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Chapter 7: Problem 42

Show that \(\nabla_{2}\left(1 / R_{12}\right)=-\nabla_{1}\left(1 / R_{12}\right)=\mathbf{R}_{21} / R_{12}^{3}\).

Short Answer

Expert verified
Question: Show that the gradient of 1/R_12 with respect to the second coordinate and the negative gradient with respect to the first coordinate are equal to R_21/R_12^3. Answer: The gradient of 1/R_12 with respect to the second coordinate is -R_21/R_12^3, and the gradient of 1/R_12 with respect to the first coordinate is R_21/R_12^3.

Step by step solution

01

Define R_12 and R_21

We first define R_12 and R_21, where R_12 is the distance between points R_1 and R_2, and R_21 is their difference vector: \(R_{12}=|R_{2}-R_{1}|\) and \(R_{21}=R_{1}-R_{2}\). Since the magnitude of R_21 is the same as the magnitude of R_12, we can write \(R_{12}=|R_{21}|\).
02

Compute gradient of 1/R_12 with respect to R_2

To compute the gradient of 1/R_12 with respect to R_2, we'll first find the gradient of R_12 with respect to R_2 and use the chain rule. We have: \(\nabla_{2} R_{12} = \nabla_{2} |R_2 - R_1| = \frac{R_2 - R_1}{R_{12}}\) Now, using the chain rule, we compute the gradient of 1/R_12 with respect to R_2: \(\nabla_{2}\left(\frac{1}{R_{12}}\right)=-\frac{1}{R_{12}^{2}}\nabla_{2} R_{12}=-\frac{R_2 - R_1}{R_{12}^3}\)
03

Compute gradient of 1/R_12 with respect to R_1

Similarly, we compute the gradient of R_12 with respect to R_1: \(\nabla_{1} R_{12} = \nabla_{1} |R_2 - R_1| = -\frac{R_2 - R_1}{R_{12}}\) Notice that the negative of the gradient of R_12 with respect to R_1 is the same as the gradient with respect to R_2, as found in step 2: \(\nabla_{1} R_{12} = - \nabla_{2} R_{12}\) Now, using the chain rule, we compute the gradient of 1/R_12 with respect to R_1: \(\nabla_{1}\left(\frac{1}{R_{12}}\right)=-\frac{1}{R_{12}^{2}}\nabla_{1} R_{12}=\frac{R_2 - R_1}{R_{12}^3}\)
04

Compare with the given expressions

We can now compare the results from steps 2 and 3 to the given expressions. We found that: \(\nabla_{2}\left(\frac{1}{R_{12}}\right)=-\frac{R_2 - R_1}{R_{12}^3}\) and \(\nabla_{1}\left(\frac{1}{R_{12}}\right)=\frac{R_2 - R_1}{R_{12}^3}\). Since \(R_{21}=R_{1}-R_{2}=-(R_2 - R_1)\), the expressions can be rewritten as: \(\nabla_{2}\left(\frac{1}{R_{12}}\right)=-\frac{R_{21}}{R_{12}^3}\) and \(\nabla_{1}\left(\frac{1}{R_{12}}\right)=\frac{R_{21}}{R_{12}^3}\). We can see that the expressions match the ones we want to show, completing the proof.

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Most popular questions from this chapter

( \(a\) ) Find \(\mathbf{H}\) in rectangular components at \(P(2,3,4)\) if there is a current filament on the \(z\) axis carrying \(8 \mathrm{~mA}\) in the \(\mathbf{a}_{z}\) direction. ( \(b\) ) Repeat if the filament is located at \(x=-1, y=2\). ( \(c\) ) Find \(\mathbf{H}\) if both filaments are present.

The magnetic field intensity is given in a certain region of space as \(\mathbf{H}=\) \(\left[(x+2 y) / z^{2}\right] \mathbf{a}_{y}+(2 / z) \mathbf{a}_{z} \mathrm{~A} / \mathrm{m} .(a)\) Find \(\nabla \times \mathbf{H} .(b)\) Find \(\mathbf{J} .(c)\) Use \(\mathbf{J}\) to find the total current passing through the surface \(z=4,1 \leq x \leq 2,3 \leq z \leq 5\), in the \(\mathbf{a}_{z}\) direction. ( \(d\) ) Show that the same result is obtained using the other side of Stokes' theorem.

Infinitely long filamentary conductors are located in the \(y=0\) plane at \(x=n\) meters where \(n=0, \pm 1, \pm 2, \ldots\) Each carries \(1 \mathrm{~A}\) in the \(\mathbf{a}_{z}\) direction. (a) Find \(\mathbf{H}\) on the \(y\) axis. As a help, $$\sum_{n=1}^{\infty} \frac{y}{y^{2}+n^{2}}=\frac{\pi}{2}-\frac{1}{2 y}+\frac{\pi}{e^{2 \pi y}-1}$$ (b) Compare your result of part \((a)\) to that obtained if the filaments are replaced by a current sheet in the \(y=0\) plane that carries surface current density \(\mathbf{K}=1 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\).

Given \(\mathbf{H}=\left(3 r^{2} / \sin \theta\right) \mathbf{a}_{\theta}+54 r \cos \theta \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m}\) in free space: \((a)\) Find the total current in the \(\mathbf{a}_{\theta}\) direction through the conical surface \(\theta=20^{\circ}, 0 \leq \phi \leq 2 \pi\), \(0 \leq r \leq 5\), by whatever side of Stokes' theorem you like the best. \((b)\) Check the result by using the other side of Stokes' theorem.

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