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danielakkerma
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Currents & Magnetic fields
Hey folks... I won't bore you with all the details(those that I deem irrelevant) but simply state my cause of privation on this issue:
(Attached, in PDF format is an illustration).
"A infinite current reaches a semi-infinite plane consisting of a well conducting medium. as a result, it disperses evenly, radially symmetrically across the plane with the total value(shown as 'I') preserved!(See graph for greater legibility)."
I am impelled to find the magnetic field at the Point A(symbol, sadly missing on the diagram, but its nonetheless conspicuous), located at the angle Theta(constant, naturally), from the original perpendicular course of the current(as demonstrated), and at a distance r, therein.
Instantly, Bio-Savart's law comes to mind:
d\vec{B} = \frac{{\mu}_0}{4 \pi}I \int \frac{d\vec{l}\times{\hat{r}}}{r^2}
You might want to throw in Ampere-Maxwell's law(though, there isn't an Amperian loop pertaining to this geometry, that could well embroider the entire current):
(With dE/dt = 0):
\oint \vec{B} \cdot {d\vec{l}}={\mu}_0 \sum I
Well, Initially, I first lingered on the subject of the inherent symmetry, evident from the statement of the problem.
I postulated, that in this case, the current is distributed evenly(as mentioned) across the various angles, through the following law: Defining a relative(angle dependent quantity)I' one gets:
dI'=\frac{I}{\pi}*d\alpha
Where one expects alpha to traverse through 0 to Pi.
Integrating it, does yield I, as expected(meaning, the total current, as predicted, is the same I to begin with).
Now, for the redefinition of Bio-Savart's law:
Since the Plane is Semi infinite, I can assume that 'current' I' starts at an arbitrary point zero(in our case, where the initial current and the plane converge) all the way to infinity(these radii, on which the current travels are straight), produce a magnetic field as if due to a simple wire(semi infinite) whose equation, I trust I don't have to prove to you is given by:
\vec{B}(I')=\frac{{\mu}_0 I'}{4 \pi r}*(\hat{I} \times \hat{r})
r- being the perpendicular from the destination to the current, and I-hat is the perceived orientation of the current(fictitious, since I is a scalar).
As you can see on your chart, on the right hand side, each r(the distance) is easily measured as the Cos(Theta+alpha), where Alpha, is measured from the boundary of the plane to Pi/2- Theta.
On the left, by employing Beta, an angle from the other axis to perpendicular to the margin of the surface carried from Theta, to Pi respectively.(But there
So, there I am, naively perhaps, thinking, that this is a mere summation;(
I simply rendered it as:
\vec{B}=\int{d\vec{B}}=\frac{{\mu}_o}{4 \pi r}*(\int^{\frac{\pi}{2}-\theta}_0\frac{{d\alpha}}{cos(\alpha+\theta)}-<br /> \int^{\pi}_{\frac{\pi}{2}-\theta}\frac{{d\beta}}{sin(\theta+\beta)})
This amounts to, after several handy reductions here and there, to a natural logarithm of Tan(Theta/2) + the hyperbolic Sin due to the integration of the Cos.
In this arena, it is inconceivable, both mathematically and physically that we should see a logarithmic decrement, especially one proportional to the angle, and plus the answer's all fumbled too, and doesn't match the one granted by the author :D(so you have Math & the drafter of the question is accord, pretty strong argument.
I leave it to you then,
With gratitude,
And thanks for your tolerance of my circumlocution,
Thank you!
Daniel
Homework Statement
Hey folks... I won't bore you with all the details(those that I deem irrelevant) but simply state my cause of privation on this issue:
(Attached, in PDF format is an illustration).
"A infinite current reaches a semi-infinite plane consisting of a well conducting medium. as a result, it disperses evenly, radially symmetrically across the plane with the total value(shown as 'I') preserved!(See graph for greater legibility)."
I am impelled to find the magnetic field at the Point A(symbol, sadly missing on the diagram, but its nonetheless conspicuous), located at the angle Theta(constant, naturally), from the original perpendicular course of the current(as demonstrated), and at a distance r, therein.
Homework Equations
Instantly, Bio-Savart's law comes to mind:
d\vec{B} = \frac{{\mu}_0}{4 \pi}I \int \frac{d\vec{l}\times{\hat{r}}}{r^2}
You might want to throw in Ampere-Maxwell's law(though, there isn't an Amperian loop pertaining to this geometry, that could well embroider the entire current):
(With dE/dt = 0):
\oint \vec{B} \cdot {d\vec{l}}={\mu}_0 \sum I
The Attempt at a Solution
Well, Initially, I first lingered on the subject of the inherent symmetry, evident from the statement of the problem.
I postulated, that in this case, the current is distributed evenly(as mentioned) across the various angles, through the following law: Defining a relative(angle dependent quantity)I' one gets:
dI'=\frac{I}{\pi}*d\alpha
Where one expects alpha to traverse through 0 to Pi.
Integrating it, does yield I, as expected(meaning, the total current, as predicted, is the same I to begin with).
Now, for the redefinition of Bio-Savart's law:
Since the Plane is Semi infinite, I can assume that 'current' I' starts at an arbitrary point zero(in our case, where the initial current and the plane converge) all the way to infinity(these radii, on which the current travels are straight), produce a magnetic field as if due to a simple wire(semi infinite) whose equation, I trust I don't have to prove to you is given by:
\vec{B}(I')=\frac{{\mu}_0 I'}{4 \pi r}*(\hat{I} \times \hat{r})
r- being the perpendicular from the destination to the current, and I-hat is the perceived orientation of the current(fictitious, since I is a scalar).
As you can see on your chart, on the right hand side, each r(the distance) is easily measured as the Cos(Theta+alpha), where Alpha, is measured from the boundary of the plane to Pi/2- Theta.
On the left, by employing Beta, an angle from the other axis to perpendicular to the margin of the surface carried from Theta, to Pi respectively.(But there
So, there I am, naively perhaps, thinking, that this is a mere summation;(
I simply rendered it as:\vec{B}=\int{d\vec{B}}=\frac{{\mu}_o}{4 \pi r}*(\int^{\frac{\pi}{2}-\theta}_0\frac{{d\alpha}}{cos(\alpha+\theta)}-<br /> \int^{\pi}_{\frac{\pi}{2}-\theta}\frac{{d\beta}}{sin(\theta+\beta)})
This amounts to, after several handy reductions here and there, to a natural logarithm of Tan(Theta/2) + the hyperbolic Sin due to the integration of the Cos.
In this arena, it is inconceivable, both mathematically and physically that we should see a logarithmic decrement, especially one proportional to the angle, and plus the answer's all fumbled too, and doesn't match the one granted by the author :D(so you have Math & the drafter of the question is accord, pretty strong argument
.I leave it to you then,
With gratitude,
And thanks for your tolerance of my circumlocution,
Thank you!
Daniel
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