Question

A helium nucleus (charge $+2e$) completes one round of a circle of radius $0.8m$ in $2sec$. Find the magnetic field at the centre of the circle.

Answer 1

The expression of magnetic field for an electron in a circular path is

B $=\frac{{\mu }_{o}2\pi NI}{4\pi r}$

where, I= current, N= no. of rotation (Here in this case)= 1

B $=\frac{{\mu }_{o}2\pi I}{4\pi r}=\frac{{\mu }_{o}I}{2r}$

We know that, Helium has 2 charges. And the formula of current I$=\frac{2e}{t}$

Also magnetic field B $=\frac{{\mu }_{0}2e}{2rt}=\frac{{\mu }_{0}2e}{rt}=\left(\frac{1.67\times {10}^{-19}}{0.8\times 2}\right)\times 4\pi \times {10}^{-7}=12.56\times {10}^{-26}T$