Magnetism

Prashant  Mital's Avatar
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5 Jan 2009 19:49:54 IST
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Please help me with the last question (Q. 61) in HC Verma Chapter 34 (Magnetic Field)
Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Physics , Magnetism

I am unable to obtain the function that must be integrated in order to get the final expression. Please help.



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Ashwini's Avatar

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6 Jan 2009 18:42:41 IST
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Hi
Kindly post complete question, for fast replies

~moderator
Prashant  Mital's Avatar

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7 Jan 2009 00:45:53 IST
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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

Prakhar Banga 's Avatar

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7 Jan 2009 14:41:25 IST
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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 = =




 





 


Hence proved.

Madmax's Avatar

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7 Jan 2009 19:15:46 IST
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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



current



Now magnetic moment =nIA (with n=1)

so





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

cheers!!

Abhishek Srivastava's Avatar

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19 Jan 2009 23:01:56 IST
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hi this  the solution if u are satisfy then plase vote for it

Abhishek Srivastava's Avatar

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24 Jan 2009 18:44:45 IST
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hey tell me are you satisfy now with this mathod of intregration.............voteme mean tell me




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