Question

A helium nucleus makes a full rotation in a circle of radius $0.8m$ 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. $2\times {10}^{-19}{\mu }_0$
• D. $\frac{2\times {10}^{-19}}{{\mu }_0}$

Answer 1

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

The current due to the revolution of the He nucleus is $I=\frac{q\omega }{2\pi }=\frac{2\times 1.6\times {10}^{-19}}{2}=1.6\times {10}^{-19}A$
Magnetic field at the center $B=\frac{{\mu }_{0}I}{2r}=\frac{{\mu }_0\times 1.6\times {10}^{-19}}{2\times 0.8}={10}^{-19}{\mu }_0$