# Along, straight, solid cylinder, oriented with its axis in the z-direc

Along, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is $\stackrel{\to }{J}$ . The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship

where the radius of the cylinder is a = 5.00 cm, ris the radial distance from the cylinder axis, b is a constant equal to 600 A/m, and $\delta$ is a constant equal to 2.50 cm.
(a) Let ${I}_{0}$ be the total current passing through the entire cross section of the wire. Obtain an expression for ${I}_{0}$ in terms of b, d, and a. Evaluate your expression to obtain a numerical value for ${I}_{0}$.
(b) Using Ampere’s law, derive an expression for the magnetic field $\stackrel{\to }{B}$ in the region $r\ge a$. Express your answer in terms of ${I}_{0}$ rather than b.
(c) Obtain an expression for the current | contained in a circular cross section of radius r... a and centered at the cylinder axis. Express your answer in terms of ${I}_{0}$ rather than b.
(d) Using Ampere’s law, derive an expression for the magnetic field $\stackrel{\to }{B}$ in the region $r\ge a.$
(e) Evaluate the magnitude of the magnetic field at r- = 6, r= a, and r= 2a.

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raefx88y

(a) Consider a long, solid cylinder which has a circular cross-section of radius a and oriented with its axis in the z- direction, the current density is not constant across the cross-section of the wire and it is given by
$\stackrel{\to }{J}=\left(\begin{array}{c}\frac{b}{r}\end{array}\right){e}^{\left(r-a\right)/\delta }\stackrel{^}{k}$
where r is the distance from the center. First we need to find an expression for the total current passing through
the entire cross section of the wire ${I}_{0}$ in terms of $b,\delta$ , and a. Consider a small segment, a ring with a width of dr and at distance from the center of the circle, the current in this element equals the area of the element multiplied by
the current density J, that is,
$dI=JdA$
the area of the element equals the area of a rectangle with a width of dr and length of $2\pi r$ (the circumference of a circle with a radius of r), that is,
$dA=2\pi rdr$
thus,
$dI=\left(\begin{array}{c}\frac{b}{r}\end{array}\right){e}^{\left(r-a\right)/\delta }2\pi rdr=2\pi b{e}^{\left(r-a\right)/\delta }dr$
the total current passing throught the wire equals the integration over the whole wire, from r = 0 to r = a, that is,
${I}_{0}={\int }_{0}^{a}dI=2\pi b{\int }_{0}^{a}{e}^{\frac{r-a}{\delta }}dr$
$=2\pi b\delta {e}^{\frac{r-a}{\delta }}{\mid }_{\left\{0\right\}}^{a}=2\pi b\delta \left(1-{e}^{-\frac{a}{\delta }}\right)$
${I}_{0}=2\pi b\delta \left(1-{e}^{-\frac{a}{\delta }}$
the radius of the cylinder is a=5.00,b is a constant equal to 600 A/m,and d is a constant equal to 2,50 cm, substitute with the givens to get,
${I}^{0}=2\pi \left(600\frac{A}{m}\right)\left(0.025m\right)\left(1-{e}^{\left(\frac{0.050}{0.025}\right)}\right)=81.5A$
${I}^{0}=81.5A$

###### Not exactly what you’re looking for?
Paul Mitchell

(b) Now for $r\ge a$, the enclosed current is ${I}^{0}$, therefore Ampere's law $\left(\oint \stackrel{\to }{B}×\stackrel{\to }{dl}={\mu }_{0}{I}_{encl}\right)$ gives,
$\oint \left\{B\right\}×\stackrel{\to }{dl}=B2\pi r={\mu }_{0}{I}_{encl}={\mu }_{0}{I}_{0}$
or,
$B=\frac{{\mu }_{0}{I}_{0}}{2\pi r}$
(c) Now we need to find the enclosed current for $r\le a$. Consider a ring with width of dr and area of $dA=2\pi dr$
the enclosed current in this area is,
$dl=JdA$
or,
$dI=2\pi b{e}^{\frac{r-a}{\delta }}d{r}^{{}^{\prime }}$
integrate from ${r}^{{}^{\prime }}=0$ to ${r}^{{}^{\prime }}=r$ to get,
$dI=2\pi b{\int }_{0}^{r}{e}^{\frac{r-a}{\delta }}dr=2\pi b\delta {e}^{\frac{{r}^{{}^{\prime }}-a}{\delta }}{\mid }_{\left\{0\right\}}^{r}$
$=2\pi b\delta \left({e}^{\frac{r-a}{\delta }}-{e}^{-\frac{a}{\delta }}\right)=2\pi b\delta {e}^{-\frac{a}{\delta }}\left({e}^{\frac{r}{\delta }}-1\right)$
but ${I}^{0}=2\pi b\delta \left(1-{e}^{-\frac{a}{\delta }}\right)$,thus
$I={I}^{0}\frac{\left({e}^{\frac{r}{\delta }}-1\right)}{\left({e}^{\frac{a}{\delta }}-1\right)}$

###### Not exactly what you’re looking for?
nick1337

(d) Now we need to use the results of part (c) to find the magnetic field for $r\le a$. Ampere's law $\left(\oint \stackrel{\to }{B}×\stackrel{\to }{dl}={\mu }_{0}{I}_{encl}\right)$, gives,
$\oint \stackrel{\to }{B}×\stackrel{\to }{dl}=B\left(r\right)2\pi r={\mu }_{0}{I}_{encl}={\mu }_{0}{I}_{0}\frac{\left({e}^{r/\delta }-1\right)}{\left({e}^{a/\delta }-1\right)}$
or,
$B=\frac{{\mu }_{0}{I}_{0}\left({e}^{r/\delta }-1\right)}{2\pi r\left({e}^{a/\delta }-1\right)}$
(e) Finally we need to evaluate the magnitude of the magnetic field at $r=\delta ,r=a,$ and $r=2a$, as,
$B\left(r=\delta \right)=\frac{{\mu }_{0}{I}_{0}\left(e-1\right)}{2\pi \delta \left({e}^{a/\delta }-1\right)}$
$=\frac{\left(4\pi ×{10}^{-7}T×m/A\right)\left(81.5A\right)}{2\pi \left(0.025m\right)}\frac{\left(e-1\right)}{\left({e}^{0.050/0.025}-1\right)}$
$=1.75×{10}^{-4}T$
$B\left(r=\delta \right)=1.75×{10}^{-4}T$
and
$B\left(r=a\right)=\frac{{\mu }_{0}{I}_{0}\left({e}^{a/\delta }-1\right)}{2\pi a\left({e}^{a/\delta }-1\right)}$
$=\frac{\left(4\pi ×{10}^{-7}T×m/A\right)\left(81.5A\right)}{2\pi \left(0.050m\right)}$
$=3.26×{10}^{-4}T$
$B\left(r=\delta \right)=3.26×{10}^{-4}T$
and
$B\left(r=2a\right)=\frac{{\mu }_{0}{I}_{0}}{2\pi a}$
$=\frac{\left(4\pi ×{10}^{-7}T×m/A\right)\left(81.5A\right)}{2\pi \left(0.100m\right)}$
$=1.63×{10}^{-4}T$
$B\left(r=2a\right)=1.63×{10}^{-4}T$