245 Induktion

Needs["ErrorBarPlots`"]

Vorversuch

a) Das Oszilloskop schlägt aus. Wird die Spule umgedreht, ist dieser Ausschlag entgegengesetzt. 

b) Ebenso.

Lässt man einen Magneten durch eine Spule fallen, so schlägt die Spannung je nach Orientierung der Spule entweder erst ins Positive, dann ins Negative oder andersherum aus. Die Amplitude des Ausschlags ist abhängig von der Fallhöhe und dem Magnetenmaterial (Aua).

Induktionsgesetz

Vorgegebene Größen

Fläche der Induktionsspule:

(Subscript[A, ind])=Quantity[41.7,(("Centimeters")^2)]//UnitConvert

Windungszahl der Induktionsspule

(Subscript[N, ind])=4000

Radius der Helmholtzspulen

(Subscript[R, helmholtz])=Quantity[295/2,"Millimeters"]//N//UnitConvert

Abstand der Helmholtzspulen

(Subscript[D, helmholtz])=Quantity[147,"Millimeters"]//N//UnitConvert

Windungszahl der Helmholtzspulen

(Subscript[N, helmholtz])=124

Messung

Messung des Spulenwiderstandes

{(Subscript[R, Spule]),(Subscript[ΔR, Spule])}=Quantity[{2.3,0.1},"Ohms"]; (Subscript[R, Spule])±(Subscript[ΔR, Spule])

Der maximale Strom darf 5 A nicht überschreiten. Also:

(Subscript[I, max])=Quantity[5,"Amperes"]

{(Subscript[U, max]),(Subscript[ΔU, max])}=UnitConvert[{(Subscript[R, Spule]),(Subscript[ΔR, Spule])}*(Subscript[I, max]),"Volts"]

Also stellen wir nicht mehr als 11 Volt ein.

Wir messen die Scheitelspannung in Abhängigkeit von der Drehfrequenz.

{(Subscript[U, real]),(Subscript[ΔU, real])}=Quantity[{11.8,0.1},"Volts"]

{(Subscript[I, real]),(Subscript[ΔI, real])}=Quantity[{4.6,0.1},"Amperes"]

(Subscript[data, 1])={
{Quantity[{3,.2},"Hertz"],(1/2)Quantity[{2.24,0.03},"Volts"]},
{Quantity[{6.1,.2},"Hertz"],(1/2)Quantity[{4.82,0.05},"Volts"]},
{Quantity[{9.1,.2},"Hertz"],(1/2)Quantity[{7.2,0.2},"Volts"]},
{Quantity[{11.9,.1},"Hertz"],(1/2)Quantity[{9.6,0.1},"Volts"]},
{Quantity[{14.9,.1},"Hertz"],(1/2)Quantity[{11.9,0.2},"Volts"]}
}

(Subscript[f, 2])=Quantity[{10,.2},"Hertz"]

(Subscript[data, 2])={
{Quantity[{.5,.1},"Amperes"],(1/2)Quantity[{1.05,0.05},"Volts"]},
{Quantity[{1,.1},"Amperes"],(1/2)Quantity[{1.89,.03},"Volts"]},
{Quantity[{1.5,.1},"Amperes"],(1/2)Quantity[{2.75,0.05},"Volts"]},
{Quantity[{2,.1},"Amperes"],(1/2)Quantity[{3.45,0.05},"Volts"]},
{Quantity[{2.5,.1},"Amperes"],(1/2)Quantity[{4.38,0.03},"Volts"]},
{Quantity[{3,.1},"Amperes"],(1/2)Quantity[{5.15,0.05},"Volts"]},
{Quantity[{3.5,.1},"Amperes"],(1/2)Quantity[{5.9,0.05},"Volts"]},
{Quantity[{4,.1},"Amperes"],(1/2)Quantity[{6.65,0.05},"Volts"]},
{Quantity[{4.5,.1},"Amperes"],(1/2)Quantity[{7.78,0.04},"Volts"]}
}

Auswertung

Zunächst bringen wir die gemessenen Daten in ein Format, mit dem wir die computergestützte Auswertung vornehmen können:

