An electron moving in a circular path of radius r makes n rotation per seconds. A the magnetic field produced at the centre has the magnitude

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\frac{μ_{0} ne}{2 r}

\frac{μ_{0} ne}{2 Πr}

\frac{μ_{0} n^{2} e}{2 r}

answer is B.

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Detailed Solution

A the magnetic field produced at the centre has the magnitude is given by

\frac{μ_{0} ne}{2 r}

Given,
Time taken =

\frac{1}{n}

The magnetic field equation is B=

\frac{µ_{0 I}}{2 r}

Where,

B = Magnetic field strength

µ_{0} = permeability of vacuum

I = current (flowing in conductor)

r = distance from conductor

I = \frac{charge}{time}

I = \frac{Q}{t}

Where,

I = current

Q = charge

t = time

e = electron

I = \frac{e}{1 / n}

B = \frac{µ_{0 n e}}{2 r}

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