Search any question & find its solution
Question:
Answered & Verified by Expert
A particle having a charge $100 \mathrm{e}$ is revolving in a circular path of radius $0.8 \mathrm{~m}$ with 1.r.p.s The magnetic field produced at the center of the circle in SI unit is ( $\mu_0$ permeability of vacuum, $\left.\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)$
Options:
Solution:
1891 Upvotes
Verified Answer
The correct answer is:
$10^{-17} \mu_0$
The magnetic field produced by a circular coil of radius $r$ at its centre is given by
$$
\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}
$$
In this case $r=0.8 \mathrm{~m}$
Frequency $\mathrm{f}=1$ r.p.s.
$$
\begin{aligned}
& \therefore \mathrm{I}=\mathrm{qf}=100 \mathrm{e} \mathrm{A} \\
& \therefore \mathrm{B}=\frac{\mu_0 \times 100 \mathrm{e}}{2 \times 0.8}=\frac{\mu_0 \times 100 \times 1.6 \times 10^{-19}}{1.6} \\
& =10^{-17} \mu_0
\end{aligned}
$$
$$
\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}
$$
In this case $r=0.8 \mathrm{~m}$
Frequency $\mathrm{f}=1$ r.p.s.
$$
\begin{aligned}
& \therefore \mathrm{I}=\mathrm{qf}=100 \mathrm{e} \mathrm{A} \\
& \therefore \mathrm{B}=\frac{\mu_0 \times 100 \mathrm{e}}{2 \times 0.8}=\frac{\mu_0 \times 100 \times 1.6 \times 10^{-19}}{1.6} \\
& =10^{-17} \mu_0
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.