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An electron revolves in a circle of radius $0.4 Ã…$ with a speed of $10^6 \mathrm{~m} / \mathrm{s}$ in a hydrogen atom. The magnetic field produced at the centre of the orbit due to the motion of the electron (in Tesla) is : $\left[\mu_0=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right.$, charge on the electron $\left.=1.6 \times 10^{-19} \mathrm{C}\right]$
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Verified Answer
The correct answer is:
$10$
$$
r=0.4 Ã…,=0.4 \times 10^{-10} \mathrm{~m}, v=10^6 \mathrm{~m} / \mathrm{s}
$$
The moving electron will constitute an electric current
Electric current $i=\frac{q}{t}$
$$
\begin{aligned}
i & =\frac{q}{\frac{2 \pi r}{v}}=\frac{q v}{2 \pi r} \\
i & =\frac{1.6 \times 10^{-19} \times 10^6}{2 \pi \times 0.4 \times 10^{-10}} \\
& =\frac{1.6 \times 10^{-3}}{0.8 \pi}
\end{aligned}
$$
Magnetic produced at the centre
$$
\begin{aligned}
B & =\frac{\mu_0 i}{2 r} \\
B & =\frac{4 \pi \times 10^{-7}}{2 \times 0.4 \times 10^{-10}} \times \frac{1.6 \times 10^{-3}}{0.8 \pi} \\
& =\frac{4 \pi \times 1.6}{0.8 \times 0.8 \pi}=\frac{6.4}{0.64} \\
& =10 \text { tesla }
\end{aligned}
$$
r=0.4 Ã…,=0.4 \times 10^{-10} \mathrm{~m}, v=10^6 \mathrm{~m} / \mathrm{s}
$$
The moving electron will constitute an electric current
Electric current $i=\frac{q}{t}$
$$
\begin{aligned}
i & =\frac{q}{\frac{2 \pi r}{v}}=\frac{q v}{2 \pi r} \\
i & =\frac{1.6 \times 10^{-19} \times 10^6}{2 \pi \times 0.4 \times 10^{-10}} \\
& =\frac{1.6 \times 10^{-3}}{0.8 \pi}
\end{aligned}
$$
Magnetic produced at the centre
$$
\begin{aligned}
B & =\frac{\mu_0 i}{2 r} \\
B & =\frac{4 \pi \times 10^{-7}}{2 \times 0.4 \times 10^{-10}} \times \frac{1.6 \times 10^{-3}}{0.8 \pi} \\
& =\frac{4 \pi \times 1.6}{0.8 \times 0.8 \pi}=\frac{6.4}{0.64} \\
& =10 \text { tesla }
\end{aligned}
$$
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