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The magnetic moment of an electron orbiting in a circulate orbit of radius $r$ with a speed $v$ is equal to
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Verified Answer
The correct answer is:
$evr/2$
Magnetic moment $\mathrm{M}=\mathrm{NiA}$
where, $\mathrm{N}=$ number of turns of the current loop and $\mathrm{i}=$ current.
Since, the orbiting electron behaves as a current loop of current $i$, we can write
$\mathrm{i}=\frac{\mathrm{e}}{\mathrm{T}}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{r}}{\mathrm{V}}}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}}$
where, $\mathrm{A}=$ area of the loop $=\pi \mathrm{r}^2$.
$\Rightarrow \quad M=(1)\left(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\right)\left(\pi \mathrm{r}^2\right)=\frac{\mathrm{evr}}{2}$
where, $\mathrm{N}=$ number of turns of the current loop and $\mathrm{i}=$ current.
Since, the orbiting electron behaves as a current loop of current $i$, we can write
$\mathrm{i}=\frac{\mathrm{e}}{\mathrm{T}}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{r}}{\mathrm{V}}}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}}$
where, $\mathrm{A}=$ area of the loop $=\pi \mathrm{r}^2$.
$\Rightarrow \quad M=(1)\left(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\right)\left(\pi \mathrm{r}^2\right)=\frac{\mathrm{evr}}{2}$
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