{"id":2087,"date":"2026-09-28T11:48:02","date_gmt":"2026-09-28T11:48:02","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=2087"},"modified":"2026-09-28T11:48:04","modified_gmt":"2026-09-28T11:48:04","slug":"eddy-currents","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/eddy-currents\/","title":{"rendered":"Eddy Currents: What They Are, What They Waste and Where They Help"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nEddy currents are loops of current that a changing magnetic field drives inside a solid conductor, circulating in the bulk metal itself with no wire to guide them. In a thin sheet they waste power in proportion to the square of the field, the square of the frequency and the square of the thickness, divided by the resistivity.\n\n<\/p><\/div>\n\n<p>Slide a strong magnet down the inside of a thick copper pipe and it refuses to fall properly. It drifts, as though the pipe were full of honey. Nothing is touching it, and copper is not magnetic in any ordinary sense.<\/p>\n\n<p>What holds it back is current \u2014 current running in circles inside the pipe wall, induced by the magnet&#8217;s own moving field and pushing back against the thing that made it. The same circulating current is why a transformer runs warm, why an induction hob heats the pan and not the glass, and why a metal core is almost never one solid lump.<\/p>\n\n<h2>What Is an Eddy Current?<\/h2>\n\n<p>An eddy current is a closed loop of current induced inside a solid conductor by a changing magnetic flux, flowing through the bulk metal itself rather than round a wire.<\/p>\n\n<p>The name is borrowed from water. Drop a stone in a stream and the flow curls into little whirlpools that close on themselves; do the equivalent to a block of metal with a changing field and the current does the same thing. Nobody chose the path. It is simply the route the metal offers.<\/p>\n\n<p>That is the part students find strange, so it is worth saying flatly. <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/electromagnetic-induction\/\">Faraday&#8217;s law of induction<\/a> does not require a wire. Every closed path you can imagine drawing inside a lump of metal encloses some flux, and if that flux changes, an EMF appears round that path.<\/p>\n\n<p>The metal&#8217;s own conductivity does the rest. An EMF round a path that is made of conductor drives a current round it, and since the block is solid, there are uncountably many such paths, nested inside one another.<\/p>\n\n<p>Two familiar laws already cover half the story. <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/lenz-law\/\">Lenz&#8217;s law<\/a>, whose own post hands this topic over rather than putting a number on it, settles which way the loops turn: always the way that opposes whatever change produced them, which is why the falling magnet is slowed rather than sped up. <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/faradays-law-formula\/\">Faraday&#8217;s law in its EMF form<\/a> settles how big the driving voltage is.<\/p>\n\n<p>Neither can tell you the one thing an engineer actually needs. How much <em>power<\/em> do the loops burn? That question needs two quantities neither law contains: how thick the metal is, and how readily it conducts.<\/p>\n\n<p>One more condition, easy to lose sight of. The flux through the metal has to be <strong>changing<\/strong>. A steady <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/ac-vs-dc-current\/\">direct current<\/a> makes a steady field and induces nothing at all; it is alternating current, or relative motion between the metal and the magnet, that does the work.<\/p>\n\n<h2>The Eddy Current Loss Formula<\/h2>\n\n<p>For a thin sheet of non-magnetic metal sitting in a uniform alternating field that lies along the sheet, the power dissipated per unit volume is given by one compact expression.<\/p>\n\n<div class=\"pf-formula\">P\/V = \u03c0\u00b2B\u00b2f\u00b2d\u00b2 \/ (6\u03c1)<\/div>\n\n<ul>\n<li><strong>P\/V<\/strong> \u2014 power dissipated per unit volume of metal, in watts per cubic metre (W\/m\u00b3). Multiply by the volume to get watts.<\/li>\n<li><strong>B<\/strong> \u2014 the <em>peak<\/em> flux density of the alternating field, in tesla (T). Not the root-mean-square value; for a sine wave the two differ by about 1.414, and therefore by a factor of two in the answer.<\/li>\n<li><strong>f<\/strong> \u2014 the frequency of the field, in hertz (Hz).<\/li>\n<li><strong>d<\/strong> \u2014 the thickness of <em>one<\/em> sheet measured across the field, in metres (m). Not the height of the stack.