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Chapter 7: Problem 33

Use an expansion in rectangular coordinates to show that the curl of the gradient of any scalar field \(G\) is identically equal to zero.

Short Answer

Expert verified
Question: Show that the curl of the gradient of any scalar field G is identically equal to zero. Answer: By applying the gradient operator (∇) to the scalar field G and calculating the curl of the gradient (∇ × (∇G)) using the vector cross product rules, we can prove that the curl of the gradient is indeed zero. This is confirmed by applying the equality of mixed partial derivatives, which simplifies the curl expression to 0i - 0j + 0k, indicating that the curl of the gradient of any scalar field G is identically equal to zero.

Step by step solution

01

Obtain the gradient of G

To find the gradient of the scalar field G, use the gradient operator (∇) and apply it to G. In rectangular coordinates, the gradient of a scalar field G can be written as: ∇G = (∂G/∂x)i + (∂G/∂y)j + (∂G/∂z)k
02

Calculate the curl of the gradient

Now, we need to find the curl of the gradient, which is written as ∇ × (∇G). Using the vector cross product rules and the rectangular coordinate system, the curl of a vector field F is given by: ∇ × F = (∂Fz/∂y - ∂Fy/∂z)i - (∂Fx/∂z - ∂Fz/∂x)j + (∂Fy/∂x - ∂Fx/∂y)k Here, F = ∇G. So, we need to replace Fx, Fy, and Fz with ∂G/∂x, ∂G/∂y, and ∂G/∂z respectively and calculate the curl.
03

Substitute the gradient components into the curl formula

Substitute the gradient components into the curl formula: ∇ × (∇G) = (∂(∂G/∂z)/∂y - ∂(∂G/∂y)/∂z)i - (∂(∂G/∂x)/∂z - ∂(∂G/∂z)/∂x)j + (∂(∂G/∂y)/∂x - ∂(∂G/∂x)/∂y)k
04

Apply the equality of mixed partial derivatives

According to the equality of mixed partial derivatives, for a scalar field G with continuous second partial derivatives, the order of differentiation doesn't matter: ∂(∂G/∂x)/∂y = ∂(∂G/∂y)/∂x ∂(∂G/∂y)/∂z = ∂(∂G/∂z)/∂y ∂(∂G/∂z)/∂x = ∂(∂G/∂x)/∂z Using these equalities, we can prove that the curl of the gradient of G is identically equal to zero.
05

Prove the curl of the gradient is zero

Using the equalities of mixed partial derivatives: ∇ × (∇G) = (∂(∂G/∂z)/∂y - ∂(∂G/∂z)/∂y)i - (∂(∂G/∂x)/∂z - ∂(∂G/∂x)/∂z)j + (∂(∂G/∂y)/∂x - ∂(∂G/∂y)/∂x)k ∇ × (∇G) = 0i - 0j + 0k Therefore, the curl of the gradient of any scalar field G is identically equal to zero.

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Most popular questions from this chapter

In spherical coordinates, the surface of a solid conducting cone is described by \(\theta=\pi / 4\) and a conducting plane by \(\theta=\pi / 2 .\) Each carries a total current I. The current flows as a surface current radially inward on the plane to the vertex of the cone, and then flows radially outward throughout the cross section of the conical conductor. \((a)\) Express the surface current density as a function of \(r ;(b)\) express the volume current density inside the cone as a function of \(r ;(c)\) determine \(\mathbf{H}\) as a function of \(r\) and \(\theta\) in the region between the cone and the plane; \((d)\) determine \(\mathbf{H}\) as a function of \(r\) and \(\theta\) inside the cone.

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 \(3.5 \mathrm{~cm}\).

A solid cylinder of radius \(a\) and length \(L\), where \(L \gg a\), contains volume charge of uniform density \(\rho_{0} \mathrm{C} / \mathrm{m}^{3}\). The cylinder rotates about its axis (the \(z\) axis) at angular velocity \(\Omega \mathrm{rad} / \mathrm{s}\). (a) Determine the current density \(\mathbf{J}\) as a function of position within the rotating cylinder. (b) Determine \(\mathbf{H}\) on-axis by applying the results of Problem 7.6. ( \(c\) ) Determine the magnetic field intensity \(\mathbf{H}\) inside and outside. \((d)\) Check your result of part ( \(c\) ) by taking the curl of \(\mathbf{H}\).

A long, straight, nonmagnetic conductor of \(0.2 \mathrm{~mm}\) radius carries a uniformly distributed current of 2 A dc. \((a)\) Find \(J\) within the conductor. (b) Use Ampère's circuital law to find \(\mathbf{H}\) and \(\mathbf{B}\) within the conductor. (c) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) within the conductor. \((d)\) Find \(\mathbf{H}\) and \(\mathbf{B}\) outside the conductor. \((e)\) Show that \(\nabla \times \mathbf{H}=\mathbf{J}\) outside the conductor.

A disk of radius \(a\) lies in the \(x y\) plane, with the \(z\) axis through its center. Surface charge of uniform density \(\rho_{s}\) lies on the disk, which rotates about the \(z\) axis at angular velocity \(\Omega \mathrm{rad} / \mathrm{s}\). Find \(\mathbf{H}\) at any point on the \(z\) axis.
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