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)

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

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..

cud u define it?

www.comsol.com/showroom/gallery/15.php



en.wikipedia.org/wiki/Helmholtz_coil

physicsx.pr.erau.edu/HelmholtzCoils/index.html

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