# 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,
$B=\frac{{\mu }_{o}}{4\pi }\frac{2\pi I}{r}\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{o}}{2r}I$
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,
$B=\frac{{\mu }_{o}}{2r}\frac{2e}{t}\phantom{\rule{0ex}{0ex}}B=\frac{{\mu }_{o}}{r}\frac{e}{t}$
On substituting the values we get,
o B=1×10-19×μo
Thus, the magnetic field at the centre of the circle is
B = μ​​​​​​o × 10-19 T

Regards

• 0
What are you looking for?