<\/li>\n<li><strong>\u03c1<\/strong> \u2014 the resistivity of the metal, in ohm metres (\u03a9 m). Copper is 1.68 \u00d7 10<sup>\u22128<\/sup> \u03a9 m and nichrome is 1.1 \u00d7 10<sup>\u22126<\/sup> \u03a9 m, the two figures used throughout this page.<\/li>\n<\/ul>\n\n<p>Every quantity in there is a property of the sheet or of the field, and there is no fitted constant anywhere \u2014 the 6 is earned in the derivation below. Take 100 cm\u00b3 of copper divided into forty sheets half a millimetre thick, put it in a 0.50 T field at 50 Hz, and the loss comes to <strong>1.530 W<\/strong>.<\/p>\n\n<p>Rearranged, the same relation answers the question a designer really asks, which is how thin a sheet has to be to keep the loss down. If you would rather not do that algebra by hand, the <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/eddy-currents\">eddy current loss calculator<\/a> solves it for the loss, the field, the frequency, the thickness or the resistivity, and prints the skin depth beside the answer so you can see whether the formula still applies.<\/p>\n\n<p>The simulator below is the picture that goes with the algebra. Set the field, the frequency, the resistivity and the number of slices the block is cut into, and it draws the current loops inside the metal, the current-density profile across one lamination, and the loss for the same fixed quantity of metal every time.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Eddy Currents Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:800px}@media(max-width:760px){.pf-sim-frame{height:1750px}}<\/style><iframe src=\"\/labs\/eddy-currents.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>How the Loss Formula Is Derived<\/h2>\n\n<p>The derivation is six lines and it is worth following, because every factor in the answer comes from somewhere you can point at.<\/p>\n\n<p>Picture a slab of non-magnetic metal of thickness d, wide and long compared with that thickness. Let the field B lie along the slab and alternate sinusoidally, and measure position x from the mid-plane outwards.<\/p>\n\n<ol>\n<li>Draw a rectangular loop running from \u2212x to +x across the thickness. Faraday&#8217;s law gives an EMF of \u2212(2xL)&nbsp;dB\/dt round it, where L is its length.<\/li>\n<li>By symmetry the electric field at \u2212x is the mirror of the one at +x, so the loop integral is just 2L times the field at x. That leaves <strong>E(x) = \u2212x&nbsp;dB\/dt<\/strong>: the driving field grows linearly outwards from the middle.<\/li>\n<li>Ohm&#8217;s law in its local form turns that into a current density, J = \u2212x&nbsp;dB\/dt&nbsp;\/&nbsp;\u03c1, and the power dissipated per unit volume at x is J\u00b2\u03c1, which grows as x\u00b2.<\/li>\n<li>Average x\u00b2 across the thickness and you pick up a factor of <strong>d\u00b2\/12<\/strong>. That 12 is the second moment of a slab about its own mid-plane, nothing more exotic.<\/li>\n<li>Average over a cycle. With B at its peak value B multiplied by sin(\u03c9t), the mean square of dB\/dt is B\u00b2\u03c9\u00b2\/<strong>2<\/strong>, the usual mean square of a cosine.<\/li>\n<li>Put \u03c9 = 2\u03c0f into what is left, so the squaring brings in 4\u03c0\u00b2. The 4 divides into that 24 and leaves 6, while the \u03c0\u00b2 stays in the numerator, giving P\/V = \u03c0\u00b2B\u00b2f\u00b2d\u00b2 \/ (6\u03c1).<\/li>\n<\/ol>\n\n<p>So the 6 is 12 \u00d7 2 with the 4 from \u03c9\u00b2 divided back out of it: the second moment of the thickness, the mean square of a cosine, and nothing else. Nothing was fitted and nothing was looked up.<\/p>\n\n<p>That is the kind of claim that deserves an independent check, and this cluster ran one. The same loss was computed a second way, by brute-force numerical integration of the dissipation over position and over a whole cycle \u2014 no closed form, no factor of 6, no pi squared anywhere in it. Across 27 different states the two routes agree to within 6.25 parts in a million.<\/p>\n\n<p>The way the agreement arrives is itself the tell. Run the numerical version with only two panels across the thickness and it is wrong by exactly 25 per cent; double the panel count and the error falls by a factor of 4.00, every single time, all the way down. A formula has no panel count, so watching that error quarter itself is real evidence about the algebra rather than a coincidence.<\/p>\n\n<h2>Why Are Transformer Cores Cut Into Thin Slices?<\/h2>\n\n<p>Because the loss goes as the square of the sheet thickness, so dividing a core into thin insulated sheets collapses the waste heat while leaving the quantity of metal untouched.<\/p>\n\n<p>One caution before the numbers, and it is the caution this whole page is built around. The block modelled here is <strong>non-magnetic<\/strong> copper, with a relative permeability of one.<\/p>\n\n<p>A real transformer core is a ferromagnetic alloy whose relative permeability runs into the thousands, which shrinks its skin depth by the square root of that factor \u2014 and that is exactly why real laminations are a fraction of a millimetre rather than centimetres. The <em>reason<\/em> for slicing transfers perfectly. The <em>numbers<\/em> below are copper&#8217;s, and no figure for any steel appears anywhere on this page, because none was sourced.<\/p>\n\n<p>Here is the same 100 cm\u00b3 of copper, 20.0 mm thick, in a 0.50 T field at 50 Hz, cut into ever more insulated slices. The fourth column is what the formula above predicts; the fifth is the exact solution of the same slab with no thin-sheet approximation in it.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr>\n<th style=\"border-bottom:2px solid #0A1628;padding:8px;text-align:left;\">Slices N<\/th>\n<th style=\"border-bottom:2px solid #0A1628;padding:8px;text-align:right;\">d (mm)<\/th>\n<th style=\"border-bottom:2px solid #0A1628;padding:8px;text-align:right;\">d in skin depths<\/th>\n<th style=\"border-bottom:2px solid #0A1628;padding:8px;text-align:right;\">Textbook P (W)<\/th>\n<th style=\"border-bottom:2px solid #0A1628;padding:8px;text-align:right;\">Exact P (W)<\/th>\n<th style=\"border-bottom:2px solid #0A1628;padding:8px;text-align:right;\">Formula high by<\/th>\n<th style=\"border-bottom:2px solid #0A1628;padding:8px;text-align:right;\">Cut vs solid<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;\">1<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">20.000<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">2.1679<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">2447.82<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">1300.08<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">88.28 %<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">1.0 x<\/td><\/tr>\n<tr><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;\">2<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">10.000<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">1.0840<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">611.95<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">579.59<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">5.58 %<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">2.2 x<\/td><\/tr>\n<tr><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;\">4<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">5.000<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.5420<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">152.99<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">152.46<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.35 %<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">8.5 x<\/td><\/tr>\n<tr><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;\">5<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">4.000<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.4336<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">97.91<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">97.77<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.14 %<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">13.3 x<\/td><\/tr>\n<tr><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;\">8<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">2.500<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.2710<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">38.25<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">38.24<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.02 %<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">34.0 x<\/td><\/tr>\n<tr><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;\">10<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">2.000<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.2168<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">24.48<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">24.48<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.01 %<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">53.1 x<\/td><\/tr>\n<tr><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;\">20<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">1.000<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.1084<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">6.12<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">6.12<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">0.00 %<\/td><td style=\"border-bottom:1px solid #D9CFB8;padding:8px;text-align:right;\">212.4 x<\/td><\/tr>\n<tr><td style=\"padding:8px;\">40<\/td><td style=\"padding:8px;text-align:right;\">0.500<\/td><td style=\"padding:8px;text-align:right;\">0.0542<\/td><td style=\"padding:8px;text-align:right;\">1.53<\/td><td style=\"padding:8px;text-align:right;\">1.53<\/td><td style=\"padding:8px;text-align:right;\">0.00 %<\/td><td style=\"padding:8px;text-align:right;\">849.8 x<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Read the table from the bottom and the formula looks perfect: from ten slices downwards the two power columns agree to the last digit printed, and at eight they sit a single hundredth of a watt apart. Read it from the top and the formula is <strong>88.28 per cent high<\/strong> on the solid block. The last column, rounded to one decimal place, tops out at 849.8 x where the simple squared rule promises 1600.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-01-768x499.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-01-768x499.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-01.jpg 1400w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1400\" height=\"910\" alt=\"Two cross-sections of the same 100 cubic centimetre copper block in a 0.50 tesla field at 50 hertz. The left block is one solid piece 20.0 millimetres across the field and carries a single large current loop, labelled 1300.08 watts. The right block is the same 20.0 millimetres divided by wine coloured insulation into eight slices of 2.500 millimetres each, every slice carrying its own narrow current loop, labelled 38.24 watts. Crossed circles on both blocks mark the magnetic field pointing into the page, lying along the slices rather than across them. A note states that both figures are the exact slab solution rather than the textbook estimate.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Same metal, same field, same frequency. The only difference is whether the current can find a long way round.<\/figcaption>\n<\/figure>\n\n<p>Now the result this page exists for. Cutting the block into forty insulated slices takes it from <strong>1300.08 W to 1.53 W<\/strong>. The squared rule says forty slices should divide the loss by 40\u00b2, which is 1600. What the metal actually delivers is a reduction of <strong>849.79 times<\/strong>.<\/p>\n\n<p>The shortfall is not a failure of lamination. Divide the two predictions, 1600 \/ 849.79 = 1.8828, and compare that with the amount by which the textbook formula overstated the solid block in the first place: 2447.82 \/ 1300.08 = 1.8828. The same number, to every figure printed.<\/p>\n\n<p>That is the whole explanation, and it is worth stating plainly: <strong>the squared rule was not wrong about the slicing. It was wrong about the block you started from.<\/strong> Why it was wrong about that block is the next section.<\/p>\n\n<h2>Where the Eddy Current Formula Stops Being True<\/h2>\n\n<p>The derivation quietly assumed the field is the same everywhere across the thickness, and the eddy currents themselves make that false: their own field opposes the applied one, so the interior of a thick piece of metal is screened.<\/p>\n\n<p>The length that decides whether you can ignore that is the <strong>skin depth<\/strong>.<\/p>\n\n<div class=\"pf-formula\">\u03b4 = sqrt( \u03c1 \/ (\u03c0f\u03bc<sub>0<\/sub>) )<\/div>\n\n<p>Here \u03bc<sub>0<\/sub> is the permeability of free space, taken throughout this cluster as 4\u03c0 \u00d7 10<sup>\u22127<\/sup> N\/A\u00b2. NIST publishes a measured <a href=\"https:\/\/physics.nist.gov\/cgi-bin\/cuu\/Value?mu0\" target=\"_blank\" rel=\"noopener\">vacuum magnetic permeability<\/a> that sits a hair away from that, far too close to move anything here: copper&#8217;s skin depth at 50 Hz comes out as 9.2255 mm on either figure.<\/p>\n\n<p>In copper the skin depth falls as the frequency rises, as the square root demands: 20.6288 mm at 10 Hz, 9.2255 mm at 50 Hz, 6.5234 mm at 100 Hz, 3.2617 mm at 400 Hz and 2.0629 mm at 1000 Hz. Resistivity pulls the other way, which is why a poorer conductor is screened less.