To calculate the magnetic vector potential \( \mathbf{A} \) due to a wire carrying an alternating current, we can use the formula:
\[ \mathbf{A} = \frac{\mu_0 I(t)}{4\pi} \int \frac{d\mathbf{l}' \times \mathbf{r}}{r^2} \]
Where:
- \( \mu_0 \) is the permeability of free space, \( \mu_0 \approx 4\pi \times 10^{-7} , \text{T m/A} \).
- \( I(t) \) is the current in the wire.
- \( d\mathbf{l}' \) is the differential length vector along the wire.
- \( \mathbf{r} \) is the vector pointing from the wire to the observation point.
- \( r \) is the distance from the wire to the observation point.
Since the current is an alternating current, we need to express \( I(t) \): \[ I(t) = I_0 \sin(\omega t) \] Where:
- \( I_0 = 10, \text{A} \)
- \( \omega = 2\pi f \) and, in our case, \( f = 30, \text{MHz} = 30 \times 10^6, \text{Hz} \).
Thus: \[ \omega = 2 \pi (30 \times 10^6) \approx 188.5 \times 10^6 , \text{rad/s} \]
The wire is oriented along the z-axis, therefore in cylindrical coordinates, the differential length vector \( d\mathbf{l}' \) can be expressed as: \[ d\mathbf{l}' = dz \hat{z} \] And the position vector to a point at a distance \( r \) from the wire in the x-y plane is:
- At a distance \( r \): \( \mathbf{r} = \sqrt{x^2 + y^2} \) (we'll use \( r = 1, \text{m} \) and \( r = 100, \text{m} \) respectively).
Step 1: Compute the Magnetic Vector Potential
Now we are interested in the vector potential at distances \( r = 1, \text{m} \) and \( r = 100, \text{m} \).
The vector \( \mathbf{r} \) from the wire at a point with radial distance \( r \) in the x-y plane is: \[ \mathbf{r} = \hat{r} r \]
Using cylindrical coordinates, the distance \( r \) can be derived from \( r^2 = x^2 + y^2 + z^2 \). However, since we're only interested in the plane radial distance at z = 0, we can simplify the problem as follows.
The magnetic vector potential in this case can be approximated since the wire is infinite, and we are asking for the potential at distances far relative to the wire length.
\[ \mathbf{A}(r) = \frac{\mu_0 I_0}{4\pi r} \hat{\phi} e^{j \omega t} \] The current \( I_0 = 10 \text{ A} \) and \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \).
Step 2: Calculate \( \mathbf{A} \) at \( r = 1, \text{m} \) and \( r = 100, \text{m} \)
At \( r = 1 , \text{m} \): \[ \mathbf{A}(1) = \frac{4\pi \times 10^{-7} \times 10}{4\pi \times 1} \hat{\phi} e^{j \omega t} \] \[ \mathbf{A}(1) = 10^{-6} \hat{\phi} e^{j \cdot 188.5 \times 10^6 t} \]
At \( r = 100 , \text{m} \): \[ \mathbf{A}(100) = \frac{4\pi \times 10^{-7} \times 10}{4\pi \times 100} \hat{\phi} e^{j \omega t} \] \[ \mathbf{A}(100) = \frac{10^{-6}}{100} \hat{\phi} e^{j \cdot 188.5 \times 10^6 t} \] \[ \mathbf{A}(100) = 10^{-8} \hat{\phi} e^{j \cdot 188.5 \times 10^6 t} \]
Final Results
Thus, the magnetic vector potential \( \mathbf{A} \) at distances \( 1 , \text{m} \) and \( 100 , \text{m} \) from the wire are:
- At \( r = 1 , \text{m}: \mathbf{A}(1) = 10^{-6} \hat{\phi} e^{j \cdot 188.5 \times 10^6 t} , \text{T m/A}\)
- At \( r = 100 , \text{m}: \mathbf{A}(100) = 10^{-8} \hat{\phi} e^{j \cdot 188.5 \times 10^6 t} , \text{T m/A} \)
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