Det kan være lurt å tenke på divergensen til \( \vec{B} \)-feltet.

Her er det nyttig å bruke at divergensen til \( \vec{B} \) må være null for å argumenter for at det ikke er noen radiell komponent. Og vi vet at \( \vec{B} \)-feltet må være en normal til strømelementene \( I d \vec{l} \), dermed kan det ikke være noen komponent langs \( z \)-aksen.

Jorden har en flytende, metallisk kjerne.

\( d B_y = -\mu_0 I \d l /(4 \pi) (z/R^3) \)

First, we do it geometrically without writing the vectors on coordinate form. We know from Biot-Savart's law that $$ \begin{equation} \d \vec{B} = \frac{\mu_0}{4 \pi} \frac{I \d \vec{l} \times \Rhat }{R^2} \; . \tag{9.1} \end{equation} $$ The cross product is $$ \begin{equation} \d \vec{l} \times \hat{R} = |\d \vec{l}| \, |\hat{R}| \, \sin \alpha \tag{9.2} \end{equation} $$ where \( \sin \alpha = z/R \). The direction of the cross product is in the negative \( y \)-direction. The contribution is therefore $$ \begin{equation} \d B_y = -\frac{\mu_0 I \d l}{4 \pi} \frac{z}{R^3} \; . \tag{9.3} \end{equation} $$ We can also find the cross product using the coordinate form of \( \hat{R} = (-x,0,z)/R \) and \( \d \vec{l} = (\d x,0,0) \). We then find that $$ \begin{equation} \d \vec{l} \times \hat{R} = (\d x,0,0) \times (-x,0,z)/R = (0,-\d x \, z,0) \tag{9.4} \end{equation} $$ The result is therefore the same as above.

In order to set up the integral, we need to insert correct boundary conditions.

For case (i) \( x \) varies from \( - \infty \) to \( + \infty \). $$ \begin{equation} B_i = -\int_{- \infty}^{\infty} \frac{\mu_0 I}{4 \pi} \frac{z}{\left( x^2 + z^2\right)^{3/2}} \d x \tag{9.5} \end{equation} $$

For case (ii) \( x \) varies from \( - \infty \) to \( 0 \). $$ \begin{equation} B_{ii} = -\int_{- \infty}^{0} \frac{\mu_0 I}{4 \pi} \frac{z}{\left( x^2 + z^2\right)^{3/2}} \d x \tag{9.6} \end{equation} $$

For case (iii) \( x \) varies from \( -L/2 \) to \( L/2 \). $$ \begin{equation} B_{iii} = -\int_{-L/2}^{L/2} \frac{\mu_0 I}{4 \pi} \frac{z}{\left( x^2 + z^2\right)^{3/2}} \d x \tag{9.7} \end{equation} $$

We find the solutions by inserting the indefinite integral.

For case (i) we use a trick where we divide by \( |x| \) above and below in the fraction, writing that \( x/|x| = \text{sign}(x) \): $$ \begin{eqnarray} B_i &=& -\frac{\mu_0 I}{4 \pi} \int_{- \infty}^{\infty} \frac{z}{\left( x^2 + z^2\right)^{3/2}} \d x \\ &=& -\frac{\mu_0 I}{4 \pi} \left[ \frac{x}{z \sqrt{ x^2 + z^2}} \right]_{-\infty}^{\infty} \\ &=& -\frac{\mu_0 I}{4 \pi} \left[ \frac{\text{sign}(x)}{z \sqrt{ 1 + (z/x)^2}} \right]_{-\infty}^{\infty} \\ &=& -\frac{\mu_0 I}{4 \pi z}2 = -\frac{\mu_0 I}{2 \pi z} \; . \end{eqnarray} $$

For case (ii): $$ \begin{eqnarray} B_{ii} &=& -\frac{\mu_0 I}{4 \pi} \int_{- \infty}^{0} \frac{z}{\left( x^2 + z^2\right)^{3/2}} \d x \\ &=& -\frac{\mu_0 I}{4 \pi} \left[ \frac{x}{z \sqrt{ x^2 + z^2}} \right]_{-\infty}^{0} \end{eqnarray} $$ Here, we need to handle the two limits differently. The limit for \( x=0 \) is zero, because of the \( x \) in the nominator. The limit for \( x\rightarrow - \infty \) is handled the same way as for case (i). Giving us $$ \begin{equation} B_{ii} = -\frac{\mu_0 I}{4 \pi z} \; . \tag{9.9} \end{equation} $$

For case (iii): $$ \begin{eqnarray} B_{iii} &=& -\frac{\mu_0 I}{4 \pi} \int_{- L/2}^{L/2} \frac{z}{\left( x^2 + z^2\right)^{3/2}}\d x \\ &=& -\frac{\mu_0 I}{4 \pi} \left[ \frac{x}{z \sqrt{ x^2 + z^2}} \right]_{-L/2}^{L/2} \\ &=& -\frac{\mu_0 I}{4 \pi} \left[ \frac{\text{sign}(x)}{z \sqrt{ 1 + (z/x)^2}} \right]_{-L/2}^{L/2} \\ &=& -\frac{\mu_0 I}{4 \pi} 2 \frac{1}{z \sqrt{ 1 + (z/(L/2))^2}} \; . \end{eqnarray} $$

In the limit when \( z \gg L \), we get that \( \sqrt{ 1 + (z/(L/2))^2} \simeq z/(L/2) = 2z/L \) and $$ \begin{equation} B_{iii} \simeq -\frac{\mu_0 I}{4 \pi} \frac{L/2}{z^2} 2 = -\frac{\mu_0 IL}{4 \pi z^2} \tag{9.10} \end{equation} $$

Divide the loop into four separate segments and find the force on each segment.

\( F = \frac{\mu_0 I_1 I_2 L}{\pi r} \)

We know that the force on a current element due to a magnetic field is $$ \begin{equation} \d \vec{F} = I \d \vec{l} \times \vec{B} \; . \tag{9.11} \end{equation} $$ It is simplest to calculate the magnetic field from the straight wire with current \( I_1 \). We may use Amperes law for a circular amperian loop with radius \( r \), getting that $$ \begin{equation} \oint_C \vec{B} \cdot \d \vec{l} = 2 \pi r B_{\phi} = \mu_0 I_1 \; \Rightarrow \; B_{\phi} = \frac{\mu_0 I }{2 \pi r} \; . \tag{9.12} \end{equation} $$ We notice that since the field is along \( \phihat \), the force on the curved elements will be zero, because here \( \d \vec{l} \) will be parallel to \( \vec{B} \) and therefore their cross product will be zero. Let us look at the two remaining segments. For the segment on the left side, the current is going downward and the magnetic field is pointing into the plane. The cross product \( I \d \vec{l} \times \vec{B} \) will therefore point to the right. The same is true for the left segment. The force on one of these segments is $$ \begin{equation} \vec{F} = I L \z \times B_{\phi} \phihat = I L B \x \tag{9.13} \end{equation} $$ where the \( x \)-axis is along the dashed line. The total force is therefore \( 2F = 2 I LB \). We insert the expression for \( B \), getting $$ \begin{equation} F = 2 I L B = 2 I_2 L \frac{\mu_0 I_1}{2 \pi r} = \frac{\mu_0 I_1 I_2 L}{\pi r} \tag{9.14} \end{equation} $$

