a hollow cylindrical conductor of inner radius 0.0109 m and outer radius 0.0202 m carries a uniform current of 6.36 a. what is the magnitude of the magnetic field at radius of 0.0188 m? t

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Accepted Answer

# Explanation:
## Step1: Calculate the current enclosed
For a hollow cylindrical conductor, the current density $J=\frac{I}{\pi(R_{outer}^2 - R_{inner}^2)}$. The current enclosed $I_{enc}$ at radius $r$ ($R_{inner}<r<R_{outer}$) is given by $I_{enc}=\frac{I(r^{2}-R_{inner}^{2})}{R_{outer}^{2}-R_{inner}^{2}}$.
Here, $I = 6.36\space A$, $R_{inner}=0.0109\space m$, $R_{outer}=0.0202\space m$, $r = 0.0188\space m$.
$$
\begin{align*}
I_{enc}&=\frac{6.36\times(0.0188^{2}-0.0109^{2})}{0.0202^{2}-0.0109^{2}}\
&=\frac{6.36\times(0.000353 - 0.000119)}{0.000408 - 0.000119}\
&=\frac{6.36\times0.000234}{0.000289}\
&=\frac{0.001488}{0.000289}\approx5.15\space A
\end{align*}
$$
## Step2: Apply Ampere's law
Ampere's law is $\oint\vec{B}\cdot d\vec{l}=\mu_{0}I_{enc}$. For a circular loop of radius $r$, $\oint\vec{B}\cdot d\vec{l}=B\times2\pi r$. So $B=\frac{\mu_{0}I_{enc}}{2\pi r}$, where $\mu_{0}=4\pi\times 10^{-7}\space T\cdot m/A$.
$$
\begin{align*}
B&=\frac{4\pi\times 10^{-7}\times5.15}{2\pi\times0.0188}\
&=\frac{2\times10^{-7}\times5.15}{0.0188}\
&=\frac{1.03\times10^{-6}}{0.0188}\approx5.48\times 10^{-5}\space T
\end{align*}
$$

# Answer:
$5.48\times 10^{-5}\space T$

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Question

Explanation:

Step1: Calculate the current enclosed

Step2: Apply Ampere's law

Answer: