Question

A particle having charge $100$ times that of an electron is revolving in a circular path of radius. $0.8m$ with $1rot/s$. The magnetic field produced at the center will be :

• A. ${10}^{-17}{\mu }_0$
• B. ${10}^{-14}{\mu }_0$
• C. ${10}^{-7}{\mu }_0$
• D. ${10}^{-5}{\mu }_0$

Answer 1

Answer: (A)

For a coil of radius $a$, carrying current $i$,
magnetic field at the centre is given -by
$B=\frac{{\mu }_{0}i}{2a}N{A}^{-1}{m}^{-1}$
Also, current $i=\frac{\text{charge}\left(q\right)}{\text{time}(t)}$ = change $\times$ frequency
Given, $q=100e=100\times 1.6\times {10}^{-19}C$
$=1.6\times {10}^{-17}Cf=1$ rot/s .
$\therefore i=1.6\times {10}^{-17}A$
Hence, $B={\mu }_0\times \frac{1.6\times {10}^{-17}}{2\times 0.8}={10}^{-17}{\mu }_0$