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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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a

\frac{B}{V}

b

\frac{v}{B}

c

\sqrt{\frac{v}{B}}

d

\sqrt{\frac{B}{v}}

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detailed solution

Correct option is C

The time period of electron moving in a circular orbit

$T = \frac{circumference of circular path}{speed} = \frac{2 π r}{v} E l e c t r i c c u r r e n t c o r r e s p o n d s t o t h e r e v o l u t i o n o f e l e c t r o n i s I = \frac{e}{T} = \frac{e}{(2 π r/v)} = \frac{ev}{2 π r} M a g n e t i c f i e l d a t c e n t r e o f c i r c l e B = \frac{μ_{0} I}{2 r} = \frac{μ_{0} ev}{4 π r^{2}} H e n c e r = \sqrt{\frac{μ_{0}}{4 π} . \frac{ev}{B}} \Rightarrow r \propto \sqrt{\frac{v}{B}}$

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