Question

An electron of charge e is going around in an orbit of radius R meters in a hydrogen atom with velocity v m / sec . The magnetic flux density associated with it at its centre is

Answer 1

\frac{\mu_0 ev}{4\pi R^2}

Answer 2

The motion of an electron in a circular orbit constitutes an electric current.

1. Define the current (I): The electron, with charge 'e', completes one revolution in time 'T'. The current due to this motion is defined as the charge passing a point per unit time: $I = \frac{e}{T}$

2. Relate time period (T) to velocity (v) and radius (R): For a circular orbit of radius R and velocity v, the distance covered in one revolution is the circumference, $2\pi R$. Therefore, the time period is: $T = \frac{2\pi R}{v}$

3. Substitute T into the current formula: $I = \frac{e}{\frac{2\pi R}{v}} = \frac{ev}{2\pi R}$

4. Calculate the magnetic field at the center: The magnetic field (magnetic flux density) B at the center of a circular loop of radius R carrying a current I is given by the formula: $B = \frac{\mu_0 I}{2R}$ where $\mu_0$ is the permeability of free space.

5. Substitute the expression for I into the magnetic field formula: $B = \frac{\mu_0}{2R} \left( \frac{ev}{2\pi R} \right)$ $B = \frac{\mu_0 ev}{4\pi R^2}$

This is the magnetic flux density associated with the electron at the center of its orbit.

Explanation of the solution: The revolving electron forms a current loop. The current is $I = \frac{\text{charge}}{\text{time period}} = \frac{e}{T}$. The time period for one revolution is $T = \frac{2\pi R}{v}$. Substituting T, the current becomes $I = \frac{ev}{2\pi R}$. The magnetic field at the center of a circular current loop is $B = \frac{\mu_0 I}{2R}$. Substituting the expression for I, we get $B = \frac{\mu_0 (ev/2\pi R)}{2R} = \frac{\mu_0 ev}{4\pi R^2}$.