(Subscript[freqs, 1])=Transpose[(Subscript[data, 1])][[1,All,1]]//QuantityMagnitude;

(Subscript[Δfreqs, 1])=Transpose[(Subscript[data, 1])][[1,All,2]]//QuantityMagnitude;

(Subscript[voltages, 1])=Transpose[(Subscript[data, 1])][[2,All,1]]//QuantityMagnitude;

(Subscript[Δvoltages, 1])=Transpose[(Subscript[data, 1])][[2,All,2]]//QuantityMagnitude;

(Subscript[data2, 1])=Table[{{(Subscript[freqs, 1])[[i]],(Subscript[voltages, 1])[[i]]},ErrorBar[(Subscript[Δfreqs, 1])[[i]],(Subscript[Δvoltages, 1])[[i]]]},{i,1,Length[(Subscript[freqs, 1])]}];

Wir führen einen linearen Fit durch, der mit den Inversen der Meßfehler gewichtet wird.

(Subscript[plot, freqs])=ErrorListPlot[(Subscript[data2, 1]),PlotRange->{{0,16},{0,6}}];

(Subscript[fit, freqs])=LinearModelFit[Transpose[{(Subscript[freqs, 1]),(Subscript[voltages, 1])}],x,x,VarianceEstimatorFunction->(1&),Weights->1/(((Subscript[Δfreqs, 1])^2)+((Subscript[Δvoltages, 1])^2))]

Show[(Subscript[plot, freqs]),Plot[(Subscript[fit, freqs])[x],{x,0,16}],AxesLabel->{"f [Hz]","U [V]"},PlotLabel->"Abbildung 1.1: Gemessene Induktionsspannung als Funktion der Drehfrequenz",ImageSize->Full]

Die Steigung ergibt sich aus dem Fit mit Fehler als:

{(Subscript[m, freqs]),(Subscript[Δm, freqs])}=Quantity[(Subscript[fit, freqs])["ParameterTableEntries"][[2,1;;2]],"Volts"/"Hertz"]

Hieraus können wir einen Meßwert für das Magnetfeld B der Helmholtzspule ermitteln (vgl. Einleitung):

{(Subscript[B, helmholtz]),(Subscript[ΔB, helmholtz])}=(({(Subscript[m, freqs]),(Subscript[Δm, freqs])})/(2π (Subscript[A, ind])(Subscript[N, ind])))//UnitConvert[#1,"Milliteslas"]&;NumberForm[(Subscript[B, helmholtz]),2]±NumberForm[(Subscript[ΔB, helmholtz]),1]

Wir vergleichen diesen Wert mit der theoretisch zu erwartenden Feldstärke der Helmholtzspulen:

{(Subscript[B, helmholtz,theor]),(Subscript[ΔB, helmholtz,theor])}=((Quantity["MagneticConstant"])/2)((8 {(Subscript[I, real]),(Subscript[ΔI, real])} ((Subscript[N, helmholtz])+(Subscript[N, helmholtz])))/((Sqrt[125])(Subscript[R, helmholtz])))//UnitConvert[#,"Milliteslas"]&;NumberForm[(Subscript[B, helmholtz,theor]),3]±NumberForm[(Subscript[ΔB, helmholtz,theor]),1]

Für eine Einordnung und einen Vergleich dieser beiden Werte vgl. die Diskussion unten. (Handschriftlich)

Dieselben Schritte bis zur Erstellung der Graphik und des Fits führen wir auch für die Meßdaten aus der zweiten Meßreihe, wo wir die Spannung in Abhängigkeit der Stromstärke in der Helmholtzspule, also effektiv als Funktion der äußeren Magnetfeldstärke ermittelt haben.

