\( \int_{I}\vec{B} \cdot \d \vec{l} = -B_0 L_{I} \), \( \int_{II}\vec{B} \cdot \d \vec{l} = 0 \), \( \int_{III}\vec{B} \cdot \d \vec{l} = B_0 L_{III} \), \( \int_{IV}\vec{B} \cdot \d \vec{l} = 0 \), \( \oint_{C}\vec{B} \cdot \d \vec{l} = 0 \)

The field is given as \( \vec{B} = B_0 \y \) We find the integral for each of the segments. We integrate in positive direction according to the right hand rule.

Segment I: Along segment I, the line element \( \d \vec{l} \) is down along the \( y \)-axis and \( \d \vec{l} = -\d y \y \). The integral is therefore $$ \begin{equation} \int_I \vec{B} \cdot \d \vec{l} = \int_I B_0 \y \cdot (-\d y \y) = -\int_I B_0 \d y = - B_0 L_I \; , \tag{10.3} \end{equation} $$ where \( L_I \) is the length of segment I.

Segment II: Along segment II, the line element \( \d \vec{l} \) is along the \( x \)-axis and \( \d \vec{l} = -\d x \x \). The integral is therefore zero since the line elements are orthogonal to the magnetic field: $$ \begin{equation} \int_{II} \vec{B} \cdot \d \vec{l} = \int_{II} B_0 \y \cdot (-\d x \x) = 0 \; , \tag{10.4} \end{equation} $$

Segment III: Along segment III, the line element \( \d \vec{l} \) is along the \( y \)-axis and \( \d \vec{l} = \d y \y \). The integral is therefore: $$ \begin{equation} \int_{III} \vec{B} \cdot \d \vec{l} = \int_{III} B_0 \y \cdot (\d y \y) = \int_{III} B_0 \d y = B_0 L_{III} \; , \tag{10.5} \end{equation} $$ where \( L_{III} \) is the length along segment III.

Segment IV: Along segment IV, the line element \( \d \vec{l} \) is along the \( x \)-axis and \( \d \vec{l} = \d x \x \). The integral is therefore zero since the line elements are orthogonal to the magnetic field: $$ \begin{equation} \int_{IV} \vec{B} \cdot \d \vec{l} = \int_{IV} B_0 \y \cdot (\d x \x) = 0 \; , \tag{10.6} \end{equation} $$ The integral is the sum of the integrals along the segments: $$ \begin{eqnarray} \int_C \vec{B} \cdot \d \vec{l} &=& \int_{I} \vec{B} \cdot \d \vec{l} + int_{I} \vec{B} \cdot \d \vec{l} + int_{I} \vec{B} \cdot \d \vec{l} + int_{I} \vec{B} \cdot \d \vec{l} \\ &=& -B_0 L_{I} + 0 + B_0 L_{III} + 0 \end{eqnarray} $$ Now, since \( L_{I} = L_{III} \) the integral is zero.

The total integral will still be zero.

In this case, none of the integrals would be zero. Instead, the integrals would be along tilted line segments \( \d \vec{l} \). However, since the integral along segment I is in the opposite direction as along segment III and the megnatic field is the same everywhere (it is uniform), these integral would be identical in magnitude, but with opposite sign. The integrals along segments I and III would cancel and similarly for the integrals along segments II and IV. The total integral will still be zero.

First, if the amperian loop was a circle, there would be no difference if we rotated the circle by an angle \( \theta \). The results would therefore be the same for the original and the tilted circle.

Second, for a circle we could still argue that for every \( \d \vec{l} \) somewhere along the curve there would be an opposite line segment \( - \d \vec{l} \) on the other side of the circle. So if the magnetic field is zero, the contributions from \( \vec{B} \cdot \d \vec{l} \) and \( \vec{B} \cdot (-\d \vec{l}) \) would cancel and the integral along the whole, closed circle would be zero.

No. We would need to know more about the magnetic field, such as its symmetries, to be able to conclude about the magnetic field.

Uniform.

We assume that the current is uniformly distributed across the cross-sectional areas so that \( \vec{J} \) is uniform in the inner cable and in the outer cable. (But is it not necessarily the same magnitude in the inner and outer cables).

