^6}{1.6\times10^{-19}\times0.26} \\ =7.1\times10^{-4} \\ =710\text{ }\mu\text{T} \end{gathered}[/tex]Final answer:
The magnitude of the magnetic field is 710 μT.