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# An electron of 10 eV is revolving around circular path in magnetic field of $1\times {{10}^{-4}}{weber}/{metr{{e}^{2}}}\;$. Find (i) Speed of electron, (ii) radius of circular path, (iii) time period.

Last updated date: 13th Jun 2024
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Hint: This question is a direct one. The speed of the electron is equal to the root value of the ratio of 2 times the kinetic energy by mass of the electron. The radius of the circular path is equal to the ratio of the product of mass and speed of the electron by the product of charge of the electron and the magnetic field. The time period is the ratio of mass by the product of charge of the electron and the magnetic field.
Formula used:
$v=\sqrt{\dfrac{2K}{m}}$
$R=\dfrac{mv}{eB}$
$T=\dfrac{2\pi m}{eB}$

From the data, the kinetic energy of the electron revolving around a circular path in the magnetic field is 10 eV. The magnitude of the magnetic field is, $1\times {{10}^{-4}}{weber}/{metr{{e}^{2}}}\;$.
Now, we will compute the speed of the electron.
The formula used to find the value of the speed of the electron is as follows.
$v=\sqrt{\dfrac{2K}{m}}$
Where K is the kinetic energy of the electron and m is the mass of the electron.
Substitute the given values in the above equation.
\begin{align} & v=\sqrt{\dfrac{2\times 10\times 1.6\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}} \\ & \Rightarrow v=5.93\times {{10}^{6}}{m}/{s}\; \\ \end{align}
Therefore, the value of the speed of the electron is $5.93\times {{10}^{6}}{m}/{s}\;$.
Now, we will compute the radius of the circular path.
The formula used to find the radius of the circular path is as follows.
$R=\dfrac{mv}{eB}$
Where v is the speed of the electron and m is the mass of the electron, e is the charge of the electron and B is the magnitude of the magnetic field.
Substitute the given values in the above equation.
\begin{align} & R=\dfrac{9.1\times {{10}^{-31}}\times 5.93\times {{10}^{6}}}{1.6\times {{10}^{-19}}\times {{10}^{-4}}} \\ & \Rightarrow R=33.73\times {{10}^{-2}}m \\ \end{align}
Therefore, the value of the radius of the circular path is $33.73\times {{10}^{-2}}m$.
Now, we will compute the time period.
The formula used to find the time period is as follows.
$T=\dfrac{2\pi m}{eB}$
Where m is the mass of the electron, e is the charge of the electron and B is the magnitude of the magnetic field.
Substitute the given values in the above equation.
\begin{align} & T=\dfrac{2\pi \times 9.1\times {{10}^{-31}}}{1.6\times {{10}^{-19}}\times {{10}^{-4}}} \\ & \Rightarrow T=35.72\times {{10}^{-8}}s \\ \end{align}
Therefore, the value of the radius of the circular path is$35.72\times {{10}^{-8}}s$.

The values are (i) Speed of electron $5.93\times {{10}^{6}}{m}/{s}\;$, (ii) radius of circular path $33.73\times {{10}^{-2}}m$, (iii) time period $35.72\times {{10}^{-8}}s$.

Note:
This question can be solved by the direct method, that is, simply substituting the values in the formulae used to find the speed of an electron, the radius of the circular path and the time period. The units of the parameters should be taken care of. All the units must be mentioned using the SI system.