(Subscript[currents, 2])=Transpose[(Subscript[data, 2])][[1,All,1]]//QuantityMagnitude;

(Subscript[Δcurrents, 2])=Transpose[(Subscript[data, 2])][[1,All,2]]//QuantityMagnitude;

(Subscript[voltages, 2])=Transpose[(Subscript[data, 2])][[2,All,1]]//QuantityMagnitude;

(Subscript[Δvoltages, 2])=Transpose[(Subscript[data, 2])][[2,All,2]]//QuantityMagnitude;

(Subscript[data2, 2])=Table[{{(Subscript[currents, 2])[[i]],(Subscript[voltages, 2])[[i]]},ErrorBar[(Subscript[Δcurrents, 2])[[i]],(Subscript[Δcurrents, 2])[[i]]]},{i,1,Length[(Subscript[currents, 2])]}];

(Subscript[plot, currents])=ErrorListPlot[(Subscript[data2, 2]),PlotRange->{{0,5},{0,4.2}}];

(Subscript[fit, currents])=LinearModelFit[Transpose[{(Subscript[currents, 2]),(Subscript[voltages, 2])}],x,x,VarianceEstimatorFunction->(1&),Weights->1/(((Subscript[Δcurrents, 2])^2)+((Subscript[Δvoltages, 2])^2))]

Show[(Subscript[plot, currents]),Plot[(Subscript[fit, currents])[x],{x,0,5}],AxesLabel->{"I [A]","U [V]"},PlotLabel->"Abbildung 1.2: Abhängigkeit der Spannung von der Stromstärke (und damit von der B-Feldstärke)",ImageSize->Full]

Induktionsspannung bei periodischem Feldstrom

Messung

Zunächst messen wir die Abhängigkeit der induzierten Spannung vom Drehwinkel der Induktionsspule.

(Subscript[Ω, 1]) stellt die initiale konstante Drehfrequenz dar.

(Subscript[Ω, 1])=Quantity[{103,1},"Hertz"]

(Subscript[data, 3])={
{{0,2}°,(1/2)Quantity[{8,0.1},"Volts"]},
{{30,2}°,(1/2)Quantity[{7.1,.1},"Volts"]},
{{60,2}°,(1/2)Quantity[{4.2,0.1},"Volts"]},
{{90,2}°,(1/2)Quantity[{.06,0.01},"Volts"]},
{{120,2}°,(1/2)Quantity[{4,0.1},"Volts"]},
{{150,2}°,(1/2)Quantity[{6.9,0.1},"Volts"]},
{{180,2}°,(1/2)Quantity[{8,.1},"Volts"]}
}

Die nächste Messung nimmt die induzierte Spannung sowie den Strom und die Spannung in der Helmholtzspule als Funktion der Frequenz auf.

(Der erste Spanungswert ist der induzierte, der zweite der in der Helmholtzspule.)

(Subscript[data, 4])={
{Quantity[{20,.1},"Hertz"],(1/2)Quantity[{5.82,0.1},"Volts"],Quantity[{1.18,0.001},"Amperes"],(1/2)Quantity[{11.4,0.8},"Volts"]},
{Quantity[{40,.1},"Hertz"],(1/2)Quantity[{7.2,0.1},"Volts"],Quantity[{.755,.001},"Amperes"],(1/2)Quantity[{11.6,0.6},"Volts"]},
{Quantity[{60.2,.1},"Hertz"],(1/2)Quantity[{7.7,0.1},"Volts"],Quantity[{.538,.001},"Amperes"],(1/2)Quantity[{11.7,0.7},"Volts"]},
{Quantity[{80.6,1},"Hertz"],(1/2)Quantity[{7.9,0.1},"Volts"],Quantity[{.411,.001},"Amperes"],(1/2)Quantity[{11.7,0.3},"Volts"]},
{Quantity[{100,.1},"Hertz"],(1/2)Quantity[{7.9,0.1},"Volts"],Quantity[{.337,.001},"Amperes"],(1/2)Quantity[{11.7,0.3},"Volts"]},
{Quantity[{119,.1},"Hertz"],(1/2)Quantity[{8,0.1},"Volts"],Quantity[{.283,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{141,.1},"Hertz"],(1/2)Quantity[{8,0.1},"Volts"],Quantity[{.240,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{160,.1},"Hertz"],(1/2)Quantity[{8,0.1},"Volts"],Quantity[{0.210,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{180,.1},"Hertz"],(1/2)Quantity[{8,0.1},"Volts"],Quantity[{.188,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{199,.1},"Hertz"],(1/2)Quantity[{8.1,0.1},"Volts"],Quantity[{.169,.001},"Amperes"],(1/2)Quantity[{11.9,0.1},"Volts"]},
{Quantity[{400,.1},"Hertz"],(1/2)Quantity[{8.1,0.1},"Volts"],Quantity[{.083,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{604,.2},"Hertz"],(1/2)Quantity[{8.1,0.1},"Volts"],Quantity[{.053,.001},"Amperes"],(1/2)Quantity[{11.9,0.2},"Volts"]},
{Quantity[{803,.3},"Hertz"],(1/2)Quantity[{8.1,0.1},"Volts"],Quantity[{.039,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{1000,10},"Hertz"],(1/2)Quantity[{8.1,0.1},"Volts"],Quantity[{.03,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{1200,10},"Hertz"],(1/2)Quantity[{8.1,0.1},"Volts"],Quantity[{.024,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{1390,10},"Hertz"],(1/2)Quantity[{8.1,0.1},"Volts"],Quantity[{.020,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{1590,10},"Hertz"],(1/2)Quantity[{8.2,0.1},"Volts"],Quantity[{.016,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{1810,10},"Hertz"],(1/2)Quantity[{8.4,0.1},"Volts"],Quantity[{.012,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]},
{Quantity[{1990,10},"Hertz"],(1/2)Quantity[{8.5,0.1},"Volts"],Quantity[{.010,.001},"Amperes"],(1/2)Quantity[{11.8,0.2},"Volts"]}
}