Vi jobber i sylinderkoordinater. Vektoren fra punktet \( (a,0) = a\rhat \) til \( (0, z) = z\z \) er $$ \begin{equation} \vec{R} = z\z - a\rhat \tag{9.15} \end{equation} $$ Lengden av denne vektoren er \( \sqrt{a^2 + z^2} \) slik at $$ \begin{align} R^2 &= a^2 + z^2 \tag{9.16}\\ \hat{\vec{R}} &= \frac{-a\rhat + z\z}{\sqrt{a^2 + z^2}} \tag{9.17} \end{align} $$

Vi begynner med å finne et uttrykk for \( \d \vec{I} \) langs strømsløyfa. Vi kjenner til at i sylinderkoordinater, så vil enhetsvektoren tilhørende asimut vinkel, altså \( \hat{\boldsymbol{\phi}} \) være tangent til en sirkel om \( z \)-aksen. Dermed har vi foreløpig \( \d \vec{I} = I\d \vec{l} = I\d l \hat{\boldsymbol{\phi}} \). Vi kan også uttrykke \( \d l \) ved hjelp av sylinderkoordinater. For elementer langs en sirkel med radius \( a \) får vi \( \d l = a\d\phi \). Dermed har vi til sammen at $$ \begin{equation} \d \vec{I} = Ia\d \phi \hat{\boldsymbol{\phi}} \tag{9.18} \end{equation} $$ Vi tar med oss herifra at for å dekke hele sløyfa, så må vi la \( \phi \) gå \( 0 \to 2\pi \), så det er dette som blir integrasjonsgrensene våre når vi i neste deloppgave skal integrere opp for å finne hele \( B \)-feltet.

Vi er dermed klare til å sette opp Biot–Savarts lov, og sette inn for uttrykkene vi har funnet i vårt system: $$ \begin{align} \d \vec{B} &= \frac{\mu_0 \d\vec{I} \times \hat{\vec{R} }}{4\pi R^2} \tag{9.19}\\ &= \frac{\mu_0 I a \d\phi \hat{\boldsymbol{\phi}} \times (-a\rhat + z\z)}{4\pi(a^2+z^2)\sqrt{a^2+z^2}} \tag{9.20}\\ &= \frac{\mu_0 I a}{4\pi}\frac{\d\phi\hat{\boldsymbol{\phi}}\times (-a\rhat + z\z)}{(a^2+z^2)^{3/2}} \tag{9.21} \end{align} $$

Vi kunne strengt tatt sagt oss fornøyde her, men siden neste deloppgave uansett ber oss om å faktisk finne \( B \)-feltet, så tar vi kryssproduktet også. sylinderkoordinater har som kartesiste og sfæriske koordinater den egenskapen at enhetsvektorene står normalt på hverandre i et høyrehåndssystem (\( \rhat, \hat{\boldsymbol{\phi}}, \z) \). Vi har derfor \( \hat{\boldsymbol{\phi}} \times \rhat = -\z \) og \( \hat{\boldsymbol{\phi}} \times \z = \rhat \). Dermed blir uttrykket vårt for det infinitesimale feltbidraget $$ \begin{equation} \d \vec{B} = \frac{\mu_0 I a}{4\pi}\frac{\d\phi(a\z + z\rhat)}{(a^2+z^2)^{3/2}} \tag{9.22} \end{equation} $$

Vi har allerede gjort det meste av jobben. Det eneste som gjenstår er å integrere opp bidragene fra hele strømsløyfen. Det er allikevel en liten ting å være obs på her. \( \rhat \) er en funksjon av \( \phi \), og vi skal integrere over \( \phi \). Heldigvis blir vi her reddet av sylindersymmetrien. Dersom det totale feltet skulle ha en komponent i en retning \( \rhat(\phi) \), så ville det implisere at å rotere systemet om \( z \)-aksen også skulle rotert denne komponenten om \( z \)-aksen. Men systemet er rotasjonssymmetrisk om \( z \)-aksen, altså ser systemet helt likt ut etter en slik rotasjon. Da kan ikke feltet endre seg under rotasjon om \( z \)-aksen, og vi kan konkludere at den totale \( \rhat(\phi) \)-komponenten av \( B \)-feltet er \( 0 \) for alle \( \phi \). Derfor integrerer vi bare \( \z \)-komponenten. $$ \begin{align} \vec{B} &= \oint_C \d \vec{B} \tag{9.23}\\ &= \int_{\phi=0}^{2\pi} \frac{\mu_0 I a}{4\pi}\frac{a\d\phi\z}{(a^2+z^2)^{3/2}} \tag{9.24}\\ &= \frac{2\pi\mu_0 I a^2}{4\pi(a^2+z^2)^{3/2}} \z \tag{9.25}\\ &= \frac{\mu_0 I a^2}{2(a^2+z^2)^{3/2}} \z \tag{9.26}\\ \tag{9.27} \end{align} $$

Vi bruker Biot-Savarts lov som gir at bidraget er gitt som $$\d \vec{B} = \frac{\mu_0}{4 \pi} \frac{I\d \vec{l} \times \vec{R}}{R^3}$$ hvor \( \vec{R} = \vec{r} - \vec{r}' = (x,y,0) - (x',0,0) = (x-x',y,0) \) og \( I\d \vec{l} = -I\d x' \x \). Vi får da at $$\d \vec{B} = \frac{I\mu_0}{4 \pi} \frac{(-1,0,0) \times (x-x',y,0)}{((x-x')^2 + y^2)^3}$$ som gir at $$\d \vec{B} = \frac{I\mu_0}{4 \pi} \frac{-y\z}{((x-x')^2 + y^2)^3}$$ (Det kan her være lurt å sjekke med høyrehåndsregelen at retningen på magnetfeltet har blitt korrekt. Enten ved å legge tommelen langs strømmen eller ved å anvende høyrehåndsregelen på kryssproduktet).

