Question

A particle having charge $100$ times that of an electron is revolving in a circular path of radius $0.8m$ with one rotation per second. Magnetic field produced at center of particle is:

• A. $10^{-17}{\mu }_0$
• B. $10^{-11}{\mu }_0$
• C. $10^{-7}{\mu }_0$
• D. $10^{-3}{\mu }_0$

Answer 1

Answer: (A)

Here : Charge on particle
$ne=100e=100\times 1.6\times 10^{-19}C$
$=1.6\times 10^{-17}C$
Radius of circular path $(r)=0.8m$
Time period $T=1$ rotation/ sec
The current associated with the particle is
$i=\frac{\text{ charge }}{\text{ time }}=\frac{1.6\times 10^{-17}}{1}$
$=1.6\times 10^{-17}A$
Hence, magnetic field produced at the centre of the coil is given by
$B={\mu }_0\times \frac{i}{2r}={\mu }_0\times \frac{1.6\times 10^{-17}}{2\times 0.8}$
$=10^{-17}{\mu }_0$