Question

An electron is moving around a proton in an orbit of radius $1A{}^\circ$and produces $16Wb{{m}^{-2}}$of magnetic field at the centre, then find the angular velocity of electron:
$\begin{aligned} & \text{A}\text{. 20}\pi \times {{10}^{16}}\dfrac{rad}{\sec } \\\ & \text{B}\text{. }{{10}^{17}}\dfrac{rad}{\sec } \\\ & \text{C}\text{. }\dfrac{5}{2\pi }\times {{10}^{16}}\dfrac{rad}{\sec } \\\ & \text{D}\text{. }\dfrac{5}{4\pi }\times {{10}^{16}}\dfrac{rad}{\sec } \\\ \end{aligned}$

Answer 1

An electron revolving around a proton will produce a circular current loop so we can apply the biot-savart’s law to calculate the magnetic field at the centre. The current due to the flow of electrons can be calculated. Which can be related to the angular frequency of the revolution of the electron around the proton. From that relation the angular frequency of revolution can be calculated.

Formulas used:
Magnetic field $d\overrightarrow{B}$ produced at the centre of the circle due to this current element is given by the Biot-savart’s law.
$d\overrightarrow{B}=\dfrac{{{\mu }_{0}}id\overrightarrow{l}\times \widehat{r}}{4\pi {{R}^{2}}}$
For an electron current, $i=\dfrac{e}{t}$
If n is the frequency of revolution then $i=en$
Frequency is related to angular frequency by relation $n=\dfrac{\omega }{2\pi }$

Complete answer:
Consider an infinitesimal length element $\overrightarrow{dl}$.Let the current produced due to the movement of electrons around the proton is $i$. Let the length element be located at the circumference of the circle of radius $R$having radial unit vector $\widehat{r}$. Magnetic field $d\overrightarrow{B}$ produced at the centre of the circle due to this current element is given by the Biot-savart’s law. So
$d\overrightarrow{B}=\dfrac{{{\mu }_{0}}id\overrightarrow{l}\times \widehat{r}}{4\pi {{R}^{2}}}$
where ${{\mu }_{0}}$ is the the permeability of free space and its value is $4\pi \times {{10}^{-7}}H{{m}^{-1}}$
The angle between every current element and the radial unit vector is $90{}^\circ$so the cross product of the line element and the radial vector will be $d\overrightarrow{l}\times \widehat{r}=\left| d\overrightarrow{l} \right|\left| \widehat{r} \right|\sin 90{}^\circ =dl$ .
So to calculate the total magnetic field, integrate $d\overrightarrow{B}$, so
$\int{d\overrightarrow{B}=B=\oint{\dfrac{{{\mu }_{0}}i}{4\pi {{R}^{2}}}}}dl=\dfrac{{{\mu }_{0}}i}{4\pi {{R}^{2}}}\oint{dl}$
Note that also see that for all points along the path and the distance to the centre is constant. So the line integral will become the circumference of the circle.So
$B=\dfrac{{{\mu }_{0}}i}{4\pi {{R}^{2}}}2\pi R=\dfrac{{{\mu }_{0}}i}{2R}$
As the current is given by the rate of flow of charge. So for an electron the current will become
$i=\dfrac{e}{t}$,where $e=1.6\times {{10}^{-19}}C$ is the charge of an electron.
If $n$ is the number of revolutions made by electron around the proton per second then the total current is given by $i=en$
Now if $\omega$is the angular velocity of the electron, then it is related to $n$by
$\begin{aligned} & \omega =2\pi n \\\ & \Rightarrow n=\dfrac{\omega }{2\pi } \\\ \end{aligned}$
So now the current can be written as angular velocity i.e.
$i=e\dfrac{\omega }{2\pi }$
So the magnetic field is now given by
$\begin{aligned} & B=\dfrac{{{\mu }_{0}}i}{2R}=\dfrac{{{\mu }_{0}}}{2R}\times e\dfrac{\omega }{2\pi } \\\ & \Rightarrow B=\dfrac{{{\mu }_{0}}}{4\pi }\times \dfrac{e\omega }{R} \\\ & \Rightarrow \omega =\dfrac{4\pi }{{{\mu }_{0}}}\times \dfrac{RB}{e} \\\ \end{aligned}$
Given the magnetic field produced is $16Wb{{m}^{-2}}$ and radius of the orbit is $1A{}^\circ$i.e.
$R={{10}^{-10}}m$. Putting all these values we get
$\omega =\dfrac{{{10}^{-10}}\times 16}{{{10}^{-7}}\times 1.6\times {{10}^{-19}}}={{10}^{17}}\text{radian}\text{.}{{\text{s}}^{\text{-1}}}$

So the correct answer is B.

Note:
An electron revolving around the proton produces a magnetic field at the centre of the loop and the direction of the produced magnetic field can be known by the application of right hand rule. As the electron’s orbit is constant the magnetic field produced will be constant so we can integrate it to get the total magnetic field.