Wir überzeugen uns mittels des Oszilloskops qualitativ von der Schwebung. Warum kommt es dazu? Dazu später mehr!

Auswertung

Zunächst bringen wir wieder die Meßdaten in eine von Mathematica verwendbare Form:

(Subscript[angles, 3])=Transpose[(Subscript[data, 3])][[1,All,1]]//QuantityMagnitude;

(Subscript[Δangles, 3])=Transpose[(Subscript[data, 3])][[1,All,2]]//QuantityMagnitude;

(Subscript[voltages, 3])=Transpose[(Subscript[data, 3])][[2,All,1]]//QuantityMagnitude;

(Subscript[Δvoltages, 3])=Transpose[(Subscript[data, 3])][[2,All,2]]//QuantityMagnitude;

(Subscript[data2, 3])=Table[{{(Subscript[angles, 3])[[i]],(Subscript[voltages, 3])[[i]]},ErrorBar[(Subscript[Δangles, 3])[[i]],(Subscript[Δvoltages, 3])[[i]]]},{i,1,Length[(Subscript[angles, 3])]}];

(Subscript[plot, angles])=ErrorListPlot[(Subscript[data2, 3]),PlotRange->{{0,π+.1},{0,4.2}},AxesLabel->{"α [rad]","U [V]"}, PlotLabel->"Abbildung 1.3: Winkelabhängigkeit der Induktionsspannung",ImageSize->Full]

Nun müssen wir auch den zweiten Datensatz präparieren.

(Subscript[freqs, 4])=Transpose[(Subscript[data, 4])][[1,All,1]]//QuantityMagnitude;

(Subscript[Δfreqs, 4])=Transpose[(Subscript[data, 4])][[1,All,2]]//QuantityMagnitude;

(Subscript[uind, 4])=Transpose[(Subscript[data, 4])][[2,All,1]]//QuantityMagnitude;

(Subscript[Δuind, 4])=Transpose[(Subscript[data, 4])][[2,All,2]]//QuantityMagnitude;

(Subscript[currents, 4])=Transpose[(Subscript[data, 4])][[3,All,1]]//QuantityMagnitude;

(Subscript[Δcurrents, 4])=Transpose[(Subscript[data, 4])][[3,All,2]]//QuantityMagnitude;

(Subscript[uhelm, 4])=Transpose[(Subscript[data, 4])][[4,All,1]]//QuantityMagnitude;

(Subscript[Δuhelm, 4])=Transpose[(Subscript[data, 4])][[4,All,2]]//QuantityMagnitude;

In diesem Schritt rechnen wir ((Subscript[U, ind])/(Subscript[U, helmholtz])) mit Fehlerfortpflanzung aus. Die Formel für den Fehler ergibt sich einfach aus der Gauß’schen Fehlerfortpflanzung.

