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. 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
$n e=100\, e =100 \times 1.6 \times 10^{-19} C$
$=1.6 \times 10^{-17} C$
Radius of circular path $(r)=0.8\, m$
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
$B=\mu_{0} \times \frac{i}{2 r}=\mu_{0} \times \frac{1.6 \times 10^{-17}}{2 \times 0.8}$
$=10^{-17} \mu_{0}$