Du kan i denne oppgaven få bruk for integralet $$\int \frac{\d u}{(u^2 + a^2)^{3/2}} = \frac{u}{a^2(u^2 + a^2)^{1/2}}$$

Vi finner \( \vec{B} \)-feltet ved å summere (integrere) bidragene fra hvert enkelt strøm-element: $$\vec{B} = \int \d \vec{B} = \int_0^{\infty} \frac{I\mu_0}{4 \pi} \frac{-y\z}{((x-x')^2 + y^2)^3} \d x'$$ Vi setter inn \( u = x' - x \) slik at \( \d u = \d x' \). Da blir grensene \( -x \) og \( \infty \) fordi \( \infty-x = \infty \) (med litt fysikk-regning på uendeligheter). Vi får da: $$\vec{B} = \int_{-x}^{\infty} \frac{I\mu_0}{4 \pi} \frac{-y\z}{(u^2 + y^2)^3} \d u$$ Dette kan vi løse med Sympy eller ved å bruke det oppgitte integralet:

import sympy as sy
y = sy.Symbol('y')
u = sy.Symbol('u')
sy.integrate(1.0/sy.sqrt(y**2 + u**2)**3,u)

1.0*u/(y^2*sqrt(u^2 + y^2))

Som er det samme som var oppgitt. Vi trekker \( y^2 \) utenfor og setter inn grensene $$\vec{B} = \frac{-I\mu_0\z}{4 \pi y} \left[ \frac{u}{(u^2 + y^2)^{1/2}}\right]_{-x}^{\infty}$$ Her kan vi bruke et triks i grensen hvor \( u \rightarrow \infty \) ved å dele på \( u \) over og under brøkstreken. Vi setter så inn \( u = \infty \). Da får vi: $$\vec{B} = \frac{-I\mu_0 y \z}{4 \pi} \left( \frac{1}{(1^2 + (y/\infty)^2)^{1/2}} - \frac{-x}{(x^2 + y^2)^{1/2}}\right)$$ som til slutt gir $$\vec{B} = \frac{-I\mu_0 \z}{4 \pi y} \left( 1 + \frac{x}{(x^2 + y^2)^{1/2}} \right)$$ Igjen sjekker vi at fortegnet er korrekt ved høyrehåndsregelen.

(Merk at det her kan være litt forvirring omkring retningen på integralet. Det er to muligheter. Vi kan velge \( I \d \vec{l} \) til å peke i negativ \( x \)-retning og så summere alle bidragene. Da integrerer vi fra \( x' = 0 \) til \( x' = \infty \). Alternativt kan vi beskrive integralet som et linjeintegral med som er parameterisert ved \( \vec{l}(s) = (-s,0,0) \) for \( s = (- \infty,0) \). Da vil \( I \d \vec{l} = -I \d s \x \) og integralet vil gå fra \( s = -\infty \) til \( s = 0 \). Etter litt mellomregning får vi det samme integralet om ovenfor. Det er i alle tilfeller lurt å sjekke at man har fått riktig retning på feltet ved høyrehåndsregelen.)

Dette er en øvelse i å oversette resultatet vi fikk ovenfor til nye koordinater: $$\vec{B} = \frac{-I\mu_0 \z}{4 \pi x} \left( 1 + \frac{y}{(y^2 + x^2)^{1/2}} \right)$$

Det blir summen av de to bidragene som blir $$\vec{B} = \frac{-I\mu_0 \z}{4 \pi} \left( \frac{1}{x}\left( 1 + \frac{y}{(y^2 + x^2)^{1/2}} \right) + \frac{1}{y}\left( 1 + \frac{x}{(y^2 + x^2)^{1/2}} \right) \right) $$

\( \vec{B}(\vec{0}) = -\frac{\mu_0 I}{4a}\left(\frac{2}{\pi} + 1 \right)\z \).

Biot-Savarts law for a thin wire is in general given by $$ \begin{equation} \vec{B} (\vec{r}) = \frac{\mu_0}{4\pi}\int \frac{I d\vec{r}' \times \vec{R}}{|\vec{R}|^3}, \tag{9.28} \end{equation} $$ where \( \vec{R} = \vec{r}-\vec{r}' \). Here \( \vec{r} \) is the point of observation (or point of measurement) and \( \vec{r}' \) is formally a parametrization of the current's spatial distribution. \( d\vec{r}' \) is an infinitesimal vector pointing along the current's path.

We can note that at point P, there's an equal distance to all points on segment \( C_1 \) and \( C_3 \). Also, using the right-hand rule for currents, we should come to the conclusion that the contribution to the magnetic field from both segments point in the same direction. Therefore we need only compute one of them and multiply the resulting integral by a factor 2.

Knowing this, we ought to pick a coordinate system. A sensible one is to set the origin at point \( P \), since the point of observation when becomes \( \vec{r} = 0 \). The simplest part of this problem is to compute the contribution from \( C_2 \), so there is where I'll begin.

The following parametrization describes the position of the half-circle (\( C_2 \)) in space: $$ \begin{equation} \vec{r}'(\theta) = a(\cos\theta \x + \sin\theta \y ) = a\rhat, \tag{9.29} \end{equation} $$ where \( \theta \) runs from \( \theta = 3\pi/2 \) to \( \theta = \pi/2 \) (the direction the current moves does matter when choosing limits). From the parametrization, we can deduce \( d\vec{r}' \): $$ \begin{equation} \frac{d\vec{r}'}{d\theta} = a(-\sin\theta \x + \cos\theta \y ) = a\uhat_\theta . \tag{9.30} \end{equation} $$ implying that \( d\vec{r}' = ad\theta \uhat_\theta \) (just multiply by \( d\theta \) on both sides of the equation). Next up is \( \vec{R} \) and its norm. $$ \begin{equation} \vec{R} = \vec{r}-\vec{r}' = -a\rhat , \tag{9.31} \end{equation} $$ and it's length is naturally just \( |\vec{R}| = R = a \).

Now we're ready to solve the integral: $$ \begin{align} \vec{B}_2 (\vec{0}) & = \frac{\mu_0 I}{4\pi}\int \frac{d\vec{r}' \times (-\vec{r}')}{R^3} \tag{9.32}\\ & = \frac{\mu_0 I}{4\pi}\int_{3\pi/2}^{\pi/2} \frac{ad\theta \uhat_\theta \times (-a\rhat )}{a^3} \tag{9.33}\\ & = -\frac{\mu_0 I}{4\pi a}\z \underbrace{\int_{\pi/2}^{3\pi/2}d\theta }_{=\pi} \tag{9.34}\\ & = -\frac{\mu_0 I}{4 a}\z . \tag{9.35} \end{align} $$

