Factor of 1/2 on the cross-product of vectors

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I was trying to derive the cross-product of vectors:



$vecB = B_theta$



$vecJ = -frac1mu_0 fracddrB_z(r)$



substituted on



$$nabla p = vecJ times vecB$$



getting



$$fracddrBigg(p+fracB_theta ^2mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



but the text says it is



$$fracddrBigg(p+fracB_theta^22mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



Where does the 1/2 factor come from in term 2?



Thanks!




vectors cross-product




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  • 3




    Note that we have $$B_thetafracdB_thetadr=frac12fracdB^2_thetadr$$
    – Mark Viola
    Aug 2 at 16:26










  • Ohhhh thanks!!!
    – rebc
    Aug 2 at 16:31










  • You're welcome! My pleasure.
    – Mark Viola
    Aug 2 at 17:05














up vote
0
down vote

favorite












I was trying to derive the cross-product of vectors:



$vecB = B_theta$



$vecJ = -frac1mu_0 fracddrB_z(r)$



substituted on



$$nabla p = vecJ times vecB$$



getting



$$fracddrBigg(p+fracB_theta ^2mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



but the text says it is



$$fracddrBigg(p+fracB_theta^22mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



Where does the 1/2 factor come from in term 2?



Thanks!




vectors cross-product




share|cite|improve this question















  • 3




    Note that we have $$B_thetafracdB_thetadr=frac12fracdB^2_thetadr$$
    – Mark Viola
    Aug 2 at 16:26










  • Ohhhh thanks!!!
    – rebc
    Aug 2 at 16:31










  • You're welcome! My pleasure.
    – Mark Viola
    Aug 2 at 17:05












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I was trying to derive the cross-product of vectors:



$vecB = B_theta$



$vecJ = -frac1mu_0 fracddrB_z(r)$



substituted on



$$nabla p = vecJ times vecB$$



getting



$$fracddrBigg(p+fracB_theta ^2mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



but the text says it is



$$fracddrBigg(p+fracB_theta^22mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



Where does the 1/2 factor come from in term 2?



Thanks!




vectors cross-product




share|cite|improve this question











I was trying to derive the cross-product of vectors:



$vecB = B_theta$



$vecJ = -frac1mu_0 fracddrB_z(r)$



substituted on



$$nabla p = vecJ times vecB$$



getting



$$fracddrBigg(p+fracB_theta ^2mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



but the text says it is



$$fracddrBigg(p+fracB_theta^22mu_0Bigg) + fracB^2_thetamu_0 r = 0$$



Where does the 1/2 factor come from in term 2?



Thanks!





vectors cross-product





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 2 at 16:24









rebc

17210




17210







  • 3




    Note that we have $$B_thetafracdB_thetadr=frac12fracdB^2_thetadr$$
    – Mark Viola
    Aug 2 at 16:26










  • Ohhhh thanks!!!
    – rebc
    Aug 2 at 16:31










  • You're welcome! My pleasure.
    – Mark Viola
    Aug 2 at 17:05












  • 3




    Note that we have $$B_thetafracdB_thetadr=frac12fracdB^2_thetadr$$
    – Mark Viola
    Aug 2 at 16:26










  • Ohhhh thanks!!!
    – rebc
    Aug 2 at 16:31










  • You're welcome! My pleasure.
    – Mark Viola
    Aug 2 at 17:05







3




3




Note that we have $$B_thetafracdB_thetadr=frac12fracdB^2_thetadr$$
– Mark Viola
Aug 2 at 16:26




Note that we have $$B_thetafracdB_thetadr=frac12fracdB^2_thetadr$$
– Mark Viola
Aug 2 at 16:26












Ohhhh thanks!!!
– rebc
Aug 2 at 16:31




Ohhhh thanks!!!
– rebc
Aug 2 at 16:31












You're welcome! My pleasure.
– Mark Viola
Aug 2 at 17:05




You're welcome! My pleasure.
– Mark Viola
Aug 2 at 17:05















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