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?

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#