in this example we are given that a flat disc
of radius r charged uniformly on its surface
at a surface charge density sigma. and about
its central axis of rotation it rotates at
an angular speed omega. and we are required
to find the magnetic moment of the disc due
to rotation of charges. now in this situation.
say if we are given with a disc. which is
of, radius r. it is charged on the surface
at sigma coulomb per meter square surface
charge density. and if the disc is rotating
or spinning at an angular speed omega. then
due to rotation of all these charges we are
required to find its magnetic moment. for
which we just consider an elemental, ring.
of radius x, which is of width d x like, which
i am drawing here in the figure. then this
ring will behave like an equivalent current
which is also spinning. and if we calculate
the charge on this element this can be written
as sigma multiplied by 2 pie x d-x. because
sigma is the surface charge density and 2
pie x d x is the area occupied by this elemental
ring. now, if we calculate the, equivalent
current. in elemental ring. then this equivalent
current d-i we can write as simply d-q omega
by 2 pie because d-q charge is. rotating at
an. angular speed omega, so its frequency
will be omega by 2 pie and current is charge
multiplied by frequency. so here using this
current we can find out the magnetic moment.
of. element, we are talking about this elemental
ring which is say d-m. so this can be written
as d-i multiplied by pie x square because
the area enclosed by this current circulating
current d-i. in this element will be pie x
square. so if i substitute the values, d-q
i can write as sigma into 2 pie x d-x. multiplied
by pie x square. so in this situation. if
i substitute, I have just left the value of
omega upon 2 pie. if we just calculate here
pie gets cancelled out, 2 also gets cancelled
out. the value of d-m we are getting is. sigma
pie, omega, x cube. d-x. and the total magnetic
moment. of disc can be calculated simply by
integrating this which is integration of d-m.
here sigma pie omega are constant so it is
integrated x cube d-x from zero to r. on integrating
you can see it’ll be x four by 4 on substituting
limits we are getting 1 by 4 sigma pie omega
r four. that will be the answer to this problem.
we also now that in case of a uniformly charge
and uniformly dens rotating body, symmetric
rotating body. the ratio of. angular momentum
to magnetic moment is always a constant so
lets use this logic to calculate it by using
an alternative method, which will continue
on the next sheet. let us solve the same problem
by an alternative method. here we can write.
as. disc is uniformly charged. we can use.
the concept that the ratio of magnetic moment
to angular momentum for uniformly charged
bodies. is always a constant which is written
as q by 2 m. where q is the uniformly distributed
charge on body and m is the mass of body.
so from this expression we can say. magnetic
moment. of, disc can be directly given as
m is equal to it is q by 2 m multiplied by
the, angular momentum. so here, angular momentum
we can directly write as i omega. for the
disc moment of inertia is half m-r square
multiplied by the angular speed. and in this
expression m gets cancelled out. and the result
we are getting is m is equal to 1 by 4. q
omega r square. and the total charge we can
write as charge of disc is, charge density
surface charge density multiplied by pie r
square on the disc. so if we substitute it
here. we are getting it 1 by 4. sigma pie
r to power 4 omega, that will be the answer
to this problem which we have directly getting
here.
