Question

A helium nucleus makes a full rotaion in a circle of radius $0.8\text{m}$ in two seconds. The value of the magnetic field $B$ at the centre of the circle will be

• A. $\frac{{10}^{19}}{{\mu }_0}$
• B. ${10}^{-19}{\mu }_0$
• C. $2\times {10}^{-10}{\mu }_0$
• D. $\frac{2\times {10}^{-10}}{{\mu }_0}$

Answer 1

The correct option is B ${10}^{-19}{\mu }_0$

Current due to motion of charge
$i=\frac{q}{T}=\frac{2\times 1.6\times {10}^{19}}{2}=1.6\times {10}^{-19}A$
Magnetic field at centre
$\therefore B=\frac{{\mu }_{0}i}{2r}=\frac{{\mu }_0\times 1.6\times {10}^{-19}}{2\times 0.8}={\mu }_0\times {10}^{-19}$