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An electron in a circular orbit of radius $0.05 \mathrm{nm}$ performs $10^{16}$ revolutions per second. The maghetic moment due to this rotation of electron is (in $\mathrm{Am}^{2}$ )
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Verified Answer
The correct answer is:
$1.26 \times 10^{-23}$
Given, $r=0.05 \mathrm{nm}=0.05 \times 10^{-9} \mathrm{m}$
$$
\begin{array}{l}
\qquad\left(\because 1 \mathrm{nm}=10^{-9} \mathrm{m}\right) \\
n=10^{16} \text { revolutions } \\
e=1.6 \times 10^{-19} \mathrm{C}
\end{array}
$$
We know that
The magnetic moment $M=A L$
$$
\begin{aligned}
M &=\pi r^{2} \times n e \\
M &=3.14 \times\left(0.05 \times 10^{-9}\right)^{2} \times 10^{16} \times 1.6 \times 10^{-19} \\
M &=0.1256 \times 10^{-18} \times 10^{16} \times 10^{-19} \\
\text {or } M &=1.26 \times 10^{-23} \mathrm{Am}^{2}
\end{aligned}
$$
$$
\begin{array}{l}
\qquad\left(\because 1 \mathrm{nm}=10^{-9} \mathrm{m}\right) \\
n=10^{16} \text { revolutions } \\
e=1.6 \times 10^{-19} \mathrm{C}
\end{array}
$$
We know that
The magnetic moment $M=A L$
$$
\begin{aligned}
M &=\pi r^{2} \times n e \\
M &=3.14 \times\left(0.05 \times 10^{-9}\right)^{2} \times 10^{16} \times 1.6 \times 10^{-19} \\
M &=0.1256 \times 10^{-18} \times 10^{16} \times 10^{-19} \\
\text {or } M &=1.26 \times 10^{-23} \mathrm{Am}^{2}
\end{aligned}
$$
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