Question

An electron is revolving round a proton, producing a magnetic field of `16 weber/m^2` in a circular orbit of radius `1Å`. Its angular velocity is?

Answer 1

Electron revolving round a proton constitute flow of current in a circular ring of radius `1Å`. We know from Biot-Savart Law that a current produces magnetic field. The situation is depicted in the figure below.

Let an infinitesimal length element `vec(dL)` of current element be located at the circumference of circle of radius `R` having radial unit vector `hatr` as shown. Magnetic field `vec(dB)` produced at the center of circle is given as

`vec(dB)=(mu_0Ivec(dL)xxhatr)/(4piR^2)`
where `mu_0` is the magnetic constant or the permeability of free space `=4pixx10^-7Hm^-1`

We see that the angle between each current element and radial unit vector `=90^@`, and `sin 90^@=1` in the cross product. To calculate total magnetic field we integrate both sides with respect to respective variables.

`B=(mu_0I)/(4piR^2)oint " "dL`

We also see that for all points along the path and the distance to the center is constant in such a case line integral is equal to the circumference of the circle.

`=>B=(mu_0I)/(4piR^2)2piR`
`=>B=(mu_0I)/(2R)` .....(1)

Current `I` due to moving electron having charge `e` is `I=e/t`.
If `f` is number of revolutions made by the electron per second, current is given as

`I=ef`

Now `f` is related to angular velocity through the expression

`omega=2pif`

Hence current in terms of angular velocity is

`I=exxomega/(2pi)`

Inserting given values in (1) we get

`16=(4pixx10^-7xx(1.60 × 10^-19)xxomega/(2pi))/(2xx10^-10)`
`=>omega=(16xx10^-10)/(10^-7xx(1.60 × 10^-19))`
`=>omega=10^17" radian"cdot" s"^-1`