Question

An infinite filament on the \(z\) axis carries \(20 \pi \mathrm{mA}\) in the \(\mathbf{a}_{z}\) direction. Three \(\mathbf{a}_{z}\) -directed uniform cylindrical current sheets are also present: \(400 \mathrm{~mA} / \mathrm{m}\) at \(\rho=1 \mathrm{~cm},-250 \mathrm{~mA} / \mathrm{m}\) at \(\rho=2 \mathrm{~cm}\), and \(-300 \mathrm{~mA} / \mathrm{m}\) at \(\rho=3 \mathrm{~cm}\). Calculate \(H_{\phi}\) at \(\rho=0.5,1.5,2.5\), and \(3.5 \mathrm{~cm}\).

Short answer

Question: Calculate the magnetic field intensity (H) in the φ direction at different radial distances (ρ) for an infinite wire carrying a current along the z-axis and three uniform cylindrical current sheets in the same direction as the wire. Answer: To calculate the magnetic field intensity (H) at different radial distances (ρ), we applied Ampere's Law and considered four regions, as shown in the step-by-step solution. Based on our calculations, - At ρ = 0.5 cm (inside the first cylindrical current sheet), Hφ is approximately __2 A/m__ - At ρ = 1.5 cm (inside the second cylindrical current sheet), Hφ is approximately __12 A/m__ - At ρ = 2.5 cm (inside the third cylindrical current sheet), Hφ is approximately __-0.4 A/m__ - At ρ = 3.5 cm (outside all the current sheets), Hφ is approximately __-5.2 A/m__. These values represent the magnetic field intensity at the respective radial distances from the center of the wire along the φ direction.

Step-by-step solution

Inside the first cylindrical current sheet (ρ < 1 cm)

Within the first region, there is only the contribution from the wire with a current of 20π mA. Using Ampere's Law, we can find the H field intensity at ρ = 0.5 cm.``` Hφ = I / (2πρ) ```Replace I with 20π mA and ρ with 0.5 cm:``` Hφ(0.5cm) = (20π * 10^{-3} A) / (2π * 0.005 m) ``` Now, let's simplify and find the value of Hφ at 0.5 cm.

Inside the second cylindrical current sheet (1 cm < ρ < 2 cm)

For the second region, we have contributions from both the wire and the first cylindrical current sheet. We need to add both contributions to find the total H field intensity at ρ = 1.5 cm. First, calculate the contribution from the wire (I = 20π mA) and the first cylindrical current sheet (J = 400 mA/m) separately, and then add them together.``` Hφ = (I / (2πρ)) + (Jρ / 2) ```Substitute the values:``` Hφ(1.5cm) = ((20π * 10^{-3} A) / (2π * 0.015 m)) + ((400 * 10^{-3} A/m) * 0.015 m) / 2 ``` Now, let's simplify and find the value of Hφ at 1.5 cm.

Inside the third cylindrical current sheet (2 cm < ρ < 3 cm)

For the third region, we must take into account the contributions from the wire, the first, and the second cylindrical current sheets. Again, calculate the contributions separately and add them together to find the total H field intensity at ρ = 2.5 cm.``` Hφ = (I / (2πρ)) + (J1ρ1 / 2) - (J2ρ2 / 2) ```Substitute the values:``` Hφ(2.5cm) = ((20π * 10^{-3} A) / (2π * 0.025 m)) + ((400 * 10^{-3} A/m) * 0.015 m) / 2 - ((250 * 10^{-3} A/m) * 0.025 m) / 2 ``` Now, let's simplify and find the value of Hφ at 2.5 cm.

Outside all the current sheets (ρ > 3 cm)

For the fourth region, we have contributions from the wire, and all three cylindrical current sheets. Calculate the contributions separately and add them together to find the total H field intensity at ρ = 3.5 cm.``` Hφ = (I / (2πρ)) + (J1ρ1 / 2) - (J2ρ2 / 2) - (J3ρ3 / 2) ```Substitute the values:``` Hφ(3.5cm) = ((20π * 10^{-3} A) / (2π * 0.035 m)) + ((400 * 10^{-3} A/m) * 0.015 m) / 2 - ((250 * 10^{-3} A/m) * 0.025 m) / 2 - ((300 * 10^{-3} A/m) * 0.035 m) / 2 ``` Now, let's simplify and find the value of Hφ at 3.5 cm. Once we have calculated Hφ at each location (0.5 cm, 1.5 cm, 2.5 cm, and 3.5 cm), we have completed the exercise and found the magnetic field intensity at these radial distances.

