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Chapter 7: Problem 14

A toroid having a cross section of rectangular shape is defined by the following surfaces: the cylinders \(\rho=2\) and \(\rho=3 \mathrm{~cm}\), and the planes \(z=1\) and \(z=2.5 \mathrm{~cm}\). The toroid carries a surface current density of \(-50 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) on the surface \(\rho=3 \mathrm{~cm}\). Find \(\mathbf{H}\) at the point \(P(\rho, \phi, z):(a) P_{A}(1.5 \mathrm{~cm}, 0\), \(2 \mathrm{~cm}) ;\left(\right.\) b) \(P_{B}(2.1 \mathrm{~cm}, 0,2 \mathrm{~cm}) ;\) (c) \(P_{C}(2.7 \mathrm{~cm}, \pi / 2,2 \mathrm{~cm}) ;\) (d) \(P_{D}(3.5 \mathrm{~cm},\), \(\pi / 2,2 \mathrm{~cm})\)

Short Answer

Expert verified
Question: Determine the magnetic field intensity (H) at points PA(1.5 cm, 0, 2 cm), PB(2.1 cm, 0, 2 cm), PC(2.7 cm, π/2, 2 cm), and PD(3.5 cm, π/2, 2 cm) for a toroid with a surface current density of -50 𝑎𝑧 A/m on the surface ρ=3 cm. Answer: The magnetic field intensity (H) at the four points are as follows: - H(PA) = 0 - H(PB) = -47.62 𝑎𝜑 A/m - H(PC) = -55.56 𝑎𝜑 A/m - H(PD) = 0

Step by step solution

01

1. Identify given parameters and the Ampere's Circuital Law formula

The given parameters are: - The toroid has a cross section of rectangular shape and is defined by the cylindrical surfaces at ρ=2 cm and ρ=3 cm, and the plane surfaces at z=1 cm and z=2.5 cm. - The toroid has a surface current density of -50 𝑎𝑧 A/m on the surface ρ=3 cm. - We need to find the magnetic field intensity at the four points: 𝑃𝐴(1.5 cm, 0, 2 cm), 𝑃𝐵(2.1 cm, 0, 2 cm), 𝑃𝐶(2.7 cm, π/2, 2 cm), and 𝑃𝐷(3.5 cm, π/2, 2 cm). According to Ampere's Circuital Law, the line integral of the magnetic field intensity (H) over a closed path, denoted as ∮𝐻·d𝐿, is equal to the net current enclosed by the path (I), which can be represented as: ∮𝐻·d𝐿 = 𝐼
02

2. Calculate the magnetic field intensity (H) at point PA

For point PA, we can see that it lies outside the toroid since ρ<2 cm. Therefore, there is no current enclosed by the closed path. By applying Ampere's Law: ∮𝐻·d𝐿 = 0 Thus, H(PA) = 0
03

3. Calculate the magnetic field intensity (H) at point PB

For point PB, we can see that it lies inside the toroid with ρ>2 cm and ρ<3 cm. Therefore, we only need to consider the current within the closed path up to ρ=2 cm, since it is the nearest surface with a current. Due to the azimuthal symmetry of the toroid, the magnetic field intensity H can be assumed to have only an ϕ component (𝐻ϕ): ∮𝐻ϕd𝐿 = -50 A/m × (2π × 2 cm) Hence, Hϕ = (-50 A/m) × (2π × 2 cm) / (2π × 2.1 cm) = -47.62 𝑎𝜑 A/m Thus, H(PB) = -47.62 𝑎𝜑 A/m
04

4. Calculate the magnetic field intensity (H) at point PC

For point PC, it lies on the surface where the toroid has a surface current density, ρ=3 cm. So, we need to consider the current enclosed by the entire cross-sectional area of the toroid: ∮𝐻ϕd𝐿 = -50 A/m × (2π × 3 cm) Since the path goes through point PC, it is equal to Hϕ × (2π × 2.7 cm): Hϕ = (-50 A/m) × (2π × 3 cm) / (2π × 2.7 cm) = -55.56 𝑎𝜑 A/m Thus, H(PC) = -55.56 𝑎𝜑 A/m
05

5. Calculate the magnetic field intensity (H) at point PD

For point PD, it lies outside the toroid since ρ>3 cm. Therefore, there is no net enclosed current in the closed path. By applying Ampere's Law: ∮𝐻·d𝐿 = 0 Thus, H(PD) = 0 Now, we have the magnetic field intensity at all four points: - H(PA) = 0 - H(PB) = -47.62 𝑎𝜑 A/m - H(PC) = -55.56 𝑎𝜑 A/m - H(PD) = 0

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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.

A cylindrical wire of radius \(a\) is oriented with the \(z\) axis down its center line. The wire carries a nonuniform current down its length of density \(\mathbf{J}=b \rho \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}^{2}\) where \(b\) is a constant. ( \(a\) ) What total current flows in the wire? \((b)\) Find \(\mathbf{H}_{i n}(0<\rhoa)\), as a function of \(\rho ;(d)\) verify your results of parts \((b)\) and \((c)\) by using \(\nabla \times \mathbf{H}=\mathbf{J}\).

Let \(\mathbf{A}=(3 y-z) \mathbf{a}_{x}+2 x z \mathbf{a}_{y} \mathrm{~Wb} / \mathrm{m}\) in a certain region of free space. (a) Show that \(\nabla \cdot \mathbf{A}=0 .(b)\) At \(P(2,-1,3)\), find \(\mathbf{A}, \mathbf{B}, \mathbf{H}\), and \(\mathbf{J}\).

A current filament on the \(z\) axis carries a current of \(7 \mathrm{~mA}\) in the \(\mathbf{a}_{z}\) direction, and current sheets of \(0.5 \mathrm{a}_{z} \mathrm{~A} / \mathrm{m}\) and \(-0.2 \mathrm{a}_{z} \mathrm{~A} / \mathrm{m}\) are located at \(\rho=1 \mathrm{~cm}\) and \(\rho=0.5 \mathrm{~cm}\), respectively. Calculate \(\mathbf{H}\) at: \((\) a \() \rho=0.5 \mathrm{~cm} ;(b) \rho=\) \(1.5 \mathrm{~cm} ;(c) \rho=4 \mathrm{~cm} .(d)\) What current sheet should be located at \(\rho=4 \mathrm{~cm}\) so that \(\mathbf{H}=0\) for all \(\rho>4 \mathrm{~cm}\) ?

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