Now the more challenging part. Since it doesn't matter which of the segments \( C_1 \) and \( C_3 \) one chooses, I'll simply pick \( C_3 \). Again, we begin by finding the parametrization of the current in space (I'll prime the coordinates of the source which in this case is a current, this is the typical convention): $$ \begin{equation} \vec{r}'(x') = x'\x + a\y , \qquad x \in [0,\infty) \tag{9.36} \end{equation} $$ from which we can deduce that \( d\vec{r}' = dx'\x \). The difference vector is simply \( \vec{R} = -\vec{r}' \). Biot-Savarts law then implore us to solve the following integral: $$ \begin{equation} \vec{B}_3(\vec{0}) = \frac{\mu_0 I}{4\pi}\int_0^\infty \frac{dx' \x \times (-x\x - a\y )}{((x'^2+a^2))^{3/2}} = -\frac{\mu_0 I a}{4\pi}\z \underbrace{\int_0^\infty \frac{dx'}{(x'^2 + a^2)^{3/2}}}_{\equiv J}, \tag{9.37} \end{equation} $$ Naturally, we can just solve \( J \) using a integral table, but instead I'll show you a neat trick to solve it. Recall (or see it for the first time) that \( \cosh^2 u - \sinh^2 u = 1 \). Looking at the denominator of \( J \), we can substitute \( x' = a\sinh u \). This leads to \( x'^2 + a^2 = a^2(\sinh^2 + 1) = a^2\cosh^2 u \). Furthermore, \( dx' = a\cosh u du \). But what about the new limits? Well, \( \sinh u = 0 \) for \( u = 0 \) and \( \sinh u \to \infty \) as \( u \to \infty \), so the limits are actually the same as before. Lucky us. Inserting into the integral then: $$ \begin{align} J & = \int_0^\infty \frac{dx'}{(x'^2+a^2)^{3/2}} = \int_0^\infty \frac{a\cosh u du}{(a^2\cosh^2 u)^{3/2}} = \int_0^\infty \frac{1}{a^2\cosh^2 u} du \tag{9.38}\\ & = \frac{1}{a^2}\tanh u\bigg|_{0}^{\infty} = \frac{1}{a^2}, \tag{9.39} \end{align} $$ where I used that \( (d/du)\tanh u = 1/\cosh^2 u \). In other words, the contribution to the magnetic field from the \( C_3 \)-segment is $$ \begin{equation} \vec{B}_3 (\vec{0}) = -\frac{\mu_0 I a}{4\pi}\z J = -\frac{\mu_0 I }{4\pi a}\z . \tag{9.40} \end{equation} $$ The \( C_1 \)-segment gives the exact same contribution to the magnetic field. Thus the magnetic field at \( P \) is simply given by $$ \begin{equation} \vec{B} (\vec{0}) = \sum_{i=1}^3 \vec{B}_i (\vec{0}) = -\frac{\mu_0 I}{4a}\left(\frac{2}{\pi} + 1 \right)\z . \tag{9.41} \end{equation} $$ Before moving on, we should check that this is physically reasonable. Using the right-hand rule for currents, you should find that the field contributions from all three segments points into the page (which means the negative \( z \)-direction). This at least increases the likelihood that the result is sensible (though, I obviously know its correct, I'm just trying to give you some good habits).

To solve this problem, we can use a simple algorithm for numerical integration which is called the trapezoidal rule. It's given by the following formula $$ \begin{equation} \int_a^b f(x) dx \approx h\left[\frac{f(a)+f(b)}{2} + \sum_{i=1}^{n-1} f(x_i) \right], \tag{9.42} \end{equation} where $h = (b-a)/n$ and $x_i = a + ih$ for $i = 0,1,..., n$. $$ Below is a simple implementation of the algorithm.

class Trapezoidal:
    def __init__(self, a, b, N):
        self.a = a
        self.b = b
        self.N = N
        self.x = np.linspace(self.a, self.b, self.N)
        self.h = (b-a)/N  #stepsize

    def Solve_SpecialCase(self, f):
        self.I = 0
        self.I = 0.5*(f(self.a)+f(self.b))
        for i in range(1, self.N):
            self.I += f(self.x[i])
        self.I *= self.h
        return self.I

Now that we have a solver, we can compute the B-field numerically. However, there's no reason to compute the integral along the \( C_2 \)-segment since it's merely an integral of a constant. Furthermore, we may still exploit the symmetry that lead to equal contribution from \( C_1 \) and \( C_3 \). The following code shows a possible implementation:

import numpy as np
from integration_solver import Trapezoidal
mu = 4*np.pi*1e-7
R = 1
I = 1
k = mu*I/(4*np.pi)
analytical_value = -(mu*I/(4*R))*(2/np.pi + 1)
a = 0
b = 100  #b = 100 is a sufficient substitute for infinity in this case.
N = 100000
def f(x):
    return -k*R*1.0/(x**2+R**2)**(3/2)
my_solver = Trapezoidal(a, b, N) #Initialize the solver
integral = 2*my_solver.Solve_SpecialCase(f) #C1 and C3 segments.
integral -= mu*I/(4*R) #C2 segments.

Here it's useful to systematically treat each segment. Instead of using numpy's built-in cross-product function, I'll compute the cross-products explicitly to reduce the number of library calls we need (this will be speed up the code in the next subproblem). I'll begin with \( C_1 \).

\( C_1 \) can be parametrized as $$ \begin{equation} \vec{r}'(x') = x'\x - a\y , \qquad d\vec{r}' = dx' \x , \tag{9.43} \end{equation} $$ Meanwhile, since we're studying the field at an arbitrary point in space, we get the following difference vector: $$ \begin{equation} \vec{R} = \vec{r} - \vec{r}' = (x-x')\x + (y+a)\y + z\z , \tag{9.44} \end{equation} $$ with corresponding norm (to the third power) $$ \begin{equation} R^3 = \left[(x-x')^2 + (y+a)^2 + z^2 \right]^{3/2}. \tag{9.45} \end{equation} $$ Biot-Savarts law the yields $$ \begin{align} \vec{B}_1(\vec{r}) & = \frac{\mu_0I}{4\pi}\int_{\infty}^0 \frac{dx' \x \times \left[(x-x')\x + (y+a)\y + z\z \right]}{\left[(x-x')^2 + (y+a)^2 + z^2 \right]^{3/2}} \tag{9.46}\\ & = -\frac{\mu_0I}{4\pi}\int_0^\infty \frac{-z\y + (y+a)\z }{\left[(x-x')^2 + (y+a)^2 + z^2 \right]^{3/2}}dx', \tag{9.47} \end{align} $$ which is the first of the three integrals we need to solve numerically.