\( B_{\phi} = \frac{\mu_0 I r}{2 \pi a^2} \)

We assume that the magnetic field has cylindrical symmetry around an axis thorugh the center of the cylinder. This means that the field cannot depend on \( \phi \). Also, we assume that the system is infinitely (or very) long. We therefore do not expect the field to depend on \( z \). Since the current is in the \( z \)-direction, Biot-Savarts law implies that the magnetic field cannot have a component in the \( z \) direction. We also know that since the divergece of the magnetic field is zero, the magnetic field cannot have a component in the \( r \)-direction. This means that $$ \begin{equation} \vec{B} = B_{\phi}(r) \; . \tag{10.7} \end{equation} $$ Given the cylindrical symmetry, we choose an Amperian loop in the form of a circle around the \( z \)-axis in the \( xy \)-plane. Amperes law is then $$ \begin{equation} \int_C \vec{B} \cdot \d \vec{l} = 2 \pi r B_{\phi} = \mu_0 I(r) = \mu_0 \int_S \vec{J} \cdot \d \vec{S} \; , \tag{10.8} \end{equation} $$ where the surface \( S \) has an area \( \pi r^2 \). The current density is uniform, \( \vec{J} = I/(\pi a^2) \z \) in this region (\( r < a \)). The integral on the right is therefore \( \pi r^2 I/(\pi a^2) \) and $$ \begin{equation} \int_C \vec{B} \cdot \d \vec{l} = 2 \pi r B_{\phi} = \int_S \vec{J} \cdot \d \vec{S} = I \frac{r^2}{a^2} \; , \tag{10.9} \end{equation} $$ which gives $$ \begin{equation} B_{\phi} = \mu_0 I \frac{r^2}{2 \pi r a^2} = \frac{\mu_0 I r}{2 \pi a^2} \; . \tag{10.10} \end{equation} $$

\( B_{\phi} = \frac{\mu_0 I}{2 \pi r} \)

We use the same approach as above. Now, the current inside the Amperian loop is \( I \). The magnetic field is therefore found from Ampere's law: $$ \begin{equation} \int_C \vec{B} \cdot \d \vec{l} = 2 \pi r B_{\phi} = \mu_0 I \; , \tag{10.11} \end{equation} $$ which gives $$ \begin{equation} B_{\phi} = \frac{\mu_0 I}{2 \pi r} \; . \tag{10.12} \end{equation} $$

\( B_{\phi} = -\frac{\mu_0 I}{2 \pi r} \frac{r^2 - b^2}{c^2 - b^2} + \frac{\mu_0 I}{2 \pi r} \)

The procedure follows the same approach as above, but we now need to include all the current flowing inside the Amperian loop. This includes both the current \( I \) and the current flowing in the outer cylinder inside a radius \( r \). The area inside a radius \( r \) is \( A(r) = \pi r^2 - \pi b^2 \) and the current density in this region is \( J = -I/(\pi c^2 - \pi b^2) \). Ampere's law therefore gives: $$ \begin{equation} \int_C \vec{B} \cdot \d \vec{l} = 2 \pi r B_{\phi} = \mu_0 I(r) = -\mu_0 I \frac{\pi r^2 - \pi b^2}{\pi c^2 - \pi b^2} + \mu_0 I \; . \tag{10.13} \end{equation} $$ This gives: $$ \begin{equation} B_{\phi} = -\frac{\mu_0 I}{2 \pi r} \frac{r^2 - b^2}{c^2 - b^2} + \frac{\mu_0 I}{2 \pi r} \; . \tag{10.14} \end{equation} $$

\( B_{\phi} = 0 \)

In this case, the net current inside the Amperian loop is zero and the field is zero.

Vi har translasjonssymmetri langs \( z \)-aksen. Vi har roasjonssymmetri om \( z-aksen \). Dette fører til at feltet er uanhengig av \( z \), og at størrelsen på feltet kun avhenger av \( r \), avstanden from sylinderaksen.

Når sylinderen roterer med en ladningsfordeling på seg kan man se på det som en strøm.

Vi ønsker oss en løkke der vi har stykkevis konstante bidrag i linjeintegralet slik at vi kan kvitte oss med integraltegnet. Det kan vi oppnå for eksempel ved å velge en løkke som går parallellt med sylinderaksen, deretter radielt ut fra sylinderaksen ut til uendeligheten, deretter tilbake igjen parallellt med sylinderaksen ute i uendeligheten, og så radielt inn igjen til startpunktet.

