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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 5.15

Place a rectangular amperian loop parallel to the $yz$-plane, with one side satisfying $z=0$ (at this point, $B=0$).

Then we have

$$\begin{array}{rl}\oint \mathbf{B}\cdot d\mathbf{l}& ={\mu }_0{I}_{\text{enc}}\\ B\cdot l& ={\mu }_0\cdot J\cdot (\text{area})\\ Bl& ={\mu }_{0}Jzl\\ B& ={\mu }_{0}Jz\end{array}$$

Or in vector form:

$$\begin{array}{}\text{(}-a<z<a\text{)}& \mathbf{B}=-{\mu }_{0}Jz\hat{y}\end{array}$$

If you don't know why it points in the $\hat{\mathbf{y}}$ direction, read the discussion presented in Example 5.8.

If $z>a$, then $I_{\text{enc}}$ is ${\mu }_{0}laJ$, giving: $$\begin{array}{}\text{(For }z>a\text{)}& \mathbf{B}=-{\mu }_{0}Ja\hat{\mathbf{y}}\end{array}$$ And opposite if $z<-a$.

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