For segment \( C_2 \), we'll use cylinder coordinates. The segment can be parametrized as $$ \begin{equation} \vec{r}'(\theta) = a(\cos\theta \x + \sin\theta\y ), \qquad d\vec{r}' = ad\theta(-\sin\theta\x + \cos\theta\y ). \tag{9.48} \end{equation} $$ and \( \vec{R} \) is then given by $$ \begin{equation} \vec{R} = (x-a\cos\theta)\x + (r-a\sin\theta)\y +z\z \tag{9.49} \end{equation} $$ with norm $$ \begin{equation} R^3 = \left[(x-a\cos\theta)^2 + (r-a\sin\theta)^2 + z^2\right]^{3/2}. \tag{9.50} \end{equation} $$ Now we'll insert it into Biot-Savarts law: $$ \begin{align} \vec{B}_2(\vec{r}) & = \frac{\mu_0I}{4\pi}\int_{3\pi/2}^{\pi/2} \frac{ad\theta(-\sin\theta\x + \cos\theta\y ) \times \left[(x-a\cos\theta)\x + (r-a\sin\theta)\y +z\z \right]}{\left[(x-a\cos\theta)^2 + (r-a\sin\theta)^2 + z^2\right]^{3/2}}d\theta \tag{9.51}\\ & = -\frac{\mu_0Ia}{4\pi}\int_{\pi/2}^{3\pi/2}\frac{z[\cos\theta\x + \sin\theta\y ] - [(x-a\cos\theta)\cos\theta + (y-a\sin\theta)\sin\theta]\z }{\left[(x-a\cos\theta)^2 + (r-a\sin\theta)^2 + z^2\right]^{3/2}}d\theta \tag{9.52} \end{align} $$

For the \( C_3 \)-segment, we have $$ \begin{equation} \vec{r}'(x') = x\x + a\y , \qquad d\vec{r}' = dx'\x , \tag{9.53} \end{equation} $$ and $$ \begin{equation} \vec{R} = (x-x')\x + (y-a)\y + z\z , \tag{9.54} \end{equation} $$ with norm $$ \begin{equation} R^3 = \left[(x-x')^2 + (y-a)^2 + z^2\right]^{3/2}. \tag{9.55} \end{equation} $$ Insertion into Biot-Savarts law yields $$ \begin{align} \vec{B}_3(\vec{r}) & = \frac{\mu_0 I}{4\pi}\int_0^\infty \frac{dx'\x \times \left[(x-x')\x + (y-a)\y + z\z \right]}{\left[(x-x')^2 + (y-a)^2 + z^2\right]^{3/2}} \tag{9.56}\\ & = \frac{\mu_0 I}{4\pi}\int_0^\infty \frac{-z\y + (y-a)\z}{\left[(x-x')^2 + (y-a)^2 + z^2\right]^{3/2}}dx'. \tag{9.57} \end{align} $$

Now we have all we need to solve the problem numerically. We do, however, need to extend the integration solver a little bit so that it can handle an arbitrary point in space. The following method/function solves this problem:

def Solve_GeneralCase(self, f, x, y, z):
      self.I = 0
      self.I = 0.5*(f(x,y,z,self.a)+f(x,y,z,self.b))
      for i in range(1, self.N):
          self.I += f(x,y,z,self.x[i])
      self.I *= self.h
      return self.I

The code below shows the actual implementation that computes the magnetic field at the origin, but with the general code:

import numpy as np
from integration_solver import Trapezoidal
mu = 4*np.pi*1e-7
R = 1
I = 1
k = mu*I/(4*np.pi)

#Functions for C1 segment
def dB1_x(x, y, z, x_source):
    return 0

def dB1_y(x, y, z, x_source):
    dB = -z
    rnorm = ((x-x_source)**2 + (y+R)**2 + z**2)**(3/2)
    return -k*dB/rnorm

def dB1_z(x, y, z, x_source):
    dB = y+R
    rnorm = ((x-x_source)**2 + (y+R)**2 + z**2)**(3/2)
    return -k*dB/rnorm

#Functions for C2 segment
def dB2_x(x,y,z,theta):
    dB = z*np.cos(theta)
    rnorm = ( (x-R*np.cos(theta))**2 + (y-R*np.sin(theta))**2 + z**2 )**(3/2)
    return -k*R*dB/rnorm

def dB2_y(x, y, z, theta):
    dB = z*np.sin(theta)
    rnorm = ( (x-R*np.cos(theta))**2 + (y-R*np.sin(theta))**2 + z**2 )**(3/2)
    return -k*R*dB/rnorm

def dB2_z(x, y, z, theta):
    dB = -((x-R*np.cos(theta))*np.cos(theta) + (y-R*np.sin(theta))*np.sin(theta))
    rnorm = ( (x-R*np.cos(theta))**2 + (y-R*np.sin(theta))**2 + z**2 )**(3/2)
    return -k*R*dB/rnorm

#Functions for C3 segment
def dB3_x(x, y, z, x_source):
    return 0

def dB3_y(x, y, z, x_source):
    dB = -z
    rnorm = ((x-x_source)**2 + (y-R)**2 + z**2)**(3/2)
    return k*dB/rnorm

def dB3_z(x, y, z, x_source):
    dB = y-R
    rnorm = ((x-x_source)**2 + (y-R)**2 + z**2)**(3/2)
    return k*dB/rnorm

#Here we solve the integral.
N = 10000
a = 0; b = 100  #(0, inf) --> (0,100)
theta0 = 0.5*np.pi; theta1 = 0.5*3*np.pi #Theta limits.
Bx = 0; By = 0; Bz = 0;
x = 0; y = 0; z = 0;

my_solver1 = Trapezoidal(a, b, N) #Solver for C1 and C3 segment.
my_solver2 = Trapezoidal(theta0, theta1, N) #Solver for C2 segment

#Add up contributions to x-direction
Bx += my_solver1.Solve_GeneralCase(dB1_x, x, y, z)
Bx += my_solver2.Solve_GeneralCase(dB2_x, x, y, z)
Bx += my_solver1.Solve_GeneralCase(dB3_x, x, y, z)
#Add up contributions to y-direction
By += my_solver1.Solve_GeneralCase(dB1_y, x, y, z)
By += my_solver2.Solve_GeneralCase(dB2_y, x, y, z)
By += my_solver1.Solve_GeneralCase(dB3_y, x, y, z)
#Add up contributions to z-direction
Bz += my_solver1.Solve_GeneralCase(dB1_z, x, y, z)
Bz += my_solver2.Solve_GeneralCase(dB2_z, x, y, z)
Bz += my_solver1.Solve_GeneralCase(dB3_z, x, y, z)

We can reuse most of the code from the last problem. We must simply extend it to loop over several values for \( y \) and \( z \), which can be done using built-in functions in numpy like meshgrid. I've used the same function definitions as shown above and the same class implementation of the trapezoidal rule. The following code shows a possible solution.

#Here we solve the integral.
N = 10000
a = 0; b = 100
theta0 = 0.5*np.pi; theta1 = 0.5*3*np.pi
x = 0
y = np.linspace(-2*R, 2*R, 11) #Several points in y-direction
z = np.linspace(-2*R, 2*R, 11) #Several points in z-direction
y, z = np.meshgrid(y,z) #Create the yz-grid.
Bx = np.zeros(np.shape(y)) #Bx has the same shape as y and z.
By = np.zeros(np.shape(y)) #By has the same shape as y and z.
Bz = np.zeros(np.shape(y)) #Bz has the same shape as y and z.


my_solver1 = Trapezoidal(a, b, N) #Solver for C1 and C3 segment.
my_solver2 = Trapezoidal(theta0, theta1, N) #Solver for C2 segment.

