Question

A helium nucleus makes a full rotation in a circle of radius $0.8$ meter in two second. The value of $B$ (magnetic field) at the center of the circle will be

Answer 1

Step 1. Given data

Helium nucleus makes a full rotation in a circle of radius $0.8$

Time taken for one full rotation of the Helium nucleus is $2$second

Step 2. Formula used

1. $\mathrm{I}=\frac{\mathrm{q\omega }}{2\mathrm{\pi }}$
2. $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{r}}$
3. $f=\frac{\omega }{2\pi }$
4. $f=\frac{1}{t}$

Where $B$ is the magnetic field, ${\mu }_0$ is the permeability of the free space, $\mathrm{I}$ is the current, $r$ is the radius, $q$ is the charge of Helium atom, $\omega$ is the angular frequency of $n^{th}$ orbit and $t$ is the time.

Step 3. Find the angular frequency

$t$$=2$ second

From equation (iii) and (iv) we get

$$\begin{gathered} \Rightarrow \frac{1}{t}=\frac{\omega }{2\pi } \\ \Rightarrow \omega =\frac{2\pi }{t} \\ \Rightarrow \omega =\pi \end{gathered}$$

Step 4. Find the current due to revolution of the electron

$\mathrm{I}$ is the current due to revolution of the electron.

$r=0.8m$

$q=2\times 1.6\times {10}^{-19C}$ is the charge of Helium atom.

Then ,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{q\omega }}{2\mathrm{\pi }} \\ =\frac{2\times 1.6\times {10}^{-19}}{2} \\ =1.6\times {10}^{-19}A \end{gathered}$$

Step 5. Find the value of magnetic field

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{r}} \\ =\frac{{\mathrm{\mu }}_0\times 1.6\times {10}^{-19}}{2\times 0.8} \\ ={\mathrm{\mu }}_0{10}^{-19} \\ =4\pi \times {10}^{-7}\times {10}^{-19} \\ =1.26\times {10}^{-25}\mathrm{tesla} \end{gathered}$$

Hence the value of the magnetic field in the center of the circle is $1.26\times {10}^{-25}\mathrm{tesla}$.