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A uniform current density is given as A/m? The corresponding vector magnetic potential is given as A = HoJo (x² + y?) = -2 Wb/m a) Find H using A b) Use J in conjunction with Ampere's law to find H.

A uniform current density is given as A/m? The corresponding vector magnetic potential is given as A = HoJo (x² + y?) = -2 Wb/m a) Find H using A b) Use J in conjunction with Ampere's law to find H.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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### Topic: Electromagnetic Fields #### Problem Statement: A uniform current density is given as: \[ \mathbf{J} = \hat{z} J_0 \quad \text{A/m}^2 \] The corresponding vector magnetic potential is given as: \[ \mathbf{A} = -\hat{z} \frac{\mu_0 J_0}{4} \left( x^2 + y^2 \right) \quad \text{Wb/m} \] **Tasks:** a) **Find the magnetic field intensity \(\mathbf{H}\) using the vector potential \(\mathbf{A}\).** b) **Use the current density \(\mathbf{J}\) in conjunction with Ampere’s law to find \(\mathbf{H}\).** #### Explanation: - **Current Density (\(\mathbf{J}\)):** This is the amount of electric current flowing per unit area and is directed along the z-axis with a magnitude \(J_0\). - **Vector Magnetic Potential (\(\mathbf{A}\)):** This is a vector field that helps describe the magnetic field \(\mathbf{B}\). It is directed along the negative z-axis and depends on the square of the distance from the origin in the xy-plane. #### Methodology: 1. **Find \(\mathbf{H}\) using \(\mathbf{A}\):** The magnetic field \(\mathbf{B}\) can be found using the curl of the vector potential: \[ \mathbf{B} = \nabla \times \mathbf{A} \] Once \(\mathbf{B}\) is calculated, the magnetic field intensity \(\mathbf{H}\) is given by: \[ \mathbf{H} = \frac{\mathbf{B}}{\mu_0} \] 2. **Use Ampere’s Law to find \(\mathbf{H}\):** Ampere's Law relates the magnetic field intensity \(\mathbf{H}\) around a closed loop to the electric current passing through the loop: \[ \oint \mathbf{H} \cdot d\mathbf{l} = \int \mathbf{J} \cdot d\mathbf{S} \] This formulation allows calculating \(\mathbf{H}\) directly when the current density \(\mathbf{J}\)
Transcribed Image Text:### Topic: Electromagnetic Fields #### Problem Statement: A uniform current density is given as: \[ \mathbf{J} = \hat{z} J_0 \quad \text{A/m}^2 \] The corresponding vector magnetic potential is given as: \[ \mathbf{A} = -\hat{z} \frac{\mu_0 J_0}{4} \left( x^2 + y^2 \right) \quad \text{Wb/m} \] **Tasks:** a) **Find the magnetic field intensity \(\mathbf{H}\) using the vector potential \(\mathbf{A}\).** b) **Use the current density \(\mathbf{J}\) in conjunction with Ampere’s law to find \(\mathbf{H}\).** #### Explanation: - **Current Density (\(\mathbf{J}\)):** This is the amount of electric current flowing per unit area and is directed along the z-axis with a magnitude \(J_0\). - **Vector Magnetic Potential (\(\mathbf{A}\)):** This is a vector field that helps describe the magnetic field \(\mathbf{B}\). It is directed along the negative z-axis and depends on the square of the distance from the origin in the xy-plane. #### Methodology: 1. **Find \(\mathbf{H}\) using \(\mathbf{A}\):** The magnetic field \(\mathbf{B}\) can be found using the curl of the vector potential: \[ \mathbf{B} = \nabla \times \mathbf{A} \] Once \(\mathbf{B}\) is calculated, the magnetic field intensity \(\mathbf{H}\) is given by: \[ \mathbf{H} = \frac{\mathbf{B}}{\mu_0} \] 2. **Use Ampere’s Law to find \(\mathbf{H}\):** Ampere's Law relates the magnetic field intensity \(\mathbf{H}\) around a closed loop to the electric current passing through the loop: \[ \oint \mathbf{H} \cdot d\mathbf{l} = \int \mathbf{J} \cdot d\mathbf{S} \] This formulation allows calculating \(\mathbf{H}\) directly when the current density \(\mathbf{J}\)
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