Question

A current filament carrying \(I\) in the \(-\mathbf{a}_{z}\) direction lies along the entire positive \(z\) axis. At the origin, it connects to a conducting sheet that forms the \(x y\) plane. (a) Find \(\mathbf{K}\) in the conducting sheet. \((b)\) Use Ampere's circuital law to find \(\mathbf{H}\) everywhere for \(z>0 ;(c)\) find \(\mathbf{H}\) for \(z<0\).

Short answer

Answer: The magnetic field \(\mathbf{H}\) at a height \(z>0\) is given by: \[\mathbf{H} = \left(-\frac{K_m s}{2} + \frac{I}{2\pi s}\right) \mathbf{a}_{\phi}\] Where \(K_m\) is the infinite magnitude of the current density, \(s\) is the radial distance from the filament, and \(\mathbf{a}_{\phi}\) is the unit vector in the azimuthal direction. The magnetic field \(\mathbf{H}\) at a height \(z<0\) is given by: \[\mathbf{H} = -\frac{K_m s}{2} \mathbf{a}_{\phi}\]

Step-by-step solution

Part (a): Calculate current density in the conducting sheet

Since the current filament conducts the current \(I\) in the \(-\mathbf{a}_{z}\) direction, the conducting sheet should carry an equal and opposite current. The current must spread uniformly over the conducting sheet. Therefore, the current density vector \(\mathbf{K}\) will be in the \(-\mathbf{a}_{z}\) direction. Since the current spreads evenly over the entire \(xy\) plane, the current density magnitude will be infinite. We can represent it as \(\mathbf{K} = -K_{m} \mathbf{a}_{z}\), where \(K_{m}\) is the infinite magnitude of the current density.

Part (b): Use Ampere's circuital law to find magnetic field H for \(z>0\)

We will use Ampere's circuital law to find the magnetic field \(\mathbf{H}\) for \(z>0\). The law states that \[\oint_{C} \mathbf{H} \cdot d\mathbf{l} = \iint_{S} \mathbf{J} \cdot d\mathbf{S} + I_{enc}\] For \(z>0\), both the current in the filament and the conducting sheet contribute to the magnetic field. Let's consider an Amperian loop of radius \(s\) and height \(z\) in the form of a cylinder. The magnetic field will be in the \(\mathbf{a}_{\phi}\) direction. We can split the line integral into two parts: one along the circular path (\(C_s\)) and the other over the conducting sheet (\(C_y\)). \[\oint_{C_s} \mathbf{H} \cdot d\mathbf{l} + \oint_{C_y} \mathbf{H} \cdot d\mathbf{l} = -K_m \iint_{S} dS + I\] Solving the integrals, we find: \[2\pi s H_z + 0 = -K_m \pi s^2 + I\] Now, we can solve for \(H_z\): \[H_z = -\frac{K_m s}{2} + \frac{I}{2\pi s}\] Since \(H_z\) is in the \(\mathbf{a}_{\phi}\) direction, the magnetic field \(\mathbf{H}\) for \(z>0\) is \[\mathbf{H} = \left(-\frac{K_m s}{2} + \frac{I}{2\pi s}\right) \mathbf{a}_{\phi}\]

Part (c): Find magnetic field H for \(z

Now, we will find the magnetic field \(\mathbf{H}\) for \(z<0\). For \(z<0\), the magnetic field is only due to the conducting sheet. We can consider an Amperian loop with radius \(s\) and height \(z\) again. The line integral will be the same as before, but the current enclosed by the loop will be zero since the current filament doesn't contribute to the field for \(z<0\). Thus, the integral equation becomes \[\oint_{C} \mathbf{H} \cdot d\mathbf{l} = -K_m \iint_{S} dS\] \[2\pi s H_z + 0 = -K_m \pi s^ 2\] Solving for \(H_z\): \[H_z = -\frac{K_m s}{2}\] Since \(H_z\) is in the \(\mathbf{a}_{\phi}\) direction, the magnetic field \(\mathbf{H}\) for \(z<0\) is \[\mathbf{H} = -\frac{K_m s}{2} \mathbf{a}_{\phi}\]

Key concepts

Current Density

Current density is a fundamental concept in the study of electromagnetism and is denoted by the symbol \( \mathbf{J} \). It represents the amount of electric current that flows through a unit area of a cross-section in a material. In the context of this exercise, where we have a current filament and a conducting sheet, the current density within the sheet can be viewed as a distributed source of current. It is crucial to note that current density has both magnitude and direction, making it a vector quantity. The direction of current density is aligned with the direction of flow of positive charge.

When the current filament carries a steady current, it induces a current density across the entire conducting sheet. It is significant to distinguish between localized current carried by the filament and the spread-out current across the sheet described by the current density vector \( \mathbf{K} \). Although our step-by-step solution indicates an infinite magnitude for the current density, in practice, the current density will be finite but still very high, as it represents current spread evenly over a thin conducting plane.

Magnetic Field

The magnetic field, often denoted by \( \mathbf{H} \) or \( \mathbf{B} \) depending on the context, is another critical element in the study of electromagnetism. It represents the magnetic influence on moving electric charges, magnetic dipoles, and other magnetic fields. The direction of the magnetic field lines indicates the path a north pole would trace out in the field, and its strength is proportional to the density of these lines.

In the given problem, Ampere's circuital law is employed to calculate the magnetic field generated by the current filament and the current density in the conducting sheet. The integral form of this law links the magnetic field surrounding a closed loop to the electric current passing through that loop. For students, understanding the direction and shape of magnetic field lines is crucial, as it helps in visualizing how magnetic fields behave around current-carrying conductors. Here, we find that the magnetic field takes on different values for regions above and below the conducting sheet, giving insight into how fields differ in the proximity of current sources.

Electromagnetic Theory

Electromagnetic theory encompasses the study of electric and magnetic fields and their interactions with matter. It is a cornerstone of classical physics and has paved the way for various modern technologies. Within this framework, Ampere's circuital law, used in our exercise, is one of the essential tools for calculating magnetic fields around conductors.

One of the intriguing aspects of electromagnetic theory is the relationship between electricity and magnetism. Moving electric charges not only create an electric field but also induce a magnetic field. This interplay is particularly evident in the exercise when determining the magnetic field for regions above and below the conducting sheet, where the presence or absence of the current filament dramatically influences the resultant magnetic field.

Maxwell's Equations

Maxwell's equations are the set of four fundamental laws that describe the behavior of electric and magnetic fields and how they develop from electric charge and current sources. These equations are the foundation of all classical electromagnetism and are vital for understanding and predicting how electric and magnetic fields interact with each other and matter.

The application of Ampere's circuital law in this exercise is essentially an application of one of Maxwell's equations, which in its integral form states that the magnetic field around a closed loop is proportional to the electric current plus the displacement current (which is absent in steady-state conditions) through the area enclosed by the loop. For students learning electromagnetism, a solid grasp of Maxwell's equations is fundamental for tackling a wide range of problems, from circuits and antennas to the propagation of light and radio waves.