MHT CET · Physics · Magnetic Effects of Current
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, \(\left.\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)\)
- A \(10^{-17} \mu_0\)
- B \(10^{-3} \mu_0\)
- C \(10^{-7} \mu_0\)
- D \(10^{-11} \mu_0\)
Answer & Solution
Correct Answer
(A) \(10^{-17} \mu_0\)
Step-by-step Solution
Detailed explanation
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.
\(
\begin{aligned}
& \therefore \mathrm{I}=\mathrm{qf}=100 \mathrm{e} \mathrm{A} \\
& \therefore \mathrm{B}=\frac{\mu_0 \times 100 \mathrm{e}}{2 \times 0.8}=\frac{\mu_0 \times 100 \times 1.6 \times 10^{-19}}{1.6} \\
& =10^{-17} \mu_0
\end{aligned}
\)
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