Vi kan umiddelbart se at \( B \) ikke kan ha noen radiell komponent. Da måtte \( \nabla \cdot \mathbf{B} \neq 0 \). Derfor har vi kun e \( z \)-komponent av \( B \)-feltet. Når vi tar med oss trajslasjonssymmetrien i systemet ser vi at \( \mathbf{B} \cdot \d \mathbf{l} \) kommer til å være konstant lik \( B_z \d l \) parallellt med sylinderaksen på innsiden av sylinderskallet, og null langs alle de andre delene av ampereløkka. For veien ut og inn er det fordi feltet står normalt, og i uendeligheten er det fordi feltet blir null.

Vi begynner med å finn strømtettheten som settes opp av den roterende sylinderen: $$ \begin{equation} \mathbf{J}_l = \rho_s (\boldsymbol{\omega}\times \mathbf{r}) \tag{10.15} \end{equation} $$

Vi setter opp Ampères lov: $$ \begin{align} \oint_C \mathbf{B} \cdot \d \mathbf{l} = \int_l \mathbf{J} \cdot \d \mathbf{l} \tag{10.16} \end{align} $$

Outside the cylinder the magnetic permeability if \( \mu_0 \). Hence, \( \vec{B}_0 = \mu_0 \vec{H} \), and \( \vec{H} = \vec{B}_0/\mu_0 \).

A loop that goes up parallel to the axis inside the cylinder and down parallel to the axis outside the cylinder.

We apply Ampere's law. There are no free currents. The only contributions to the integral are from the two paths along the axis. The other two paths do not contribute, since the \( \vec{H} \)-field is directed along the axis. We get $$ \begin{equation} H \Delta l - (B_0/\mu_0) \Delta l = 0 \, \Rightarrow \, H = B_0/\mu_0 \; . \tag{10.18} \end{equation} $$

\( \vec{H} \) is the same on both sides. There is no normal component of either \( \vec{H} \) or \( \vec{B} \) on either side.

\( \vec{M} = \chi_m \vec{H} = \chi_m \vec{B}_0/\mu_0 \) where the sign depends on the sign of \( \chi_m \).

The magnetic field \( \vec{B} = \mu \vec{H} = (\mu/\mu_0) \vec{B}_0 = (1+\chi_m) \vec{B}_0 \). The results therefore depend on the sign of \( \chi_m \). For a diamagnetic material \( \chi_m < 0 \) and \( B \) is smaller than \( B_0 \). For a paramagnetic material \( \chi_m>0 \) and \( B \) is larger than \( B_0 \).

Vi vet fra Biot-Savarts lov at magnetfeltet må være normalt på strømretningen. Derfor kan ikke magnetfeltet ha en komponent i \( x \)-retningen som er strømretningen.

Fordi systemet har translasjonssymmetri i \( xy \)-planet, kan feltet ikke avhenge av \( x \) eller \( y \), slik at \( \vec{B} = \vec{B}(z) \). Dessuten er systemet identisk om vi roterer det \( \pi \) om \( x \)-aksen. Det betyr at \( B_z(z) = - B_z(-z) \) og \( B_y(z) = - B_y(-z) \).

Vi vet at fluksen av \( \vec{B} \)-feltet ut av en lukket overflate må være null. Hvis vi ser på fluksen ut av et rektangulært prisme, så må fluksen ut gjennom sidene normalt på \( x \)-aksen være null fordi feltet er null i denne retningen. Netto fluks ut av prismet gjennom flatene normalt på \( y \)-aksen vil være null, fordi feltet ikke avhenger av \( y \). Netto fluks ut gjennom flatene normalt på \( z \)-aksen vil være \( 2 B_z(z) S \) fordi feltet er symmetrisk om \( xy \)-planet. Denne fluksen må være null og dermed må \( B_z \) være null.

