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NEET PHYSICSEasy

An electron moving in a circular orbit of radius rr makes nn rotations per second. The magnetic field produced at the centre has magnitude:

A

μ0ne2πr\frac{\mu_0 n e}{2\pi r}

B

Zero

C

n2er\frac{n^2 e}{r}

D

μ0ne2r\frac{\mu_0 n e}{2r}

Step-by-Step Solution

An electron moving in a circular path creates an equivalent current loop . The electric current II is defined as the amount of charge passing a point per unit time, I=q/tI = q/t . Since the electron (charge ee) completes nn rotations per second, its frequency is nn, and the time for one rotation (period TT) is 1/n1/n . Therefore, the equivalent current is I=e/(1/n)=neI = e / (1/n) = ne . According to the sources, the magnitude of the magnetic field BB at the centre of a circular current loop of radius rr is given by B=μ0I2rB = \frac{\mu_0 I}{2r} . Substituting I=neI = ne into this formula gives B=μ0ne2rB = \frac{\mu_0 n e}{2r}.

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