<\/p>\n\n<p>Nichrome makes that concrete. The same solid 20.0 mm block at 0.50 T and 50 Hz loses 1300.08 W in copper and only <strong>37.38 W<\/strong> in nichrome \u2014 about 35 times less in the <em>more<\/em> resistive metal. Its skin depth is 74.650 mm, which dwarfs the block, so for nichrome the textbook formula is exact. Our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/electrical-resistance\/\">electrical resistance and resistivity<\/a> covers where those material figures come from.<\/p>\n\n<p>Solve the slab exactly, with no thin-sheet assumption at all, and a second closed form drops out. Write s for the thickness measured in skin depths, s = d\/\u03b4, and \u03bc for the permeability of the metal, which equals \u03bc<sub>0<\/sub> here because the metal is non-magnetic.<\/p>\n\n<div class=\"pf-formula\">P\/V = \u03c1B\u00b2(sinh s \u2212 sin s) \/ ( \u03bc\u00b2d\u03b4(cosh s + cos s) )<\/div>\n\n<p>Compare the two and you get a boundary rather than a warning. The textbook formula first reads 0.5 per cent high at <strong>0.5929<\/strong> skin depths, 1 per cent high at <strong>0.7051<\/strong>, 5 per cent at <strong>1.0545<\/strong>, 10 per cent at <strong>1.2542<\/strong> and 50 per cent at <strong>1.8782<\/strong>. Those were scanned in steps of 0.0001 skin depths, not estimated.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-02-768x564.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-02-768x564.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-02.jpg 1400w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1400\" height=\"1028\" alt=\"A line chart of how far the textbook eddy-current formula reads high, plotted against lamination thickness measured in skin depths. The horizontal axis is linear from 0 to 3.2 skin depths; the vertical axis is logarithmic from 0.01 to 1000 per cent, each gridline ten times the one below it. Seven measured points rise steeply and are joined by a dark line: 0.02 per cent at 0.25 skin depths, 0.25 at 0.50, 1.28 at 0.75, 4.05 at 1.00, 20.43 at 1.50, 64.17 at 2.00 and 313.59 at 3.00. Two wine dashed guides mark the measured boundaries, one at 0.7051 skin depths where the formula first reads 1 per cent high and one at 1.2542 where it first reads 10 per cent high. Notes state that the points are joined only to show the trend, that nothing is fitted or extrapolated, and that at 0.10 skin depths the overestimate rounds to 0.00 per cent and falls below the bottom of the scale.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">The vertical scale is logarithmic. Under about two-thirds of a skin depth the error is invisible; past one skin depth it runs away.<\/figcaption>\n<\/figure>\n\n<p>Here is the practical version of that chart. A real lamination sits far inside the 1 per cent line \u2014 the 0.500 mm copper sheet in the table above is only 0.0542 skin depths thick. A solid block usually does not.<\/p>\n\n<p>How badly does it fail? At 400 Hz that same 20.0 mm copper block is 6.132 skin depths thick, and the textbook formula claims 156.7 kW where the metal delivers 4.062 kW \u2014 out by a factor of <strong>38.56<\/strong>. Wind the frequency up to 1000 Hz and make the metal more conductive still and the gap widens to a factor of 330.7, which is the worst state the simulator above can be driven to.<\/p>\n\n<h2>Why a Thicker Block Can Lose Less Than a Thinner One<\/h2>\n\n<p>Past about 2.25 skin depths, making a piece of metal thicker <em>lowers<\/em> its loss per unit volume, because the extra metal sits in the screened interior and carries almost no current.<\/p>\n\n<p>The peak sits at <strong>2.2542<\/strong> skin depths, found by scanning 200,000 thicknesses. By ten skin depths the loss per unit volume has fallen to 23.97 per cent of that peak. The textbook formula does the opposite and grows as d\u00b2 for ever, which is precisely how the gap in the last section opened out to a factor of 330.7.<\/p>\n\n<p>The clean way to see it is the total loss per unit of surface area, which <em>saturates<\/em>. For copper at 1 T and 50 Hz it reads 1.16354 \u00d7 10<sup>6<\/sup> W\/m\u00b2 at five skin depths, 1.15333 \u00d7 10<sup>6<\/sup> at ten, and 1.15319 \u00d7 10<sup>6<\/sup> at both twenty and fifty. Past a few skin depths, adding metal adds volume and no extra loss whatsoever.