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?

1 Answer
Jan 3, 2018

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 $\vec{\mathrm{dL}}$ of current element be located at the circumference of circle of radius $R$ having radial unit vector $\hat{r}$ as shown. Magnetic field $\vec{\mathrm{dB}}$ produced at the center of circle is given as

$\vec{\mathrm{dB}} = \frac{{\mu}_{0} I \vec{\mathrm{dL}} \times \hat{r}}{4 \pi {R}^{2}}$
where ${\mu}_{0}$ is the magnetic constant or the permeability of free space $= 4 \pi \times {10}^{-} 7 H {m}^{-} 1$

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

$B = \frac{{\mu}_{0} I}{4 \pi {R}^{2}} \oint \text{ } \mathrm{dL}$

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.

$\implies B = \frac{{\mu}_{0} I}{4 \pi {R}^{2}} 2 \pi R$
$\implies B = \frac{{\mu}_{0} I}{2 R}$ .....(1)

Current $I$ due to moving electron having charge $e$ is $I = \frac{e}{t}$.
If $f$ is number of revolutions made by the electron per second, current is given as

$I = e f$

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

$\omega = 2 \pi f$

Hence current in terms of angular velocity is

$I = e \times \frac{\omega}{2 \pi}$

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))
$\implies \omega = {10}^{17} {\text{ radian"cdot" s}}^{-} 1$