#Here we solve the field component-wise
Bx += my_solver1.Solve_GeneralCase(dB1_x, x, y, z)
Bx += my_solver2.Solve_GeneralCase(dB2_x, x, y, z)
Bx += my_solver1.Solve_GeneralCase(dB3_x, x, y, z)

By += my_solver1.Solve_GeneralCase(dB1_y, x, y, z)
By += my_solver2.Solve_GeneralCase(dB2_y, x, y, z)
By += my_solver1.Solve_GeneralCase(dB3_y, x, y, z)

Bz += my_solver1.Solve_GeneralCase(dB1_z, x, y, z)
Bz += my_solver2.Solve_GeneralCase(dB2_z, x, y, z)
Bz += my_solver1.Solve_GeneralCase(dB3_z, x, y, z)

plt.streamplot(y, z, By, Bz) #Creates streamlines.
plt.xlabel("y")
plt.ylabel("z")
plt.title("Streamlines in the yz-plane at x = 0")
plt.show()

which gave the following figure.

\( \vec{B} = \frac{2}{3}\mu_0 \sigma \omega a \z \)

We'll work in spherical coordinates with the origin placed at the sphere's center. Biot-Savarts law for a surface charge is $$ \begin{equation} \vec{B} = \frac{\mu_0}{4\pi}\int \frac{\vec{J}\times \vec{R}}{R^3}dS', \tag{9.58} \end{equation} $$ where the surface current density is given by \( \vec{J} = \sigma \vec{v} = \sigma \omega a\sin\theta \uhat_\phi \) (recall that \( v = \omega r \), and the distance from the axis is \( r = a\sin\theta \). Furthermore, I've used the fact that \( \vec{v} \) is moving about the z-axis which is the direction of \( \uhat_\phi \)). Furthermore we have that \( \vec{R} = \vec{r} - \vec{r}' = -a\rhat \), and \( R^3 = a^3 \). We also need an expression for \( \vec{J} \times \vec{R} \): $$ \begin{equation} \vec{J} \times \vec{R} = \sigma \omega a \sin\theta \uhat_\phi \times (-a\rhat) = -\sigma\omega a^2 \sin\theta \uhat_\theta, \tag{9.59} \end{equation} $$ where \( \uhat_\theta = \cos\phi\cos\theta\x + \sin\phi\cos\theta\y - \sin\theta\z \). Now, we're integrating over all angles, and \( \int_0^{2\pi}\sin\phi d\phi = \int_0^{2\pi}\cos\phi d\phi = 0 \), we can ignore the parts that points along \( \x \) and \( \y \). Recall that \( dS = a^2\sin\theta d\phi d\theta \) in spherical coordinates. Inserting the relevant parts into Biot-Savarts law, then, gives $$ \begin{align} \vec{B} & = \frac{\mu_0}{4\pi}\z\int_0^\pi\int_0^{2\pi}\frac{\sigma\omega a^2\sin^2\theta}{a^3}a^2\sin\theta d\phi d\theta \tag{9.60}\\ & = \frac{\mu_0 \sigma \omega a}{2}\z\int_0^\pi d\theta\sin^3\theta = \frac{\mu_0 \sigma \omega a}{2}\z\int_0^\pi d\theta\sin\theta(1-\cos^2\theta) \tag{9.61}\\ & = \frac{\mu_0 \sigma \omega a}{2}\z\int_{-1}^1 du(1-u^2) = \frac{2}{3}\mu_0 \sigma \omega a \z. \tag{9.62} \end{align} $$

We'll use Monte-Carlo integration to solve this problem. But first we need an explicit expression for the integrals we are to solve. The current density is still given by $$ \begin{equation} \vec{J} = \sigma\omega a \sin\theta \uhat_\phi = \sigma\omega a \sin\theta(-\sin\phi\x + \cos\phi\y) \tag{9.71} \end{equation} $$

Furthermore, for an arbitrary point in space we get $$ \begin{equation} \vec{R} = (x-a\cos\phi \sin\theta)\x + (y-a\sin\phi\sin\theta)\y + (z-a\cos\theta)\z \tag{9.72} \end{equation} $$ Now we need the cross-product between the current density and the difference vector: $$ \begin{align} \vec{J}\times \vec{R} & = \sigma\omega a \sin\theta(-\sin\phi\x + \cos\phi\y) \times [(x-a\cos\phi \sin\theta)\x \tag{9.73}\\ & \qquad \qquad \qquad + (y-a\sin\phi\sin\theta)\y + (z-a\cos\theta)\z ] \tag{9.74}\\ & = \sigma\omega a \sin\theta[\cos\phi (z-a\cos\theta)\x + \sin\phi(z-a\cos\theta)\y \tag{9.75}\\ & \qquad - (\cos\phi(x-a\cos\phi\sin\theta) + \sin\phi(y-a\sin\phi\sin\theta)) \z], \tag{9.76} \end{align} $$ and the norm $$ \begin{equation} R^3 = \left[(x-a\cos\phi \sin\theta)^2 + (y-a\sin\phi\sin\theta)^2 + (z-a\cos\theta)^2\right]^{3/2}. \tag{9.77} \end{equation} $$ For a sphere we get \( dS = a^2 d\theta d\phi \sin\theta \equiv a^2d\Omega \). From Biot-Savarts law, we then get the following integral to approximate: $$ \begin{align} \vec{B} & = \frac{\mu_0}{4\pi}\int \frac{\vec{J}\times \vec{R}}{R^3}a^2d\Omega \tag{9.78}\\ & = \frac{\mu_0 \sigma \omega a^3}{4\pi}\int\limits_0^\pi\int\limits_0^{2\pi} d\Omega\sin\theta \bigg\{ \x \underbrace{\frac{\cos\phi(z-a\cos\theta)}{R^3}}_{\equiv f_x(\theta, \phi)} + \y \underbrace{\frac{\sin\phi(y-a\cos\theta)}{R^3}}_{\equiv f_y(\theta, \phi)} \tag{9.79}\\ & \qquad\qquad\qquad\qquad\qquad - \z\underbrace{\frac{\cos\phi(x-a\cos\phi\sin\theta) + \sin\phi(y-a\sin\phi\sin\theta)}{R^3}}_{\equiv -f_z(\theta, \phi)} \bigg\} \tag{9.80}\\ & = \frac{\mu_0 \sigma \omega a^3}{4\pi}\int\limits_0^\pi\int\limits_0^{2\pi}d\phi d\theta\sin^2\theta \bigg\{ \x f_x(\theta, \phi) + \y f_y(\theta, \phi) + \z f_z(\theta, \phi) \bigg\} \tag{9.81}\\ & = \underbrace{(2\pi-0)(\pi-0)}_{\prod_j (b_j-a_j)}\cdot\frac{\mu_0 \sigma \omega a^3}{4\pi}\int\limits_0^{1}\int\limits_0^{1}\sin^2 \theta(u) dudv \bigg\{\x f_x(\theta(u), \phi(v)) \tag{9.82}\\ & \qquad \qquad \qquad\qquad\qquad\qquad\qquad\quad + \y f_y(\theta(u), \phi(v)) + \z f_z(\theta(u), \phi(v)) \bigg\} \tag{9.83}\\ & \approx \frac{\mu_0 \sigma \omega a^3 \pi}{2N}\sum_{i=1}^N \sin^2 \theta(u_i) \bigg\{\x f_x(\theta(u_i), \phi(v_i)) + \y f_y(\theta(u_i), \phi(v_i)) \tag{9.84}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad\qquad\qquad\qquad + \z f_z(\theta(u_i), \phi(v_i)) \bigg\}, \tag{9.85} \end{align} $$ where \( u \) and \( v \) are sampled from uniform distributions on \( (0,1) \) and \( \theta(u) = \pi u \) and \( \phi(v) = 2\pi v \). The code below shows a possible implementation to solve this problem.

