/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A slab and a sheet \(* *\) A v... [FREE SOLUTION] | 91Ó°ÊÓ

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A slab and a sheet \(* *\) A volume current density \(\mathbf{J}=J \hat{\mathbf{z}}\) exists in a slab between the infinite planes at \(x=-b\) and \(x=b\). (So the current is coming out of the page in Fig. 6.37.) Additionally, a surface current density \(\mathcal{J}=2 b J\) points in the \(-\hat{\mathbf{z}}\) direction on the plane at \(x=b\) (a) Find the magnetic field as a function of \(x\), both inside and outside the slab. (b) Verify that \(\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}\) inside the slab. (Don't worry about the boundaries.)

Short Answer

Expert verified
The magnetic field inside the slab is \(B_{in} = \frac{1}{2}\mu_{0}J*x\), outside the slab the magnetic field is zero. Ampere's law is verified inside the slab.

Step by step solution

01

Magnetic Field inside the Slab

To find the magnetic field inside, we choose an Amperian loop as a rectangle with sides parallel to the z and y axes. The length of the side parallel to the z axis is \(dx\) and on the y axis is \(L\). Using Ampere’s Law, the integral of \(B.dl\) around the loop equals \(\mu_0*I_{enc}\). The enclosed current \(I_{enc}\) is \(J*dx*L\), thus, \(B_{in} * 2L = \mu_{0}*J*dx*L\). Solving this for \(B_{in}\) results in \(B_{in} = \frac{1}{2}\mu_{0}*J*x\).
02

Magnetic Field outside the Slab

To find the magnetic field outside, we choose an Amperian loop of length \(L\) from \(x = b\) (inside the slab) to \(x = b + d\) (outside the slab). Again applying Ampere’s Law, \(B_{out} * 2L = \mu_{0}*J*2b*L – \mu_{0}*2b*J*L\). This gives \(B_{out} = 0\). Thus, the magnetic field outside the slab is 0.
03

Verify Ampere's Law inside the Slab

To verify Ampere's law inside the slab we need to compute the curl of B inside the slab and check if it equals \(\mu_{0} J\). The curl of B inside the slab (\(\nabla \times B_{in}\)) is 0 except for the j component. The j component of \(\nabla \times B_{in}\) equals \(-dB_{in}/dx\) which is \(-\mu_{0}*J/2\). However, in the question we have \(\mathbf{J} = J \hat{\mathbf{z}}\), thus, \(\mu_{0} * \mathbf{J} = \mu_{0}*J*-\hat{\mathbf{j}}\). This is equal to the Curl B, therefore Ampere's Law is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ampere's Law
Understanding Ampere's Law is crucial for studying the behavior of magnetic fields in the presence of electric currents. Ampere's Law, a foundational principle in magnetostatics, is articulated as the integral form of the Maxwell-Ampere equation. It states that the line integral of the magnetic field \textbf{B} around a closed loop is proportional to the electric current passing through the loop. In equation form, this is expressed as: \[\begin{equation}\oint\mathbf{B}\cdot d\mathbf{l} = \mu_0 I_{\text{enc}}\end{equation}\]where \(\mu_0\) is the permeability of free space, and \(I_{\text{enc}}\) represents the enclosed current. In our exercise, the use of Ampere's Law enables us to solve for the magnetic field inside a current-carrying slab by considering an Amperian loop and equating the magnetic field's line integral around it to the product of \(\mu_0\) and the current enclosed by the loop. This leads to understanding the relationship between the magnetic field and the volume current density within the material.
Volume Current Density
Volume current density, denoted as \(\mathbf{J}\), is a vector quantity that tells us how much electric current flows through a unit area of a three-dimensional volume. It is defined as the current per unit cross-sectional area being perpendicular to the current's flow direction. Mathematically, it is given by:\[\begin{equation}\mathbf{J} = \frac{dI}{dA}\end{equation}\]In the context of our exercise, the volume current density is constant and points in the \(\hat{\mathbf{z}}\) direction within the slab located between the planes at \(x=-b\) and \(x=b\). This volume current density can be used to calculate the total current enclosed by the Amperian loop, which is a step in determining the magnetic field inside the slab using Ampere's Law.
Surface Current Density
In contrast to the volume current density, the surface current density (\textmathcal{K} or \textmathcal{J}) refers to the current flow across a surface. It represents the amount of current per unit length along the surface and is typically expressed in amperes per meter (A/m). The surface current density is given by:\[\begin{equation}\mathcal{J} = \frac{I}{L}\end{equation}\]where \(I\) is the current and \(L\) is the length in the direction perpendicular to the flow of current. For the given exercise, the surface current density exists on the plane at \(x = b\) and has a direction opposite to \(\mathbf{J}\), effectively acting as a sheet of current atop the slab. This surface current density plays a pivotal role in calculating the magnetic field at and beyond the boundary of the slab.
Magnetic Field Calculations
Magnetic field calculations involve the determination of the magnetic field's magnitude and direction due to various sources of currents, such as wires, loops, or slabs. In our situation, we calculate the magnetic field inside and outside a current-carrying slab. Inside the slab, the magnetic field varies linearly with \(x\), as deduced through Ampere's Law, which allows us to set up the integral over our selected Amperian loop to find the enclosed current and, subsequently, the magnetic field. Outside the slab, however, considerations must include the effect of surface current density, which affects field calculations at the boundary. Through careful setup of the appropriate Amperian loop, it is found that the magnetic field outside the slab is zero, indicating that the surface current effectively cancels the field produced by the volume current at and outside the edge of the slab.
Verification of Ampere's Law
To verify Ampere's Law, we must show that the relationship between the magnetic field and the electric current density is upheld. Ampere's Law, when considered in its differential form through Maxwell's equations, connects the curl of the magnetic field \(abla \times \mathbf{B}\) to the volume current density \(\mathbf{J}\). Mathematically, this is presented as:\[\begin{equation}abla \times \mathbf{B} = \mu_0 \mathbf{J}\end{equation}\]In the exercise solution, we calculated the magnetic field inside the slab and arrived at the conclusion that the magnetic field's curl matched the given volume current density when multiplied by \(\mu_0\). Such a verification exercise ensures that our magnetic field results are consistent with theoretical expectations and Maxwell's equations. Through this verification process, we deepen our understanding of the interplay between electric currents and the resultant magnetic fields.

