Physics - Magnetism Question with Solution | TestHub

Charge of the particle, $q=100e=100\times 1.6\times 10^{-19}\text{ C}=1.6\times 10^{-17}\text{ C}$

Frequency of rotation, $f=1\text{ Hz}$

Equivalent current, $I=qf=1.6\times 10^{-17}\times 1=1.6\times 10^{-17}\text{ A}$

Magnetic field at the centre of the circular path is given by $B=\frac{{\mu }_{0}I}{2r}$

Substituting the values, we get:

$B=\frac{{\mu }_0\times 1.6\times 10^{-17}}{2\times 0.8}$

$B=\frac{{\mu }_0\times 1.6\times 10^{-17}}{1.6}$

$B=10^{-17}{\mu }_0$

Answer: $10^{-17}{\mu }_0$