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Magnetism

anchit saini's Avatar
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2 Jul 2008 07:35:47 IST
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Verify that on the axis of helmholtz coil , midway between the coils , the first 3 derivatives of B with respect to x vanish.


 


(This is the complete qn)


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gokul subramanian's Avatar

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2 Jul 2008 09:17:43 IST
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il try out subjectively:


 


 code: let m be 'mew not'


 


field=B={m*n*i(cos@1-cos@2)}/2


 


where @ is the angle made between the pt of consideration(o) of the field wid the end pt of the solenoid..


 


so x=Rcot@ where x is dist along hor of endpts from O..


 


as the pt u r considering is midpt @1=@2 +90


 


now diff B wrt @ (or x) and chk up..il try to post the whole thing soon


did for solenoid so not applicable for the q


 


 

gokul subramanian's Avatar

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2 Jul 2008 09:26:26 IST
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the formula is pretty easy to derive..


jus consider rings such tht number of turns/unit length is n then is length dx u will have num of turns =ndx and so on..

gokul subramanian's Avatar

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2 Jul 2008 09:29:39 IST
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cud u define it?

anchit saini's Avatar

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abhishek sinha's Avatar

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2 Jul 2008 16:40:32 IST
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Wikipedia has given the correct approach .


First calculate B(x) on the common axis . ( to do this just superpose two fields of two coils by putting  x= x ; and x= ( h-x) = ( R-x) respectively . )


Now evaluate the derivatives at x=R/2;


But for first two derivatives , if u know vector calculus , it will directly follow from the differential version of Maxwell's law .


We always have div B = 0 so B , at the common axis being a fn of only x we get dB/dx= 0 ............( 1)


again curl B = 0 ( since at the midpoint J = 0  and B here a static field )


so curl curl B = 0


so  del squared B = 0 ( as div.B = 0 )


i.e. d2B/dx2 = 0




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