Question

An electron moves around the nucleus in a circular orbit of radius \(r\) and makes \(n\) revolutions per second. The value of equivalent current in the orbit is :

• A. \(\dfrac{e}{n}\)
• B. \(ne\)
• C. \(\dfrac{ne}{r}\)
• D. \(\dfrac{e}{nr}\)

Hint:

Whenever a charge \(q\) completes \(f\) revolutions per second, the equivalent current is \[ I=qf. \] Here \(q=e\) and \(f=n\), therefore \(I=en\).

Answer 1

The Correct Option is B

Concept: Electric current is defined as the rate of flow of charge. Mathematically, \[ I=\frac{Q}{t} \] where \[ Q=\text{charge flowing}, \] and \[ t=\text{time taken}. \] An electron revolving around the nucleus constitutes a circulating charge. Such a moving charge behaves like a current loop and therefore produces a magnetic field.

Step 1: Determine the charge passing a point in one revolution. The charge carried by one electron is \[ e=1.6\times10^{-19}\,\text{C}. \] Whenever the electron completes one revolution, the charge \(e\) effectively passes a given point once. Thus, \[ Q=e. \]

Step 2: Relate frequency to time period. The electron makes \(n\) revolutions per second. Therefore, \[ n=\frac{1}{T} \] where \(T\) is the time period. Hence, \[ T=\frac{1}{n}. \]

Step 3: Use the definition of current. Current is \[ I=\frac{Q}{T}. \] Substituting \[ Q=e \] and \[ T=\frac{1}{n}, \] we get \[ I=\frac{e}{1/n}. \] \[ I=en. \]

Step 4: Interpret the result physically. The current increases if the electron completes more revolutions per second because charge passes the observation point more frequently. Thus the equivalent current associated with the orbiting electron is \[ \boxed{I=ne}. \]

Step 5: Select the correct option. Comparing with the given options, \[ \boxed{\text{(B) }ne} \] is the correct answer.