Search any question & find its solution
Question:
Answered & Verified by Expert
An electron makes a full rotation in a circle of radius $0.8 \mathrm{~m}$ in one second. The magnetic field at the centre of the circle is
$$
\left(\mu_0=4 \pi \times 10^{-7} \text { SI units }\right)
$$
Options:
$$
\left(\mu_0=4 \pi \times 10^{-7} \text { SI units }\right)
$$
Solution:
2470 Upvotes
Verified Answer
The correct answer is:
$4 \pi \times 10^{-26} \mathrm{~T}$
$$
\begin{aligned}
& \omega=\frac{2 \pi}{T} \\
& I=\frac{q \omega}{2 \pi}=\frac{1.6 \times 10^{-19} \times 2 \pi}{2 \pi} \quad \ldots .\left(\because I=\frac{q}{t}\right) \\
& I=1.6 \times 10^{-19} A
\end{aligned}
$$
$\therefore \quad$ The magnetic field at the centre of the circle is:
$$
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.8} \\
& \mathrm{~B}=4 \pi \times 10^{-26} \mathrm{~T}
\end{aligned}
$$
\begin{aligned}
& \omega=\frac{2 \pi}{T} \\
& I=\frac{q \omega}{2 \pi}=\frac{1.6 \times 10^{-19} \times 2 \pi}{2 \pi} \quad \ldots .\left(\because I=\frac{q}{t}\right) \\
& I=1.6 \times 10^{-19} A
\end{aligned}
$$
$\therefore \quad$ The magnetic field at the centre of the circle is:
$$
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 1.6 \times 10^{-19}}{2 \times 0.8} \\
& \mathrm{~B}=4 \pi \times 10^{-26} \mathrm{~T}
\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.