Question

A helium nuclues is moving in a circular path of radius $0.8m$. If it takes $2$ sec to complete one revolution. Find out magnetic field produced at the centre of the circle.

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

Answer 1

The correct option is A ${\mu }_0\times {10}^{-19}T$

Given: r=0.8 m, t=2 sec

Solution: Formula used for magnetic field of moving particle in circular path is:

$B=\frac{{\mu }_{0}Ni}{2r}=\frac{{\mu }_{0}Nq}{2rt}=\frac{{\mu }_{0}N\left(ne\right)}{2rt}=\frac{{\mu }_0\times 1\times 2\times 1.6\times {10}^{-19}}{2\times 2\times 0.8}$

Finally we get,

$B={\mu }_0\times {10}^{-19}$T

Hence the correct option is A