Kindly post complete question, for fast replies
The question is:
Consider a solid sphere of radius 'r' and mass 'm' which has a charge 'q' distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed 'ω'. Show that the magnetic moment 'μ', and the angular momentum 'l' of the sphere are related as :
μ = ql / 2m
Let the sphere's centre be at the origin and it be rotating about the z axis.
So angular momentum = [ Assuming that it rotates counter clockwise as seen from above ]
Let there be a disk whose z coordinate is a and thickness da.
Let there be a ring of radius x on this disk and thickness dx.
Area of ring = .
Volume of ring = 2 x dx da.
Charge on ring = 2 x dx da / (4/3 ) Q = .
As time period of a complete revolution is 2 / , the current in this ring is / (2 / ).
Area vector of the loop of the current is .
So of the ring is * / (2 / )
Integrating the expression w.r.t. x to find of the disk and then keeping the limits 0 and ( the z coordinate of the disk is a so its radius is ), we get the of the disk as .
To get the of the sphere, we have to integrate this and put the limits -R to R (as a varies from -R to R).
Expansion of the square, integration and putting the limits leaves us with = =
One thing I'd like to point out the following fact:
ALL SYMMETRIC SYSTEMS WITH CHARGE EVENLY SPREAD CAN BE EQUATED TO A ELECTRON ROTATING AROUNG THE NUCLEUS
Tht's wht I remember...... and this fact makes such problems very easy
Now magnetic moment =nIA (with n=1)
Now as mvr =L and area=
This reduces calculation time by a massive amount!! And mind you, this is not just a coincidence.its a proven fact.
But if you wanna do it by the integration method, 3774's method is perfect!!
hope this helps
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