  def compute_Bfield_MonteCarlo(x,y,z, n):
      s_x = 0; s_y = 0; s_z = 0
      for i in range(n):
          u = np.random.uniform(0, 1)
          theta = np.pi*u
          v = np.random.uniform(0, 1)
          phi = 2*np.pi*v
          cos_phi = np.cos(phi)
          sin_phi = np.sin(phi)
          cos_theta = np.cos(theta)
          sin_theta = np.sin(theta)
          rnorm = ((x-a*cos_phi*sin_theta)**2 + (y-a*sin_phi*sin_theta)**2\
                      + (z-a*cos_theta)**2)**1.5
          f_x = cos_phi*(z-a*cos_theta)/rnorm
          f_y = sin_phi*(z-a*cos_theta)/rnorm
          f_z = -(cos_phi*(x-a*cos_phi*sin_theta)\
                      + sin_phi*(y-a*sin_phi*sin_theta))/rnorm
          s_x += f_x*sin_theta**2
          s_y += f_y*sin_theta**2
          s_z += f_z*sin_theta**2

      k = 0.5*mu*sigma*omega*a**3*np.pi
      s_x *= k/n
      s_y *= k/n
      s_z *= k/n
      return s_x, s_y, s_z


  a = 1
  mu = 4*np.pi*1e-7
  sigma = 1
  omega = 1
  k = mu*sigma*omega*a/(4*np.pi)
  y = np.linspace(-2*a, 2*a, 101)
  z = np.linspace(-2*a, 2*a, 101)
  x = 0
  y, z = np.meshgrid(y, z)
  Bx = np.zeros_like(y)
  By = np.zeros_like(y)
  Bz = np.zeros_like(y)
  Bx, By, Bz = compute_Bfield_MonteCarlo(x,y,z, n=10000)
  phi = np.linspace(0,2*np.pi, 1001)
  x_circle = a*np.cos(phi)
  y_circle = a*np.sin(phi)
  plt.plot(x_circle, y_circle, "-r")
  plt.streamplot(y,z, By, Bz)
  plt.title("Fieldlines in the yz-plane")
  plt.xlabel(r"$y/a$", size=14)
  plt.ylabel(r"$z/a$", size=14)
  plt.axis("equal")

There's not much new mathematics to be done to solve this problem. We need to set \( a \to r \) and \( \sigma \to \rho \). This means that \( \vec{J} = \rho \vec{v} = \rho \omega r \sin \theta \uhat_\phi \) and \( dV = r^2drd\Omega \). Biot-Savarts law for a volume current density is then given as (to arrive at this, take the expression you found in the last problem and replace \( a \to r \) under the integral sign and set \( \sigma \to \rho \)): $$ \begin{align} \vec{B} & = \frac{\mu_0}{4\pi}\int \frac{\vec{J}\times \vec{R}}{R^3}dV \tag{9.86}\\ & = \frac{\mu_0\rho \omega}{4\pi}\int\limits_{0}^{a} \int\limits_{0}^{\pi} \int\limits_{0}^{2\pi} r^3\sin\theta dr d\Omega \bigg\{ \x \overbrace{\frac{\cos\phi (z-r\cos\theta)}{R^3}}^{\equiv f_x(r,\theta, \phi)} + \y \overbrace{\frac{\sin\phi (z-r\cos\theta)}{R^3}}^{\equiv f_y(r,\theta,\phi)} \tag{9.87}\\ & \qquad\qquad\qquad - \z \underbrace{\frac{\cos\phi(x-r\cos\phi\sin\theta) + \sin\phi(y-r\sin\phi\sin\theta)}{R^3}}_{\equiv -f_z(r,\theta,\phi)} \bigg\} \tag{9.88}\\ & = \frac{\mu_0\rho \omega}{4\pi}\int\limits_{0}^{a} \int\limits_{0}^{\pi} \int\limits_{0}^{2\pi}r^3\sin^2\theta dr d\theta d\phi \bigg\{\x f_x(r,\theta, \phi) + \y f_y(r,\theta, \phi) + \z f_z(r, \theta, \phi) \bigg\} \tag{9.89}\\ & = 2a\pi^2\frac{\mu_0\rho \omega}{4\pi}\int\limits_0^1\int\limits_0^1\int\limits_0^1 r(u)^3\sin^2\theta(v)du dv dw \bigg\{\x f_x(r(u),\theta (v), \phi (w)) \tag{9.90}\\ & \qquad \qquad \qquad + \y f_y(r(u),\theta(v), \phi(w)) + \z f_z(r(u), \theta(v), \phi(w)) \bigg\} \tag{9.91}\\ & \approx \frac{\mu_0 \rho \omega \pi a}{2N}\sum_{i=1}^N r(u_i)^3 \sin^2\theta(v_i) \bigg\{\x f_x(r(u_i),\theta (v_i), \phi (w_i)) \tag{9.92}\\ & \qquad \qquad \qquad + \y f_y(r(u_i),\theta(v_i), \phi(w_i)) + \z f_z(r(u_i), \theta(v_i), \phi(w_i)) \bigg\} \tag{9.93} \end{align} $$ The code belows shows how this can be implemented.

