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

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Solution

## 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 $2$secondStep 2. Formula used$\mathrm{I}=\frac{\mathrm{q\omega }}{2\mathrm{\pi }}$$\mathrm{B}=\frac{{\mathrm{\mu }}_{0}\mathrm{I}}{2\mathrm{r}}$$f=\frac{\omega }{2\pi }$$f=\frac{1}{t}$Where $B$ is the magnetic field, ${\mu }_{0}$ is the permeability of the free space, $\mathrm{I}$ is the current, $r$ is the radius, $q$ is the charge of Helium atom, $\omega$ is the angular frequency of ${n}^{th}$ orbit and $t$ is the time.Step 3. Find the angular frequency$t$$=2$ secondFrom equation (iii) and (iv) we get $⇒\frac{1}{t}=\frac{\omega }{2\pi }\phantom{\rule{0ex}{0ex}}⇒\omega =\frac{2\pi }{t}\phantom{\rule{0ex}{0ex}}⇒\omega =\pi \phantom{\rule{0ex}{0ex}}$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=2×1.6×{10}^{-19C}$ is the charge of Helium atom.Then ,$\mathrm{I}=\frac{\mathrm{q\omega }}{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=\frac{2×1.6×{10}^{-19}}{2}\phantom{\rule{0ex}{0ex}}=1.6×{10}^{-19}A$Step 5. Find the value of magnetic field $\mathrm{B}=\frac{{\mathrm{\mu }}_{0}\mathrm{I}}{2\mathrm{r}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu }}_{0}×1.6×{10}^{-19}}{2×0.8}\phantom{\rule{0ex}{0ex}}={\mathrm{\mu }}_{0}{10}^{-19}\phantom{\rule{0ex}{0ex}}=4\pi ×{10}^{-7}×{10}^{-19}\phantom{\rule{0ex}{0ex}}=1.26×{10}^{-25}\mathrm{tesla}$Hence the value of the magnetic field in the center of the circle is $1.26×{10}^{-25}\mathrm{tesla}$.

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