Vi velger en amperesløyfe fra \( -z \) til \( z \) og fra \( -y \) til \( y \) som er orientert samme vei som strømmen, altså med overflatenormal i positiv \( x \)-retning. Da blir Amperes lov: $$\oint_C \vec{B} \cdot \d \vec{l} = \mu_0 \int_S \vec{J} \cdot \d \vec{S}$$ Vi velger først at \( z \le h \) slik at hele amperesløyfen er inne i magnetfeltet. Da er $$(-2y)B_y(z)+(2y)B_y(-z) = \mu_0 (2y) \int_{-z}^{z} J_0 |z| \d z$$ Vi løser for \( B_y(z) \) når \( z>0 \) og finner at $$-2B_y(z) = \mu_0 J_0 z^2$$ Da er $$B_y = -\frac{\mu_0 J_0}{2}z^2$$ Dette gjelder for \( 0\ge z \ge h \). For negativ \( z \) har feltet motsatt retning fordi \( B_y(z) = -B_y(-z) \), slik at $$B_y = - \frac{\mu_0 J_0}{2}z^2$$. Dette kan vi skrive kompakt som $$B_y = -\frac{\mu_0 J_0}{2}|z|z$$ Men hva blir feltet utenfor platen? Vi velger den samme amperesløyfen, men nå går det kun strøm gjennom en del av overflaten, slik at $$\int_S \vec{J} \cdot \d \vec{S} = \mu_0 (2y) \int_{-h}^{h} J_0 |z| \d z = \mu_0 J_0 h^2$$ Vi setter dette inn og finner at for \( z > h \) så er $$-2B_y(z) = \mu_0 J_0 h^2$$ og dermed $$B_y = - \frac{\mu_0 J_0 h^2}{2}$$ og tilsvarende, men med motstatt fortegn for \( z < -h \).

Vi bruker Amperes lov og velger en strømløkke i en sirkel rundt linjen med radius \( r \). Da gir Amperes lov $$\oint_C \vec{B} \cdot \d \vec{l} = 2 \pi r B_{\phi} = \mu_0 I$$ Det gir at $$B_{\phi} = \frac{\mu_0 I}{2 \pi r}$$ I punktet \( (x,y,0) \) er \( r = y \) og \( \phihat = \z \) slik at $$\vec{B}(x,y,0) = \frac{\mu_0 I}{2 \pi y} \z$$

Bidraget er gitt av Biot-Savarts lov på differensial form: $$\d \vec{B} = \frac{\mu_0 }{4 \pi} \frac{I \d \vec{l} \times \vec{R}}{R^3}$$ hvor \( I \d \vec{l} \) = I \d x' \x$ og \( \vec{R} = (0,y,0) - (x',0,0) = (-x',y,0) \) slik at $$\d \vec{B} = \frac{\mu_0 }{4 \pi} \frac{I \d x' \x \times (-x',y,0)}{((x')^2 + y^2)^{3/2}} = \frac{\mu_0 }{4 \pi} \frac{I \d x' y \z }{((x')^2 + y^2)^{3/2}}$$

Vi setter inn integralet og finner $$\int \d \vec{B} = \int_{-a/2}^{a/2} \frac{\mu_0 }{4 \pi} \frac{I \d x' y \z }{((x')^2 + y^2)^{3/2}} = \frac{I \mu_0 y}{4 \pi} \left[ \frac{x'}{y^2\sqrt{y^2 + (x')^2}} \right]_{x'=-a/2}^{x'=a/2} = \frac{I \mu_0}{2 \pi y} \frac{1}{\sqrt{1 + \frac{y^2}{(a/2)^2}}}$$

Gauss' law: $$ \begin{equation} \oint_S \vec{D}\cdot d\vec{S} = Q_{\text{free}} \; . \tag{10.19} \end{equation} $$ Ampere's law (in the stationary case): $$ \begin{equation} \oint_C \vec{H} \cdot d\vec{l} = \int_S \vec{J}_{\text{free}} \cdot d\vec{S} \; . \tag{10.20} \end{equation} $$

\( Q = v \rho = \pi a^2 L \rho \), \( \vec{D} = \frac{\rho a^2}{2 r} \rhat \)

We notice that we have rotation and translation symmetry, therefore we can use cylindrical coordinates. We do not expect the field to have a \( \phi \)-dependence due to the rotational symmetry or a \( z \)-dependence due to the translational symmetry.

What about the components? We do not expect the field to have a \( z \)-component because we can rotate the system \( 180^{\circ} \) and get the same system back. The \( \vec{D} \)-field has the same symmetry as the electric field \( \vec{E} \). We do not exect the \( \vec{E} \)-field to have a \( \phi \)-component because \( \oint \vec{E} \cdot \d \vec{l} = 0 \). We therefore expect \( \vec{E} = E_r(r)\rhat \) and therefore \( \vec{D} = D_r(r) \rhat \).

