Question

A particle having charge $100$ times that of an electron is revolving in a circular path of radius. $0.8\, m$ with $1\, rot/s$. 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}{2\,a}N{{A}^{-1}}{{m}^{-1}} $
Also, current $ i=\frac{\text{charge}\left( q \right)}{\text{time}\,(t)} $ = change $ \times $ frequency
Given, $ q=100\,e=100\times 1.6\times {{10}^{-19}}C $
$ =1.6\times {{10}^{-17}}C f=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}} $