biot savart question..

**transgalactic** · 22nd August 2009 01:50 PM

here is a question:
http://i27.tinypic.com/14tsro1.gif

a thin ring with uniform resistance $\rho$
with radius $r_0$ connected to current I.
the ring is divided into two areas.
calculate the total magnetic field in the center

by current divider:
$I_1=I\frac{l_2}{l_1+l_2}\\$
$I_2=I\frac{l_1}{l_1+l_2}\\$
by biot savart law :
$dB=\frac{\mu _0 I ds\times \hat{r}}{4\pi r^2}\\$
$B_1=\int_{0}^{l_1}\frac{\mu _0 I ds }{4\pi r^2}(+\hat{k})\\$
$B_2=\int_{0}^{l_2}\frac{\mu _0 I ds }{4\pi r^2}(-\hat{k})$
$B=B_1+B_2$

and this is the formal solution
http://i29.tinypic.com/2moc8ds.jpg
they have r^3 in the denominator
why??
the crosst prodct of ds with r is 1
why they take cross product of theta with r??
i cant understand the solution
how they built this integral
??

**MrAl** · 26th August 2009 03:55 PM

Originally Posted by transgalactic

here is a question:
http://i27.tinypic.com/14tsro1.gif

a thin ring with uniform resistance $\rho$
with radius $r_0$ connected to current I.
the ring is divided into two areas.
calculate the total magnetic field in the center

by current divider:
$I_1=I\frac{l_2}{l_1+l_2}\\$
$I_2=I\frac{l_1}{l_1+l_2}\\$
by biot savart law :
$dB=\frac{\mu _0 I ds\times \hat{r}}{4\pi r^2}\\$
$B_1=\int_{0}^{l_1}\frac{\mu _0 I ds }{4\pi r^2}(+\hat{k})\\$
$B_2=\int_{0}^{l_2}\frac{\mu _0 I ds }{4\pi r^2}(-\hat{k})$
$B=B_1+B_2$

and this is the formal solution
http://i29.tinypic.com/2moc8ds.jpg
they have r^3 in the denominator
why??
the crosst prodct of ds with r is 1
why they take cross product of theta with r??
i cant understand the solution
how they built this integral
??

Hi there,

Where are you getting these 'formal' solutions from anyway?
Take a look on the hyperphysics web site as they have
some good information that will help.

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