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NEET PHYSICSMOVING CHARGES AND MAGNETISMEasy

Question

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}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMelectronmovingcircularradiusrotations

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