  def compute_Bfield_MonteCarlo(x, y, z, n):
      s_x = 0; s_y = 0; s_z = 0
      for i in range(n):
          u = np.random.uniform(0, 1)
          theta = np.pi*u
          v = np.random.uniform(0, 1)
          phi = 2*np.pi*v
          w = np.random.uniform(0, 1)
          r = a*w
          cos_phi = np.cos(phi)
          sin_phi = np.sin(phi)
          cos_theta = np.cos(theta)
          sin_theta = np.sin(theta)
          rnorm = ((x-r*cos_phi*sin_theta)**2 + (y-r*sin_phi*sin_theta)**2\
                      + (z-r*cos_theta)**2)**1.5
          f_x = cos_phi*(z-r*cos_theta)/rnorm
          f_y = sin_phi*(z-r*cos_theta)/rnorm
          f_z = -(cos_phi*(x-r*cos_phi*sin_theta)\
                      + sin_phi*(y-r*sin_phi*sin_theta))/rnorm
          s_x += r**3*sin_theta**2*f_x
          s_y += r**3*sin_theta**2*f_y
          s_z += r**3*sin_theta**2*f_z
      k = 0.5*mu*rho*omega*np.pi*a
      s_x *= k/n
      s_y *= k/n
      s_z *= k/n
      return s_x, s_y, s_z
  a = 1
  mu = 4*np.pi*1e-7
  rho = 1
  omega = 1
  k = mu*rho*omega/(4*np.pi)
  y = np.linspace(-4*a, 4*a, 101)
  z = np.linspace(-4*a, 4*a, 101)
  x = 0
  y, z = np.meshgrid(y, z)
  Bx = np.zeros_like(y)
  By = np.zeros_like(y)
  Bz = np.zeros_like(y)
  Bx, By, Bz = compute_Bfield_MonteCarlo(x,y,z, n=10000)
  phi = np.linspace(0,2*np.pi, 1001)
  x_circle = a*np.cos(phi)
  y_circle = a*np.sin(phi)
  plt.plot(x_circle, y_circle, "-r")
  plt.streamplot(y,z, By, Bz)
  plt.title("Fieldlines in the yz-plane")
  plt.xlabel(r"$y/a$", size=14)
  plt.ylabel(r"$z/a$", size=14)
  plt.axis("equal")

Vi har translasjonssymmetri i \( xy \)-planet, siden strømtettheten ser lik ut om vi flytter den i \( xy \)-planet. Vi har også \( \pi/ \) rad rotasjonssymmetri om \( x \)-aksen. I denne sammenhengen er det viktig å se forskjellen på \( \pi/2 \) rad rotasjonssymmetri og speilsymmetri, siden vi skal gjøre kryssprodukter. Dersom vi setter en høyrehånd på oversiden av planet og deretter speiler det og roterer det tilbake, så blir dette til en venstrehånd. I denne sammenhengen blir det derfor enklere å jobbe med rotasjonssymmetri enn med speilsymmetri. Det å velge \( \pi/2 \) rotasjonssymmetri føre til at vi bevarer høyrehåndssystemer.

Dette fører til at \( \mathbf{B} \)-feltet må være uniformt innenfor alle plan som er parallelle med \( xy \)-planet, altså \( \mathbf{B} = \mathbf{B}(z) \). Rotasjonssymmetrien fører i tillegg til at \( B_y(z) = -B_y(-z) \).

Denne oppgaven kan vi løse med Ampères lov og symmetriargumentene fra forrige deloppgave.

Først kan vi se rett ut fra Biot–Savarts lov at \( \mathbf{B} \)-feltet ikke kan ha noen komponent i \( x \)-retning. Det skyldes at alle våre strømelementer går i \( x \)-retning, og magnetfeltbidragene står normalt på strømelementene. Altså \( B_x = 0 \). Så \( z \)-komponenten. Fra Biot-Savarts lov ser vi at det å skifte retning på strømmen også skifter retningen på alle feltbidragene, og dermed hele feltet. Dersom vi hadde hatt en \( z \)-komponent ville det å snu strømmen føre til at feltet byttet mellom å peke mot og å peke vekk fra platen. Men det å snu strømmen er det samme som å rotere \( \pi/2 \) om z-aksen, noe som ikke skulle påvirket retningen på \( z \)-komponentene. Altså kan feltet ikke ha noen komponent i \( z \)-retning. Vi har dermed konkludert med $$ \begin{equation} \mathbf{B}(z) = B_y(z) \y = -B_y(-z)\y \tag{9.94} \end{equation} $$

Derfor vil vi lage oss en amperesløyfe som er et rektangel i \( yz \)-planet med \( z \in [-a ,a] \) og \( y \in [-b, b] \). Dette rektangelet vil ha to linjestykker med lengde \( 2b \) som er parallelle med \( \y \), og to linjestykker med lengde \( 2a \) som står normalt på \( \y \). Vi setter opp Ampères lov. $$ \begin{align} \oint_C \mathbf{B} \cdot \d \mathbf{l} &= \mu_0 I_\text{encl} \tag{9.95}\\ 2b (-\y) \cdot \mathbf{B}(a) + 2b\y\cdot \mathbf{B}(-a) &= \mu_0 2b \mathbf{J}_s \cdot \x \tag{9.96}\\ -4b B_y(a) &= 2b\mu_0 J_s \tag{9.97}\\ B_y(a) = \frac{-\mu_0 J_s}{2} \tag{9.98}\\ \tag{9.99} \end{align} $$ Merk at vi i integralet har valgt fortegnene på \( \y \) slik at vi integrerer slik at omløpsretninger følger av høyrehåndsregelen med tommelen i retningen til strømtettheten \( \mathbf{J}_s \).

Siden vi valgte et vilkårlig område \( (-a, a) \) for høyden av rektangelet kan vi bytte ut \( a \) med \( z \). Om vi i tillegg legger på at \( B_y(z) \y = -B_y(-z)\y \) får vi $$ \begin{equation} \mathbf{B}(z) = \frac{-\mu_0 J_s}{2}\text{sign}(z) \y \tag{9.100} \end{equation} $$

Ledningen føler bare en kraft hvor magnetfeltet virker. Et infinitesimalt segment av ledningen føler kraften \( d \mathbf{F} = I \d\mathbf{l} \times \mathbf{B} \). \( y \)-komponenten av denne kraften er gitt ved $$ \begin{equation} dF_y = |d\mathbf{F}|\sin \theta \tag{9.101} \end{equation} $$ der \( |d\mathbf{F}| = IB dl = IBR d\theta \) slik at størrelsen til den totale \( y \)-komponenten av kraften er gitt ved $$ \begin{equation} F_y = \int dF_y = IBR \int_0^{\pi} \sin \theta d\theta = 2IBR. \tag{9.102} \end{equation} $$

\( x \)-komponenten er gitt ved $$ \begin{equation} F_x = IBR \int_0^\pi \cos \theta d\theta = 0 \tag{9.103} \end{equation} $$

noe vi også kunne sett ut i fra symmetrien til problemet. Videre må kraften peke i negativ \( y \)-retning (jmf. høyrehåndsregelen). Dermed $$ \begin{equation} \mathbf{F} = -2IBR \hat{\mathbf{j}} \tag{9.104} \end{equation} $$