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)
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ISBN:9780133923605
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Publisher:Robert L. Boylestad
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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 \(\mathbf{J}\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F132c2b50-c806-45f9-8baf-5357fa898a21%2F70db13dd-21f9-4a34-8d73-63f4e4ce4a8b%2F0rx4vss_processed.jpeg&w=3840&q=75)
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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