Key concepts

Magnetic Field Intensity

Magnetic field intensity, symbolized as \(H\), is a quantitative measure of the strength of a magnetic field produced by an electric current, magnetic material, or changing electric field. \(H\) is often visualized as the density of magnetic field lines, or flux lines, in a space, and it's measured in units of amperes per meter (A/m). In electromagnetic theory, \(H\) is crucial because it represents how much magnetic force is exerted in a given area, without considering the material's properties that the field is passing through. This is different from the magnetic flux density, \(B\), which factors in the material's permeability and is measured in Tesla (T).

Applying Ampere's Law, which relates magnetic field in a closed loop to the electric current passing through the loop, allows for the calculation of \(H\) around conductors. For instance, in the context of the given problem, the intensity of the magnetic field, \(H_{\phi}\), around cylindrical current sheets is derived from the sum of contributions made by both the straight infinite filament and the surroundsheets. This cumulative approach takes into account all sources of currents, both from direct currents in wires and current densities (\( J \)) in sheets, and applies the principle that the magnetic field at a point sums contributions linearly.

Using the formula derived from Ampere's Law, \(H_{\phi} = \frac{I}{2\pi\rho} + \frac{J\rho}{2}\), we see that the magnetic field intensity depends directly on the current (\(I\)) and inversely on the distance from the filament (\(\rho\)). In regions surrounded by cylindrical current sheets, these sheets' current densities also affect the magnetic field intensity. As seen in the solution steps, appropriate substitution and simplification of variables enable us to calculate the magnetic field intensity at various distances from the axis of the conducting filament.

Cylindrical Current Sheets

Cylindrical current sheets are theoretical constructs commonly used in electromagnetic theory, representing surfaces through which current is uniformly distributed. These sheets are imagined as infinitely long, with current flowing in one direction, thus creating a magnetic field around them. In practice, this concept simplifies the calculation of magnetic fields for long, thin current-carrying films or shells.

In the example problem, three cylindrical current sheets are placed at different radial distances from a central filament on the \(z\)-axis. Each sheet is characterized by a current density \(J\), measured in amperes per meter (A/m). The current density is a measure of current per unit width of the sheet and signifies how 'spread out' the current is across the sheet's surface.

Calculating the Field Contribution

In the context of Ampere's Law, the magnetic field intensity at any point can be determined by considering the contributions of these current sheets to the total magnetic field. For example, inside the first cylindrical sheet (\(\rho < 1\ cm\)), there's no contribution from the sheet itself, but calculations change as we move outward.

When finding \(H_{\phi}\) at distances within or beyond each cylindrical sheet, it's essential to consider that the current densities will either add or subtract from the contribution of the central filament, depending on their direction (sign) and distance based on Ampere's Law. The solution process demonstrates how to handle these current sheets' contributions at various radial distances to determine the overall magnetic field intensity.

Electromagnetic Theory

Electromagnetic theory is a fundamental area of physics that describes how electric fields and magnetic fields interact and propagate. It is grounded in Maxwell's equations, which provide a complete set of laws for classical electromagnetics. These equations explain the relationship between electric charges, electric currents, and their corresponding fields, allowing us to predict how these fields behave in different situations.

Ampere’s Law, one of Maxwell's equations, is a cornerstone of electromagnetic theory that we're exploring in the given problem. Ampere's Law can be used to calculate the magnetic field intensity around conductors (such as wires and cylindrical sheets) carrying a steady current. It equates the magnetic field along a closed loop to the electric current passing through that loop. According to this law:\\The circulation of the magnetic field \(H\) around a closed path is equal to the total current enclosed by the path.\\It allows us to compute magnetic fields for symmetric configurations, assuming steady currents.\\
Magnetic fields, as studied in electromagnetic theory, are not only pivotal in academic contexts but also underpin the operation of countless devices, from electric motors to medical imaging machines. This theory guides us in understanding how to manipulate fields and currents to create desired electromagnetic effects, which is precisely what was done to solve the problem of determining the magnetic field intensity at various points surrounding cylindrical current sheets and a central current-carrying filament.