Question

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.

Short answer

Question: Calculate the magnetic field intensity (H) and the magnetic flux density (B) everywhere due to a conducting filament carrying a current of 12 A in the z direction. Answer: For ρ < 1 cm: H(ρ) = 0, B(ρ) = 0. For 1 cm < ρ < 2 cm: H(ρ) = 12 / (2πρ), B(ρ) = (72μ₀) / (2πρ). For ρ > 2 cm: H(ρ) = 12 / (2πρ), B(ρ) = (12μ₀) / (2πρ).

Step-by-step solution

Apply Ampere's circuital law

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 \(\mathbf{B}\).

Calculate H for \(\rho < 1 cm\)

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 \(\rho < 1 cm\).

Calculate H for \(1 cm < \rho < 2 cm\)

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 \(1 cm < \rho < 2 cm\).

Calculate H for \(\rho > 2 cm\)

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

Calculate \(\mathbf{B}\) for all regions

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.

Key concepts

Ampere's circuital law

Ampere's circuital law is a fundamental principle in electromagnetics that helps us understand how magnetic fields relate to electric currents. It states that the line integral of the magnetic field intensity \(\mathbf{H}\) around a closed path is equal to the total current \(I_{enc}\) enclosed by that path. Mathematically, it's expressed as \(\oint \mathbf{H} \cdot d\mathbf{l} = I_{enc}\). In simple terms, it means that the total magnetic effect around a loop is proportional to the amount of current that flows through it.

This law is particularly useful when dealing with symmetrical situations such as around a long, straight current-carrying wire or a circular loop of wire. It's a cornerstone in determining magnetic field patterns created by currents, much like how Gauss's law is used for electric fields.

In our specific problem, we use a circular path centered around the current-carrying filament, enabling us to apply Ampere's law effectively to calculate the magnetic field intensity \(\mathbf{H}\). By choosing the path that simplifies the integration, we easily compute \(\mathbf{H}\) across different media with varied permeability.

Magnetic field intensity

Magnetic field intensity, denoted as \(\mathbf{H}\), describes the strength and direction of a magnetic field in a given region. It is a vector quantity and is crucial in understanding how the magnetic field interacts with different materials.

The idea of magnetic field intensity becomes clear when we use Ampere's law to find \(\mathbf{H}\). As we've seen, the current enclosed determines the behavior of \(\mathbf{H}\). If no current is enclosed, as in the region where \(\rho < 1 \text{ cm}\), \(\mathbf{H}\) is zero. However, for regions where the current is enclosed, the intensity is calculated based on the current flowing through the path, like \(H(\rho) = \frac{12}{2\pi\rho}\).

This tells us that the magnetic field weakens as we move farther from the source of current. Having a clear understanding of \(\mathbf{H}\) helps in determining the effect of a magnetic field across different distances from the current source.

Magnetic flux density

Magnetic flux density, labeled as \(\mathbf{B}\), essentially depicts how much magnetic field passes through a given area. It's linked to magnetic field intensity by the material's permeability, given by the formula \(\mathbf{B} = \mu \mathbf{H}\).

This relationship reflects how material properties can affect magnetic fields. For example, in regions where the permeability (\(\mu\)) is higher, such as when \(1 \leq \rho \leq 2 \text{ cm}\), \(\mathbf{B}\) is stronger because \(\mu\) is a significant factor. Thus, \(\mathbf{B} = \frac{72\mu_0}{2\pi\rho}\mathbf{a}_{\phi}\) compared to regions where no additional permeability effect exists (\(\rho > 2 \text{ cm}\) or \(\rho<1 \text{ cm}\)).

Magnetic flux density is crucial in applications like magnetic storage devices and transformers, where understanding the magnetic field's behavior in materials directly translates to their performance and design.

Permeability

Permeability is a measure of how easy it is for a magnetic field to pass through a material. It is denoted by \(\mu\) and is a product of the relative permeability (\(\mu_{r}\)) and the vacuum permeability (\(\mu_{0}\)). Mathematically, \(\mu = \mu_{r} \times \mu_{0}\).

In our exercise, permeability plays a key role in determining the behavior of the magnetic field in the different regions around the filament. The relative permeability varies with regions: \(\mu_{r}=1\) for \(\rho<1 \text{ cm}\) and \(\rho>2 \text{ cm}\), and \(\mu_{r}=6\) for \(1<\rho<2 \text{ cm}\). This variance affects both \(\mathbf{H}\) and \(\mathbf{B}\).

Understanding how permeability affects magnetic fields aids in designing and working with materials in various electromagnetic applications, as it impacts how magnetic lines propagate through different mediums.

Current-carrying filament

A current-carrying filament is essentially a thin wire through which electric current flows. This current creates a magnetic field around the filament, which is characterized by certain fundamental principles of electromagnetism.

In the exercise we discussed, a filament holds a steady current of 12 A along the \(\mathbf{a}_{z}\) direction at \(z=0\). This setup is essential to measuring the magnetic effects using Ampere's law. The uniform flow in a given direction ensures a predictable circular magnetic field pattern around it.

Filaments serve as simple models for more complex systems like electromagnets and coils, helping understand how current distributions can shape magnetic fields. Therefore, current-carrying filaments are a useful concept for both theoretical studies and practical applications, such as in the design of inductors and transformers.