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To my surprise I couldn't find any estimate on how many amperes of electrical current are flowing in total in the Earth's core to be able to create the magnetic field strength $5\cdot 10^{-5}$ T on it's surface.

How can I estimate this electrical current?
How would the current be spatially distributed?

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3 Answers 3

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We can do an estimate using Ampere's law $$ \oint_A{\vec B}\cdot d{\vec l}~=~\int_A\vec j\cdot d\vec a. $$ I am going to ball park numbers here. We think of the current along a wire at the outer core of the Earth with the wire representing an average current. The wire then has a radius between $1220$ km $3400$ km for the radius of the outer core. We set this at about $2000$ km. Now assume the magnetic field thorugh this loop is constant or we are concerned with the average so that $$ 2\pi|B|R~=~\pi|j|R^2, $$ so the current density has magnitude $|J|~\simeq~2|B|/R$. NOw consider the current density as distributed in the inner core, where I think the currents are. This is a volume $V~=~\frac{4\pi}{3}(R_1^3~-~R_2^3)$ with $I~\simeq~|J|V$. Now put all this together and estimate $B~=~10^{-4}T$ and $V~\simeq~3.7\times 10^{18}m^3$ we then have as an approximation $$ I~\simeq~2\times (10^{-4}T)(3.7\times 10^{18}m^3)/2\times 10^{6}m~=~3.7\times 10^8amps. $$

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  • $\begingroup$ This I kind of estimate I was looking for. I have the feeling that the value of $10^8$ could be even (much) larger considering the Taylor columns? $\endgroup$
    – Gerard
    Jul 3, 2017 at 8:04
  • $\begingroup$ Is $B$ in this calculation is the field strength at the position of the core itself, or at Earth's surface? According to news.berkeley.edu/2010/12/16/earth-magnetic-field one should perhaps use $B=25$T giving $10^{16} amps$ .... $\endgroup$
    – Gerard
    Jul 3, 2017 at 14:30
  • $\begingroup$ This is $25$ or $50$ Gauss, which means $5\times 10^{-3}T$. That would jump my figure to $10^9$ to $10^{10}$ amps. $\endgroup$ Jul 3, 2017 at 21:25
  • $\begingroup$ Of course ..... In another article I found experts talking about current densities 0.04 A/m^2 or 0.004 A/m^2. I believe this does not contradict your estimate. $\endgroup$
    – Gerard
    Jul 4, 2017 at 7:18
  • $\begingroup$ The density is per area. I think this reflects the hypothesis that the current is at the outer and inner core boundary. That area would be about $2\times 10^{13}m^2$ which suggests $10^{10}$ 10 $10^{11}$amps, which a bit more. I do think we have narrowed in on the order of magnitude estimate. $\endgroup$ Jul 4, 2017 at 11:27
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According to the dynamo theory,

[...] the magnetic field is induced and constantly maintained by the convection of liquid iron in the outer core. A requirement for the induction of field is a rotating fluid. Rotation in the outer core is supplied by the Coriolis effect caused by the rotation of the Earth. The Coriolis force tends to organize fluid motions and electric currents into columns (see also Taylor column) aligned with the rotation axis.

To compute the magnetic field, or the electric current density, one should then be able to solve the extremely complex nonlinear magnetohydrodynamic (MHD) equations for an electrically conducting fluid undergoing thermal convection in a rapidly rotating spherical shell.

This is quite an ambitious goal: it is because of this that you couldn't find any estimate of the electric current flowing in the Earth's core (more exactly, it would be current density).

It must be mentioned, though, that there have been some important numerical studies, one of which was even able to predict the geomagnetic field reversal.

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  • $\begingroup$ I understand it is extremely complex, I was looking for some kind of estimate like in L.B. Crowell's attempt. How do you assess his answer? $\endgroup$
    – Gerard
    Jul 3, 2017 at 8:00
  • $\begingroup$ The Nature article of Glatzmaier/Roberts contains no numerical estimates of currents. I have the feeling that the value of $10^8$ should be even (much) larger considering the Taylor columns. $\endgroup$
    – Gerard
    Jul 3, 2017 at 8:03
  • $\begingroup$ @Gerard I know that the article contains no numerical estimate of the current (even if Glatzmaier and Robert would probably be able to give you one). As for L.B. Cromwell's answer, there are some very big problems with an estimate of that kind, such as: 1) We cannot at all assume that the magnetic field is uniform. How can it be uniform if we are integrating so close to the source? Also, take a look at the representation in the cited article: that doesn't look uniform to me 2) We know that the currents are localized in the outer core, not in the inner core. $\endgroup$
    – valerio
    Jul 3, 2017 at 8:14
  • $\begingroup$ 3) Most important: we know that the currents are organized in Taylor columns, and not uniformly distributed in the inner core. We don't know the structure of these columns, and personally I cannot even think of a way of taking it into account approximately. $\endgroup$
    – valerio
    Jul 3, 2017 at 8:15
  • $\begingroup$ You see, the point is that since the currents are organized in a columnar structure their magnetic fields will interfere with each other and partially "cancel out", so it is difficult to estimate the magnetic field without knowing this structure... $\endgroup$
    – valerio
    Jul 3, 2017 at 8:19
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Using the expression of a dipolar field, dimension of the Earth and dimension of the core: $ B= \frac{\mu_0}{4 \pi} \frac{3 \mu.u - \mu}{r^3} $ with $\mu_0= {4 \pi} 10^{-7}$ uSI (international system units), $r=6200$ km, $B=4.6 \cdot 10^{-5} T ($u is a unit radial vector) We obtain the magnetic moment of order $\mu = 4.6\cdot 10^{-5}/10^{-7}*(6.2\cdot 10^6)^3 = 10^{23}$ uSI $= I S$ where S is the surface of the current loop. From the inner core dimension of order 1000 km, the order of magnitude of S is $S=(10^6)^2=10^{12} m^2$ and the current is of order $I=\mu/S= 4.6\cdot 10^{-5}/10^{-7}*(6.2\cdot 10^6)^3/(10^6)^2 = 10^{11} A$

The corresponding average current density is of order $j = I/S = 4.6\cdot 10^{-5}/10^{-7}*(6.2\cdot 10^6)^3/(10^6)^4 = 0.1 A/m^2$

These are orders of magnitude.

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