Q91 / 120 · VITEEE 2011 · Physics
A particle carrying a charge to 100 times the charge on an electron is rotating per second in a circular path of radius $0.8 \mathrm{~m}$. The value of the magnetic field produced at the centre will be $\left(\mu_{0}=\right.$ permeabil ity for vacuum $)$
- A$\frac{10^{-7}}{\mu_{0}}$
- Correct answerB$10^{-17} \mu_{0}$
- C$10^{-6} \mu_{0}$
- D$10^{-7} \mu_{0}$
Solution
Current, i $=\frac{q}{t}=100 \times \mathrm{e}$
$$
\begin{array}{l}
\text { Magnetic field, } \mathrm{B}_{\text {centre }}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi i}{r} \\
=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \times 100 e}{r} \\
\frac{\mu_{0} \times 200 \times 1.6 \times 10^{-19}}{4 \times 0.8}=10^{-17} \mu_{0}
\end{array}
$$
Hence, the correct answer is option: 'B'.