We will use Gauss' law on a cylinder with radius \( r \). What is the free charge inside such a cylinder? If \( r > a \), the charge inside is the volume of the cylinder times the volume charge density, which is \( Q = v \rho = \pi a^2 L \rho \). If \( r < a \), the charge inside the volume of the cylinder times the volume charge density, which in this case is \( Q = v \rho = \pi r^2 L \rho \).

Now, we apply Gauss' law to a cylinder of radius \( r \) where \( r > a \): $$ \begin{equation} \int_S \vec{D} \cdot \d \vec{S} = Q_{\text{free}} = \pi a^2 L \rho \; . \tag{10.21} \end{equation} $$ There is no flux into or out of the top and bottom planar surfaces, because the field has no \( z \)-component. For the curved cylinder syrface \( \vec{D} \) points in the same direction as the surface normal and \( \vec{D} \cdot \rhat \) is constant in the integral because \( \vec{D} \) only depends on \( r \): $$ \begin{equation} \int_S \vec{D} \cdot \d \vec{S} = D_r(r) 2 \pi r L = \pi a^2 L \rho \; , \tag{10.22} \end{equation} $$ which gives $$ \begin{equation} D_r = \frac{\rho a^2}{2r} \tag{10.23} \end{equation} $$ and $$ \begin{equation} \vec{D} = \frac{\rho a^2}{2 r} \rhat \; . \tag{10.24} \end{equation} $$

We have the same symmetries as the previous exercise, but because the magnetic field can't have divergence we know it will point in \( \phi \)-direction. The free current is given by the current density, and we get the field: $$ \begin{equation*} \vec{H}(r) = \frac{1}{2}\frac{a^2}{r}|\vec{J}|phihat \end{equation*} $$

The arguments are the same as in the part (b), but we now use the charge inside a cylinder of radius \( r \), where \( r < a \): \( Q = v \rho = \pi r^2 L \rho \). We apply Gauss' law and get: $$ \begin{equation} \int_S \vec{D} \cdot \d \vec{S} = D_r(r) 2 \pi r L = \pi r^2 L \rho \; , \tag{10.25} \end{equation} $$ which gives $$ \begin{equation} D_r = \frac{\rho r}{2} \tag{10.26} \end{equation} $$ and $$ \begin{equation} \vec{D} = \frac{\rho r}{2} \rhat \; . \tag{10.27} \end{equation} $$

Symmetrien i systemet tilsier at feltet må være \( \mathbf{B} = B(r) \hat{\boldsymbol{\phi}} \). Det kommer av rotasjonssymmetri om sylinderens akse, og translasjonssymmetri langs sylinderens akse, samt at \( \nabla \cdot \mathbf{B} = 0 \). Vi legger derfor en Ampèresløyfe som er en sirkel normalt på sylinderaksen. $$ \begin{align} \oint_C \mathbf{B} \cdot \d \mathbf{l} = \mu_0 \int_S \mathbf{J} \cdot \d \mathbf{S} \tag{10.28}\\  2\pi r B(r) = \frac{\mu_0 Ir^2}{a^2} \tag{10.29}\\ B(r) = \frac{\mu_0 Ir}{2\pi a^2} \tag{10.30}\\ \mathbf{B}(r) = \frac{\mu_0Ir}{2\pi a^2}\hat{\boldsymbol{\phi}} \tag{10.31} \end{align} $$ for \( r \in [0, a] \).

For \( r > a \) har vi en fast innesluttet strøm \( I \), slik at \( B \)-feltet blir \( \mathbf{B}(r) = \frac{\mu_0 I}{2\pi r}\hat{\boldsymbol{\phi}} \)

Vi bruker samme Ampèresløyfe som i forrige oppgave. Andelen strøm som er innesluttet i en avstand \( r \in [b, c] \) er \( \frac{\pi (r^2-b^2)}{\pi (c^2 - b^2)} = \frac{r^2-b^2}{c^2 - b^2} \), slik at \( I_{encl} = I\frac{r^2-b^2}{c^2 - b^2} \). $$ \begin{align} \oint_C \mathbf{B} \cdot \d \mathbf{l} = \mu_0 \int_S \mathbf{J} \cdot \d \mathbf{S} \tag{10.32}\\  2\pi r B(r) = \mu_0 I\frac{r^2-b^2}{c^2 - b^2} \tag{10.33}\\  \mathbf{B}(r) = \mu_0 I\frac{r^2-b^2}{2\pi r(c^2 - b^2)} \hat{\boldsymbol{\phi}} \tag{10.34} \end{align} $$ For \( r > c \) har vi fast innesluttet strøm \( I \), slik at \( \mathbf{B}(r) = \frac{\mu_0 I}{2\pi r}\hat{\boldsymbol{\phi}} \). For \( r < b \) har vi ingen innesluttet strøm. Når vi legger på det vi vet om symmetriene i systemet vet vi da at feltet der inne på være \( \mathbf{0} \).

