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\heading{Magnetic Field}
\auname{I.D. 039120068}
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{\bf The problem:}
The coaxial cable is built from two concentric cables. In the internal cable there is a current density of $\vec{J}=J_0\frac{r}{a}\hat{z}$
and in the external cable there is a current density of $\vec{J}=J_0\frac{r}{b}\hat{z}$.
Find the force of the magnetic field inside and outside the coaxial cable.
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\includegraphics{e_45_4_101_p0.jpg}
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{\bf The solution:}
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In general, we can calculate the magnetic field made by a thick current wire or a cylinder with the Ampere's law. Due to the symmetry the field is only in $\hat{\varphi}$.
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\begin{eqnarray}
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\oint_L{\vec{B}\cdot\vec{dl}}&=&4\pi\kappa\int{\vec{J}\cdot\vec{ds}}=4\pi\kappa 2\pi\int{J(r)rdr}\hat{\varphi}\\
\vec{B}&=&\frac{4\pi\kappa}{r}\int{J(r)rdr}\hat{\varphi} \\
\kappa&=&\frac{\mu _0}{4\pi}\\
J(r)&=&
\begin{cases}
\frac{J_0r}{a}&r\leq R_1\\
0&R_1R_3)
\end{cases}
\end{eqnarray}
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In order to calculate the magnetic field in the whole space we will use superposition of the magnetic field made by the inner cable $(\vec{B_1})$ and the outer cable $(\vec{B_2})$ \\
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a) for $r\leq R_1$:
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\begin{eqnarray}
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\vec{B_1}&=&\frac{4\pi\kappa}{r}\int_0^r{J(r)rdr}=\frac{4\pi\kappa J_0}{ar}\int_0^r{r^2dr}=\frac{4\pi\kappa r^2J_0}{3a}\hat{\varphi} \\
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\vec{B_2}&=&0
\end{eqnarray}
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b) for $R_1\leq r\leq R_2$:
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\begin{eqnarray}
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\vec{B_1}&=&\frac{4\pi\kappa}{r}\int_0^{r}{J(r)rdr}
=\frac{4\pi\kappa}{r}\left(\int_0^{R_1}{\frac{J_0}{a}r^2dr}+\int_{R_1}^r{0\ rdr}\right)
=\frac{4\pi\kappa R_1^3J_0}{3ar}\hat{\varphi} \\
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\vec{B_2}&=&0
\end{eqnarray}
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c) for $R_2\leq r\leq R_3$:
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\begin{eqnarray}
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\vec{B_1}&=&\frac{4\pi\kappa}{r}\int_0^{r}{J(r)rdr}
=\frac{4\pi\kappa}{r}\left(\int_0^{R_1}{\frac{J_0}{a}r^2dr}+\int_{R_1}^r{0\ rdr}\right)
=\frac{4\pi\kappa R_1^3J_0}{3ar}\hat{\varphi} \\
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\vec{B_2}&=&\frac{4\pi\kappa}{r}\int_0^{r}{J(r)rdr}
=\frac{4\pi\kappa}{r}\left(\int_0^{R_2}{0\ rdr}+\int_{R_2}^r{\frac{J_0}{b}r^2dr}\right)
=\frac{4\pi\kappa J_0(r^3-R_2^3)}{3br}\hat{\varphi} \\
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\vec B_{\tbox{total}}&=&4\pi\kappa J_0\left(\frac{R_1^3}{3ar}+\frac{r^3-R_2^3}{3br}\right)\hat{\varphi}
\end{eqnarray}
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d) for $r\geq R_3$:
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\begin{eqnarray}
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\vec{B_1}&=&\frac{4\pi\kappa}{r}\int_0^{r}{J(r)rdr}
=\frac{4\pi\kappa}{r}\left(\int_0^{R_1}{\frac{J_0}{a}r^2dr}+\int_{R_1}^r{0\ rdr}\right)
=\frac{4\pi\kappa R_1^3J_0}{3ar}\hat{\varphi} \\
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\vec{B_2}&=&\frac{4\pi\kappa}{r}\int_0^{r}{J(r)rdr}
=\frac{4\pi\kappa}{r}\left(\int_0^{R_2}{0\ rdr}+\int_{R_2}^{R_3}{\frac{J_0}{b}r^2dr}+ \int_{R_3}^ {r}{0\ rdr}\right)
=\frac{4\pi\kappa(R_3^3-R_2^3)J_0}{3br}\hat{\varphi} \nonumber \\
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\vec B_{\tbox{total}}&=&\frac{4\pi\kappa J_0}{3r}\left(\frac{R_1^3}{a}+\frac{R_3^3-R_2^3}{b}\right)\hat{\varphi}
\end{eqnarray}
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\Dn
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Here is the plot of the magnetic field strength:
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\includegraphics[scale=0.5]{e_45_4_101_s_p1.jpg}%
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