0
$\begingroup$

I am having trouble with the following problem, although with similar problems I run into the same issue when it comes to validation. The example problem below is a problem for Amperes Law.

enter image description here

The derivation I can get through, the issue I have is when it comes to validating the two basic laws of magnetostatics as described below:

enter image description here

enter image description here

Whenever I try to validate these two laws the equation is always slightly mismatched with $J$ with the $a$, $b$, and $\rho$ terms (specifically in the curl one). What is the process for validating both of the laws algebraically? An example using my derivations would be appreciated.

Below are my derivations of the magnetic field for reference:

enter image description here

enter image description here

Improve this question
$\endgroup$
2
  • $\begingroup$ What is $\rho$, $J_0$ and $a_z$? $\endgroup$
    – Mauricio
    Commented Nov 4, 2021 at 9:04
  • $\begingroup$ @Mauricio I get the impression that $\rho$ is radius, $J_0$ is a constant, and $\mathbf{a}_z$ is the $z$ unit vector. $\endgroup$
    – DanDan面
    Commented Nov 4, 2021 at 9:10

2 Answers 2

Reset to default
1
$\begingroup$

The solution looks valid to me. We have the following magnetic field in space: $$\mathbf{B}=\mu_0J_0a\hat{\mathbf{\varphi}}\begin{cases}0&\rho\leq a\\\frac{\rho-a}{\rho}&a<\rho<b\\\frac{b-a}{\rho}&\rho\geq b\end{cases}$$ We now consider the form of the relevant differential operators in cylindrical coordinates. Since our $\mathbf{B}$ only has a nonzero $\varphi$ component, we ignore all terms concerning $\mathbf{B}_\rho$ and $\mathbf{B}_z$.

Taking the divergence, we have that: $$\nabla\cdot\mathbf{B}=\frac{1}{\rho}\frac{\partial\mathbf{B}_\varphi}{\partial\varphi}=0$$ This is zero since the magnetic field doesn't depend on angle. Now taking the curl, we have that: $$\nabla\times\mathbf{B}=-\frac{\partial\mathbf{B}_\varphi}{\partial z}\hat{\mathbf{\rho}}+\frac{1}{\rho}\frac{\partial(\rho\mathbf{B}_\varphi)}{\partial\rho}\hat{\mathbf{z}}$$ Note that the magnetic field doesn't depend on the $z$ coordinate either. Hence, only the second term survives: $$\nabla\times\mathbf{B}=\frac{\hat{\mathbf{z}}}{\rho}\frac{\partial}{\partial\rho}\left(\mu_0J_0a\begin{cases}0&\rho\leq a\\\rho-a&a<\rho<b\\b-a&\rho\geq b\end{cases}\right)=\mu_0\begin{cases}\hat{\mathbf{z}}J_0\left(\frac{a}{\rho}\right)&a<\rho<b\\0&\text{otherwise}\end{cases}=\mu_0\mathbf{J}$$ So, Maxwell's equations are indeed satisfied, and we are happy.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Is there an explanation for why multiplying by p/p is necessary for the curl? How come the law doesn't validate if we take the partial of just B without multiplying by p/p? $\endgroup$
    – Jayson Okhman
    Commented Nov 4, 2021 at 15:07
  • $\begingroup$ Because then you aren't taking the curl correctly in cylindrical coordinates. You can look at the linked Wikipedia page in my answer to learn a little bit more about using differential operators in curvilinear coordinates, $\endgroup$
    – DanDan面
    Commented Nov 4, 2021 at 17:01
0
$\begingroup$

For $a<\rho<b$

Trying to use the integral form $\oint \vec B \cdot d\vec l=\mu_0\iint \vec J\cdot d\vec A $ $$\iint \vec J.d\vec A=\int_a^\rho J_0\frac{a}{\rho}2\pi \rho d\rho=2\pi J_0a(\rho-a)$$ Symmetry gives us $B_z=0 $ and $B_{\rho}=0$ $$\oint \vec B \cdot d \vec l=\int_0^{2\pi}B_{\varphi}\rho d\varphi=2\pi \rho B_\varphi $$ Giving $$B_\varphi=\mu_0 J_0a\biggl (1-\frac{a}{\rho}\biggl)$$ $$\vec \nabla \times \vec B=\frac{1}{\rho} \frac{\partial (\rho B_\varphi)}{\partial\rho}\hat k=\mu_0 J_0 \frac{a}{\rho}\hat k=\mu_0 \vec J$$ So it validates for $a<\rho<b$ . I think it will be easy to do it for $\rho>b$

Improve this answer
$\endgroup$
1
  • $\begingroup$ For p> b simply change change top surface integral bound to be b, since p>b the enclosed current is constant. and then just solve the same way $\endgroup$
    – jensen paull
    Commented Nov 4, 2021 at 10:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.