A$\mathbf{1}\mathbf{.50}\mathbf{-}\mathbf{m}$cylinder of radius $\mathbf{1}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{cm}$is made of a complicated mixture of materials. Its resistivity depends on the distance $\mathit{x}$from the left end and obeys the formula $\mathbf{\rho }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{+}{\mathbf{bx}}^{\mathbf{2}}$, where aand bare constants. At the left end, the resistivity is $\mathbf{2}\mathbf{.25}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\Omega }$m, while at the right end it is localid="1668398251768" $\mathbf{8}\mathbf{.50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\Omega }\mathbf{\cdot }\mathbf{m}$. (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries 1.75 - Acurrent? (c) If we cut the rod into two 75.0 - cmhalves, what is the resistance of each half?

We know, $\rho \left(0\right)=a$, so

$\begin{array}{r}\rho \left(0\right)=a=2.25\times {10}^{-8}\mathrm{\Omega }\cdot m\\ \rho \left(L\right)=8.50\times {10}^{-8}\mathrm{\Omega }\cdot m\\ And,\rho \left(L\right)=2.25\times {10}^{-8}\mathrm{\Omega }\cdot m+b(1.50m{)}^2\end{array}$

So, for b localid="1668398625029" $\begin{array}{r}b=\frac{8.50\times {10}^{-8}\mathrm{\Omega }\cdot m-2.25\times {10}^{-8}\mathrm{\Omega }\cdot m}{(1.50m{)}^2}\\ b=2.78\times {10}^{-8}\mathrm{\Omega }/\mathrm{m}\end{array}$

Now, cross-sectional area of the cylinder:

$\begin{array}{r}A=\pi r^2\\ A=\pi (0.011\mathrm{m}{)}^2\\ A=3.80\times {10}^{-4}{\mathrm{m}}^2\end{array}$

Therefore,

$\begin{array}{r}R=\frac{\left(2.25\times {10}^{-8}\mathrm{\Omega }\cdot \mathrm{m}\right)\left(1.50\mathrm{m}\right)}{3.80\times {10}^{-4}{\mathrm{m}}^2}+\frac{\left(2.78\times {10}^{-8}\mathrm{\Omega }/\mathrm{m}\right)(1.50\mathrm{m}{)}^3}{3\left(3.80\times {10}^{-4}{\mathrm{m}}^2\right)}\\ R=1.71\times {10}^{-4}\mathrm{\Omega }\end{array}$

Thus, the resistance of the rod is $1.71\times {10}^{-4}\mathrm{\Omega }$

$\begin{array}{r}E=\rho J\\ Weknow,E=\left(a+b{x}^2\right)\frac{I}{A}\end{array}$

At midpoint of the cylinder

$E=\frac{l\left(a+b{L}^2/4\right)}{A}$

Substitute the values in the equation

$\begin{array}{r}E=\frac{\left(1.75\mathrm{A}\right)\left[\left(2.25\times {10}^{-8}\mathrm{\Omega }\cdot \mathrm{m}\right)+\left(2.78\times {10}^{-8}\mathrm{\Omega }/\mathrm{m}\right)(1.50\mathrm{m}{)}^2/4\right]}{3.80\times {10}^{-4}{\mathrm{m}}^2}\\ E=1.76\times {10}^4\mathrm{V}/\mathrm{m}\end{array}$

Therefore, the electric field at the midpoint of the cylinder is

For the left half:

Integrate dR in the limits of $x=0\to L/2$

$\begin{array}{r}{R}_{LH}=\int dR=\frac{1}{A}{\int }_0^{L/2} \left(a+b{x}^2\right)dx\\ R_{LH}=\frac{1}{A}{\left[ax+\frac{b{x}^3}{3}\right]}_0^{L/2}\\ R_{LH}=\frac{aL}{2A}+\frac{b{L}^3}{24A}\end{array}$

Substitute the given values

$\begin{array}{r}{R}_{LH}=\frac{\left(2.25\times {10}^{-8}\mathrm{\Omega }\cdot m\right)\left(1.50\mathrm{m}\right)}{2\times \left(3.80\times {10}^{-4}{\mathrm{m}}^2\right)}+\frac{\left(2.78\times {10}^{-8}\mathrm{\Omega }/\mathrm{m}\right)\left(1.50{\mathrm{m}}^3\right)}{24\left(3.80\times {10}^{-4}{\mathrm{m}}^2\right)}\\ R_{LH}=5.47\times {10}^{-5}\mathrm{\Omega }\end{array}$

For the right half:

To determine the resistance of the right half, subtract the resistance of left half from the total resistance. So,

$\begin{array}{r}{R}_{RH}=R-R_{LH}\\ R_{RH}=1.71\times {10}^{-4}\mathrm{\Omega }-5.47\times {10}^{-5}\mathrm{\Omega }\\ R_{RH}=1.16\times {10}^{-4}\mathrm{\Omega }\end{array}$

Therefore, the resistance of the left half of the cylinder is $5.47\times {10}^{-5}\mathrm{\Omega }$, and the resistance of the right half of the cylinder is $1.16\times {10}^{-4}\mathrm{\Omega }$