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Most popular questions from this chapter

A rotating solid cylinder ** (a) A very long cylinder with radius \(R\) and uniform volume charge density \(\rho\) spins with frequency \(\omega\) around its axis. What is the magnetic field at a point on the axis? (b) How would your answer change if all the charge were concentrated on the surface?

Right-angled wire ** A wire carrying current \(I\) runs down the \(y\) axis to the origin, thence out to infinity along the positive \(x\) axis. Show that the magnetic field at any point in the \(x y\) plane (except right on one of the axes) is given by $$ B_{z}=\frac{\mu_{0} I}{4 \pi}\left(\frac{1}{x}+\frac{1}{y}+\frac{x}{y \sqrt{x^{2}+y^{2}}}+\frac{y}{x \sqrt{x^{2}+y^{2}}}\right) $$

Field at the tip of a cone \(* *\) A hollow cone (like a party hat) has vertex angle \(2 \theta\), slant height \(L\), and surface charge density \(\sigma\). It spins around its symmetry axis with angular frequency \(\omega\). What is the magnetic field at the tip?

Field at different radii \(*\) A current of 8000 amperes flows through an aluminum rod \(4 \mathrm{~cm}\) in diameter. Assuming the current density is uniform through the cross section, find the strength of the magnetic field at \(1 \mathrm{~cm}\), at \(2 \mathrm{~cm}\), and at \(3 \mathrm{~cm}\) from the axis of the rod.

Field in the plane of a ring ** A ring with radius \(R\) carries a current \(I\). Show that the magnetic field due to the ring, at a point in the plane of the ring, a distance \(a\) from the center (either inside or outside the ring), is given by $$ B=2 \cdot \frac{\mu_{0} I}{4 \pi} \int_{0}^{\pi} \frac{(R-a \cos \theta) R d \theta}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}} $$ Hint: The easiest way to handle the cross product in the BiotSavart law is to write the Cartesian coordinates of \(d l\) and \(\mathbf{r}\) in terms of an angle \(\theta\) in the ring. This integral can't be evaluated in closed form (except in terms of elliptic functions), but it can always be evaluated numerically if desired. For the special case of \(a=0\) at the center of the ring, the integral is easy to do; verify that it yields the result given in

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