A conducting filament at \(z=0\) carries \(12 \mathrm{~A}\) in the \(\mathbf{a}_{z}\) direction. Let \(\mu_{r}=1\) for \(\rho<1 \mathrm{~cm}, \mu_{r}=6\) for \(1<\rho<2 \mathrm{~cm}\), and \(\mu_{r}=1\) for \(\rho>2 \mathrm{~cm} .\) Find: (a) H everywhere; \((b) \mathbf{B}\) everywhere.

Using Ampere's circuital law, \(\oint \mathbf{H} \cdot d\mathbf{l} = I_{enc}\), where \(I_{enc}\) is the current enclosed by the integration path. Since the filament is at \(z=0\), let's consider a circular integration path in the \(x-y\) plane with radius \(\rho\). The magnetic field intensity is \(\rho\)-dependent but not angle-dependent. Therefore, \(\mathbf{H} = H(\rho) \mathbf{a}_{\phi}\). Also, \(d\mathbf{l} = d\phi \mathbf{a}_{\phi} \rho\). Now, we compute the integral around the circle in different regions of \(\rho\): 1. For \(\rho < 1 cm\), \(\mu_r = 1\). 2. For \(1cm < \rho < 2 cm\), \(\mu_r = 6\). 3. For \(\rho > 2 cm\), \(\mu_r = 1\). We will find \(H(\rho)\) separately for each region and then use that to calculate \(\mathbf{B}\).

Since \(\rho < 1cm\), we have no current enclosed by the integration path. Therefore, \(I_{enc} = 0\). Applying Ampere's circuital law: \begin{align*} \oint \mathbf{H} \cdot d\mathbf{l} &= 0 \\ H(\rho)\oint d\phi \rho &= 0 \\ H(\rho) 2\pi\rho &= 0 \end{align*} Thus, \(H(\rho) = 0\) for \(\rho < 1 cm\).

In this region, the integration path encloses the current-carrying filament, so the enclosed current is \(I_{enc} = 12 A\). Applying Ampere's circuital law: \begin{align*} \oint \mathbf{H} \cdot d\mathbf{l} &= I_{enc} \\ H(\rho) 2\pi\rho &= 12 \\ H(\rho) &= \frac{12}{2\pi\rho} \end{align*} Thus, \(H(\rho) = \frac{12}{2\pi\rho}\) for \(1 cm < \rho < 2 cm\).

The enclosed current is still \(I_{enc} = 12 A\). Similar to step 3, we have: \(H(\rho) = \frac{12}{2\pi\rho}\) for \(\rho > 2 cm\).

The magnetic flux density can be calculated using the relation, \(\mathbf{B} = \mu \mathbf{H}\). Now, we evaluate \(\mathbf{B}\) in all three regions: 1. For \(\rho < 1 cm\): \(\mu = \mu_r \mu_0 = 1 \cdot \mu_0\) Thus, \(\mathbf{B}(\rho) = 0 \cdot \mathbf{a}_{\phi} = 0\) for \(\rho < 1 cm\). 2. For \(1 cm < \rho < 2 cm\): \(\mu = \mu_r \mu_0 = 6 \cdot \mu_0\) Thus, \(\mathbf{B}(\rho) = (6 \cdot \mu_0) \cdot \frac{12}{2\pi\rho} \mathbf{a}_{\phi} = \frac{72\mu_0}{2\pi\rho}\mathbf{a}_{\phi}\) for \(1cm <\rho < 2cm\). 3. For \(\rho > 2 cm\): \(\mu = \mu_r \mu_0 = 1 \cdot \mu_0\) Thus, \(\mathbf{B}(\rho) = \mu_0 \cdot \frac{12}{2\pi\rho} \mathbf{a}_{\phi} = \frac{12\mu_0}{2\pi\rho}\mathbf{a}_{\phi}\) for \(\rho > 2 cm\). We have calculated both the magnetic field intensity (\(\mathbf{H}\)) and the magnetic flux density (\(\mathbf{B}\)) for all the regions around the conducting filament.