A helium nucleus is completing one round of a circle of radius 0.8m in 2 seconds. Compute the magnetic field at the centre of circle.

Dear Student,

The expression for the magnetic field for an electron revolving around a nucleus is given as,
 $B=\frac{{\mu }_o}{4\pi }\frac{2\pi NI}{r}$
As the number of rotation is 1 there we have,
 $$\begin{gathered} B=\frac{{\mu }_o}{4\pi }\frac{2\pi I}{r} \\ B=\frac{{\mu }_o}{2r}I \end{gathered}$$
we know that current is defined as charge per unit time ($I=\frac{q}{t}$) and for helium atom there are only two charges therefore the current for helium nucleus is given as,
 $I=\frac{2e}{t}$
On substituting the above equation in magnetic field equation we get,
 $$\begin{gathered} B=\frac{{\mu }_o}{2r}\frac{2e}{t} \\ B=\frac{{\mu }_o}{r}\frac{e}{t} \end{gathered}$$
On substituting the values we get,
 $B=\left(\frac{1.67\times {10}^{-19} C}{0.8 m\times 2 s}\right)\mu$_o B=1×10^-19×μ_o
Thus, the magnetic field at the centre of the circle is
B = μ_{​​​​​​o} × 10^-19 T 

Regards