Question

the figure below shows the cross section of a long cylindrical conducting cylinder of radius ( a = 0.0820 mathrm{~m} ). the current density in the cross section is given by ( j=(3.30 \times 10^{6} mathrm{~a} / mathrm{m}^{3}) r ). where inside the cylinder does the magnetic field have a magnitude of ( 1.18 \times 10^{-3} mathrm{~t} )?

Explanation:

Step1: Recall Ampere's Law for a Cylinder

For a long - straight cylindrical conductor with current density \(J\), the current \(I_{enc}\) enclosed within a radius \(r\) (\(r\leq a\)) is \(I_{enc}=\int_{0}^{r}J\cdot 2\pi r'dr'\). Given \(J = 3.30\times10^{6}\text{ A/m}^3\), then \(I_{enc}=2\pi J\int_{0}^{r}r'dr'\).
Using the integral formula \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \(\int_{0}^{r}r'dr'=\frac{r^{2}}{2}\). So \(I_{enc}=\pi Jr^{2}\).
Ampere's Law is \(\oint\vec{B}\cdot d\vec{l}=\mu_{0}I_{enc}\). For a circular loop of radius \(r\) (symmetry implies \(B\) is constant around the loop), \(\oint\vec{B}\cdot d\vec{l}=B\cdot 2\pi r\).

Step2: Solve for \(B\)

Substitute \(I_{enc}=\pi Jr^{2}\) into Ampere's Law \(B\cdot 2\pi r=\mu_{0}\pi Jr^{2}\).
Solve for \(B\): \(B=\frac{\mu_{0}Jr}{2}\).
We know \(\mu_{0}=4\pi\times 10^{-7}\text{ T}\cdot\text{m/A}\), \(J = 3.30\times10^{6}\text{ A/m}^3\), and \(B = 1.18\times10^{-3}\text{ T}\).
Rearrange the formula for \(r\): \(r=\frac{2B}{\mu_{0}J}\).
Substitute the values: \(r=\frac{2\times1.18\times 10^{-3}}{4\pi\times10^{-7}\times3.30\times10^{6}}\).
First, calculate the denominator \(4\pi\times10^{-7}\times3.30\times10^{6}=4\times3.14\times3.30\times10^{-1}\approx4.1448\).
Then \(r=\frac{2\times1.18\times 10^{-3}}{4.1448}\approx5.74\times10^{-4}\text{ m}\).

Answer:

\(5.74\times 10^{-4}\text{ m}\)