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An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to
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The correct answer is:
$\sqrt{\frac{v}{B}}$
The time period of electron moving in a circular orbit
$T=\frac{\text { circumference of circular path }}{\text { speed }}$
$=\frac{2 \pi r}{v}$
and equivalent current due to electron flow
$I=\frac{e}{T}=\frac{e}{(2 \pi r / v)}=\frac{e v}{2 \pi r}$
Magnetic field at centre of circle
$B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 e v}{4 \pi r^2}$
$\Rightarrow r \propto \sqrt{\frac{v}{B}}$
$T=\frac{\text { circumference of circular path }}{\text { speed }}$
$=\frac{2 \pi r}{v}$
and equivalent current due to electron flow
$I=\frac{e}{T}=\frac{e}{(2 \pi r / v)}=\frac{e v}{2 \pi r}$
Magnetic field at centre of circle
$B=\frac{\mu_0 I}{2 r}=\frac{\mu_0 e v}{4 \pi r^2}$
$\Rightarrow r \propto \sqrt{\frac{v}{B}}$
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