Question

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

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

Answer 1

The correct option is A ${10}^{-17}{\mu }_0$

Here change on particle

$q=ne=100e=100\times 1.6\times {10}^{-13}C$

$=1.6\times {10}^{-17}C$

Radius of circular path r = 0.8 m

Time period T = 1 rotation/s

The current associated with the particle is

$i=\frac{charge}{time}$

$=1.6\times {10}^{-17}A$

Magnetic field produced at the centre of the coil is given by

$B={\mu }_0\times \frac{i}{2r}$

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

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