- #1
Pengwuino
Gold Member
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- 20
The construction of and excitation of an infinite, straight, right circular solenoid of radius a, with N turns per unit length, are such taht tis current I(t) is the same everywehre along its length and is changed very slowly in time. Show that the fields far from the solenoid are approximately:
[tex]E \approx \frac{{ - \mu _0 Na^2 }}{{2\rho }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi [/tex]
Attempt:
We know from previous work that for a localized charge distribution far from the origin for slowly changing currents with no charge density and only dipole terms is:
[tex]
E = \frac{{\mu _0 }}{{4\pi r^2 }}\hat r \times \frac{{\partial m(t - \frac{r}{c})}}{{\partial t}}[/tex]
So it should be a simple matter of computing the dipole moment per unit length. Now I've been second guessing myself continually on the charge density but I've kinda come to the conclusion that it the multipole moment should be
[tex]
\begin{array}{l}
m = \frac{1}{2}\int {I(t - \frac{\rho }{c})N\delta (\rho ' - a)\rho '^2 d\rho 'd\phi '(\hat \rho ' \times \hat \phi ')} \\
= 2\pi a^2 I(t - \frac{\rho }{c})N\hat z \\
\end{array}[/tex]
However, simply plugging this making the assumption that due to the symmetry, r becomes [tex]\rho[/tex], gives:
[tex]
E = \frac{{ - \mu _0 a^2 N}}{{2\rho ^2 }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi [/tex]
which is off by that [tex]\rho^2[/tex] instead of [tex]\rho[/tex]. Is there something obvious I am missing here?
[tex]E \approx \frac{{ - \mu _0 Na^2 }}{{2\rho }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi [/tex]
Attempt:
We know from previous work that for a localized charge distribution far from the origin for slowly changing currents with no charge density and only dipole terms is:
[tex]
E = \frac{{\mu _0 }}{{4\pi r^2 }}\hat r \times \frac{{\partial m(t - \frac{r}{c})}}{{\partial t}}[/tex]
So it should be a simple matter of computing the dipole moment per unit length. Now I've been second guessing myself continually on the charge density but I've kinda come to the conclusion that it the multipole moment should be
[tex]
\begin{array}{l}
m = \frac{1}{2}\int {I(t - \frac{\rho }{c})N\delta (\rho ' - a)\rho '^2 d\rho 'd\phi '(\hat \rho ' \times \hat \phi ')} \\
= 2\pi a^2 I(t - \frac{\rho }{c})N\hat z \\
\end{array}[/tex]
However, simply plugging this making the assumption that due to the symmetry, r becomes [tex]\rho[/tex], gives:
[tex]
E = \frac{{ - \mu _0 a^2 N}}{{2\rho ^2 }}\frac{{\partial I(t - \frac{\rho }{c})}}{{\partial t}}\hat \phi [/tex]
which is off by that [tex]\rho^2[/tex] instead of [tex]\rho[/tex]. Is there something obvious I am missing here?