(Subscript[uratio, 4])=((Subscript[uind, 4])/(Subscript[uhelm, 4]));

(Subscript[Δuratio, 4])=Sqrt[(((((Subscript[Δuind, 4])/(Subscript[uhelm, 4]))))^2)+(((-(((Subscript[uind, 4]) (Subscript[Δuhelm, 4]))/((Subscript[uhelm, 4])^2))))^2)];

Weiter rechnen wir die Widerstände ((Subscript[U, helmholtz])/I) aus und wenden wieder die Fehlerfortpflanzung an.

(Subscript[res, 4])=((Subscript[uhelm, 4])/(Subscript[currents, 4]));

(Subscript[Δres, 4])=Sqrt[(((((Subscript[Δuhelm, 4])/(Subscript[currents, 4]))))^2)+(((-(((Subscript[uhelm, 4]) (Subscript[Δcurrents, 4]))/((Subscript[currents, 4])^2))))^2)];

(Subscript[data2, 4])=Table[{{(Subscript[freqs, 4])[[i]],(Subscript[uratio, 4])[[i]]},ErrorBar[(Subscript[Δfreqs, 4])[[i]],(Subscript[Δuratio, 4])[[i]]]},{i,1,Length[(Subscript[freqs, 4])]}];

(Subscript[data3, 4])=Table[{{(Subscript[freqs, 4])[[i]],(Subscript[res, 4])[[i]]},ErrorBar[(Subscript[Δfreqs, 4])[[i]],(Subscript[Δres, 4])[[i]]]},{i,1,Length[(Subscript[freqs, 4])]}];

(Subscript[plot, quotients,1])=ErrorListPlot[(Subscript[data2, 4]),PlotRange->{{0,2000},{0,1}},AxesLabel->{"f [Hz]","((Subscript[U, ind])/(Subscript[U, helmholtz]))"},PlotLabel->"Abbildung 1.4: Frequenz(un)abhängigkeit des Spannungsverhältnisses",ImageSize->Full]

(Subscript[plot, quotients,2])=ErrorListPlot[(Subscript[data2, 4]),PlotRange->{{0,101},{0,1}},AxesLabel->{"f [Hz]","((Subscript[U, ind])/(Subscript[U, helmholtz]))"},PlotLabel->"Abbildung 1.5: Frequenzabhängigkeit des Spannungsverhältnisses bei f ≤ 100 Hz",ImageSize->Full]

(Subscript[plot, res])=ErrorListPlot[(Subscript[data3, 4]),PlotRange->{{0,2012},{0,700}},AxesLabel->{"f [Hz]","R [Ω]"},PlotLabel->"Abbildung 1.6: Widerstand der Helmholtzspule",ImageSize->Full];

(Subscript[fit, res])=LinearModelFit[Transpose[{(Subscript[freqs, 4]),(Subscript[res, 4])}],x,x,VarianceEstimatorFunction->(1&),Weights->1/((Subscript[Δfreqs, 4])+(Subscript[Δres, 4]))]

Show[(Subscript[plot, res]),Plot[(Subscript[fit, res])[x],{x,0,2012}]]

Offensichtlich erachtet die Fit-Funktion, die Datenpunkte mit dem Inversen der Summe ihrer Fehler gewichtet, die letzten Punkte mit den großen Fehlern für nicht fittenswert.

Die Steigung des Fits beträgt:

{(Subscript[m, res]),(Subscript[Δm, res])}=Quantity[(Subscript[fit, res])["ParameterTableEntries"][[2,1;;2]],"Volts"/("Amperes"*"Hertz")]

Wir erhalten also für die Induktivität:

{(Subscript[L, helmholtz]),(Subscript[ΔL, helmholtz])}=(({(Subscript[m, res]),(Subscript[Δm, res])})/(2π))//UnitConvert[#,"Millihenries"]&;NumberForm[(Subscript[L, helmholtz]),{3,1}]±NumberForm[(Subscript[ΔL, helmholtz]),{1,1}]

Bestimmung des Erdmagntfeldes durch Kompensation

Messung

Frequenz ohne Kompensation:

{(Subscript[f, ohne]),(Subscript[Δf, ohne])}=Quantity[{15.2,0.2},"Hertz"]