Vi har funnet feltet fra hver enkelt av lederene i koaksialkabelen i de forrige deloppgavene. Nå trenger vi kun å sette dem sammen. Vi setter strømretningen i den indre lederen til å være positiv. Om vi kaller feltet i oppgave (a) for \( \mathbf{B}_{indre} \) og feltet fra (b) for \( \mathbf{B}_{ytre} \), så skal det totale feltet, ved superposisjon, være $$ \begin{equation} \mathbf{B}(r) = \mathbf{B}_{indre} - \mathbf{B}_{indre} \tag{10.35} \end{equation} $$ der det negative fortegnet kommer av at det er motsatt strømretning i denne oppgaven sammenlignet med i oppgave b. Vi deler dette opp i 4 regioner: \( r < a \), \( r \in (a, b) \) og \( r \in [b, c] \) og \( r > c \): $$ \begin{align} \mathbf{B}(r) = \left\{ \begin{array}{cc} \frac{\mu_0 I r}{2\pi a^2} \hat{\boldsymbol{\phi}} & \text{ for } r < a \\ \frac{\mu_0 I}{2\pi r} \hat{\boldsymbol{\phi}} & \text{ for } r\in (a, b) \\ I\frac{r^2-b^2}{2\pi(c^2 - b^2)} \hat{\boldsymbol{\phi}} & \text{ for } r\in(b, c) \\ \mathbf{0} & \text{ for } r > c \end{array} \right. \tag{10.36} \end{align}$$

1. Figur IV utelukkes pga brutte feltlinjer, altså at den ikke overholder \( \nabla \cdot \mathbf{B} = 0 \)
2. Figur V utelukkes fordi den viser et asimut symmetrisk felt fra en asymmetrisk strømfordeling.
3. Figur I og VI utelukkes fordi ytterlederen ikke skal påvirke feltet på innsiden. (Kan vises med symmetrieargumenter, slik som i (b))
4. Figur III utelukkes fordi det ikke er noe tegn til at ytterlederen påvirker feltet.

Bruk Ampères lov. Symmetriargumentet for en solenoide kan du ta for gitt, dvs. du kan anta at feltet er i \( \z \)-retning (langs aksen), og at det er neglisjerbart utenfor solenoiden.

Fluksen er den samme gjennom alle deler av kretsen. La oss indekseree størrelsene med \( i \) for å representere delkrets \( i \). Da har vi \( \Phi_m = B_1S_1 = B_2S_2 = \cdots = B_nS_n \), så \( \Phi = B_iS_i \) for \( i = 1, 2, ..., n \). Amperes lov gir oss $$ \begin{equation} \oint_C \boldsymbol{H}\cdot d\boldsymbol{l} = NI, \tag{10.39} \end{equation} $$ og venstre siden kan omskrives slik: $$ \begin{equation} \begin{split} \oint_C \boldsymbol{H}\cdot d\boldsymbol{l} & = \sum_i \int_{C_i}\boldsymbol{H}\cdot d\boldsymbol{l} = \sum_i H_i l_i \\ & = \sum_i \frac{l_i}{\mu_i}B_i = \frac{l_i}{\mu_i S_i}\underbrace{B_i S_i}_{=\Phi_m} = \Phi_m\sum_i\underbrace{\frac{l_i}{\mu_i S_i}}_{R_{mi}}. \end{split} \tag{10.40} \end{equation} $$ Setter vi dette inn i Amperes lov får vi $$ \begin{equation} \Phi_m\sum_iR_{mi} = NI, \tag{10.41} \end{equation} $$ som medfører $$ \begin{equation} \Phi_m=\frac{NI}{\sum_iR_{mi} }, \tag{10.42} \end{equation} $$ som var det vi skulle vise.

\( 2.36 \cdot 10^{-8} \) Wb

7% feil.