Induktionsspannung ohne Kompensation ((1/2)-mal Spitze-Spitze):

{(Subscript[U, ohne]),(Subscript[ΔU, ohne])}=(1/2)Quantity[{158,5},"Millivolts"]//N//UnitConvert[#,"Volts"]&

Stromstärke des Kompensationsstroms:

{(Subscript[I, comp]),(Subscript[ΔI, comp])}=Quantity[{-0.061,0.001},"Amperes"]

Induktionsspannung mit Kompensation ((1/2)-mal Spitze-Spitze):

{(Subscript[U, mit]),(Subscript[ΔU, mit])}=(1/2)Quantity[{30,2},"Millivolts"]//N//UnitConvert[#,"Volts"]&

Drehfrequenz nochmals aufgenommen bei der Messung mit Kompensation:

{(Subscript[f, mit]),(Subscript[Δf, mit])}=Quantity[{15.0,0.5},"Hertz"]

Auswertung

Ohne Kompensation

Wir ermitteln wieder das Verhältnis U/f und den Fehler darauf:

q=((Subscript[U, ohne])/(Subscript[f, ohne]))

Δq=Sqrt[(((((Subscript[ΔU, ohne])/(Subscript[f, ohne]))))^2)+((((((Subscript[U, ohne]) (Subscript[Δf, ohne]))/((Subscript[f, ohne])^2))))^2)]

Anschließend nutzen wir die Formel wie oben:

{(Subscript[B, abs]),(Subscript[ΔB, abs])}=(({q,Δq})/(2π (Subscript[A, ind])(Subscript[N, ind])))//UnitConvert[#1,"Milliteslas"]&;NumberForm[(Subscript[B, abs]),{2,3}]±NumberForm[(Subscript[ΔB, abs]),{1,3}]

Mit Kompensation

Wir ermitteln die Vertikalkomponente aus dem gemessenen Kompensationsstrom: Ihr Betrag muß so groß sein wie der des Helmholtz-Feldes.

{(Subscript[B, vert]),(Subscript[ΔB, vert])}=((Quantity["MagneticConstant"])/2)((8 {(Subscript[I, comp]),(Subscript[ΔI, comp])} ((Subscript[N, helmholtz])+(Subscript[N, helmholtz])))/((Sqrt[125])(Subscript[R, helmholtz])))//Abs//UnitConvert[#,"Milliteslas"]&;NumberForm[(Subscript[B, vert]),{3,4}]±NumberForm[(Subscript[ΔB, vert]),{1,4}]

Die Horizontalkomponente ergibt sich wieder wie oben:

q2=((Subscript[U, mit])/(Subscript[f, mit]))

Δq2=Sqrt[(((((Subscript[ΔU, mit])/(Subscript[f, mit]))))^2)+((((((Subscript[U, mit]) (Subscript[Δf, mit]))/((Subscript[f, mit])^2))))^2)]

{(Subscript[B, hor]),(Subscript[ΔB, hor])}=(({q2,Δq2})/(2π (Subscript[A, ind])(Subscript[N, ind])))//UnitConvert[#1,"Milliteslas"]&;NumberForm[(Subscript[B, hor]),{2,3}]±NumberForm[(Subscript[ΔB, hor]),{1,3}]

Damit ergibt sich der Inklinationswinkel in Radian zu:

α=ArcTan[((Subscript[B, vert])/(Subscript[B, hor]))]

Δα=Sqrt[((((((Subscript[B, hor])(Subscript[ΔB, vert]))/(((Subscript[B, hor])^2)+((Subscript[B, vert])^2)))))^2)+((((((Subscript[ΔB, hor])(Subscript[B, vert]))/(((Subscript[B, hor])^2)+((Subscript[B, vert])^2)))))^2)]

Das entspricht in Grad:

NumberForm[(α/(2π))*360,3]±NumberForm[((Δα)/(2π))*360,1]

Der Literaturwert für Heidelberg beträgt indes etwa 65° (vgl. ftp://ftp.ngdc.noaa.gov/geomag/images/I_map_mf_2005_large.jpeg, 30. 4. 2012, 00:45).