<\/p>\n\n<figure style=\"margin:32px auto;max-width:720px;text-align:center;\">\n  <img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-03-768x599.jpg\" srcset=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-03-768x599.jpg 768w, https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/eddy-currents-diagram-03.jpg 1400w\" sizes=\"auto, (max-width: 760px) 100vw, 720px\" width=\"1400\" height=\"1092\" alt=\"A line chart of current density against position across one lamination. The horizontal axis runs from minus one at one face through zero at the mid-plane to plus one at the other face, measured in half-thicknesses; the vertical axis runs from 0 to 1.0 and is labelled current density, normalised, with no unit. Two curves share that one vertical scale. The dark curve, for a lamination six skin depths thick, is almost flat across the middle and rises into a sharp spike at each face, reaching 0.9953 at the surface. The gold curve, for a lamination 0.2 skin depths thick, is a straight ramp from zero at the mid-plane to only 0.1414 at each face. Two wine dashed vertical lines mark one skin depth in from each face for the dark curve. Notes state that both curves use the same vertical scale and that for the thin curve one skin depth lies five thicknesses outside the sheet, so it cannot be marked on this frame.\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;border-radius:4px;\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Without this picture the peak is just an odd number. With it, the reason is obvious: in the thick sheet almost all the current is crowded into the two faces.<\/figcaption>\n<\/figure>\n\n<p>Once a sheet is that thick, the familiar scaling rules quietly stop applying. At two skin depths in copper, doubling the frequency multiplies the exact loss by <strong>1.8762<\/strong> rather than by 4. Doubling the thickness multiplies it by <strong>0.6478<\/strong> \u2014 the loss goes <em>down<\/em>. Doubling the resistivity multiplies it by <strong>0.7067<\/strong> instead of halving it.<\/p>\n\n<p>None of that contradicts the formula. It just means the formula was never describing that sheet, and the only honest way to quote it is with a thickness in skin depths beside it.<\/p>\n\n<h2>Real-World Examples of Eddy Currents<\/h2>\n\n<p>Eddy currents show up wherever metal meets a changing field. Sometimes an engineer is working hard to suppress them; just as often, somebody is paying good money to produce them on purpose.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/09\/a-laminated-core-with-its-sheet-edges-visible-or-an-induction-hob-in-use.jpg\"\n       alt=\"The stacked sheet edges of a laminated core, the standard way of cutting eddy currents in an alternating field\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" width=\"1920\" height=\"1080\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">Every visible line in the stack is an insulated boundary the eddy currents cannot cross.<\/figcaption>\n<\/figure>\n\n<h3>Transformer and motor cores, where the loss is pure waste<\/h3>\n\n<p>This is the calculation that fixes a lamination thickness. Thinner sheets mean less waste heat and a higher price, more insulation and less metal per centimetre of stack, so the choice is an economic one made against exactly the numbers in the table above.<\/p>\n\n<p>Core loss is also a continuous draw whenever equipment is energised, doing useful work or not. One warning before those watts turn into money. The core loss a manufacturer quotes is two mechanisms added together: the eddy-current share this page computes, and hysteresis, the price of reversing the magnetisation of a ferromagnetic core twice in every cycle.<\/p>\n\n<p>Hysteresis grows roughly with the first power of the frequency where the eddy-current loss grows with its square, so a quoted core-loss figure contains both and will not balance against anything computed here. Nothing on this page models it. Our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/electromagnetism\/electricity-cost\/\">working out the cost of electricity<\/a> turns a standing loss in watts into a bill.<\/p>\n\n<h3>Induction hobs and induction heating, where the loss is the point<\/h3>\n\n<p>Flip the sign of the design goal and everything reverses. A coil under the glass drives a field into the pan, the eddy currents in the pan base do the cooking, and the hob stays comparatively cool because glass conducts nothing. Instead of thin sheets you want a thick, reasonably resistive base, and the frequency is chosen to put the skin depth where the heat is wanted.<\/p>\n\n<h3>Eddy-current brakes and damping<\/h3>\n\n<p>Drag a conductor past a magnet and Lenz&#8217;s law guarantees a retarding force with nothing in contact and nothing to wear out. Trains and roller coasters use it at scale; the spinning aluminium disc in an old electricity meter and the settling needle of a moving-coil instrument use it in miniature.<\/p>\n\n<p>It is also why the magnet crawls down that copper pipe. This page deliberately puts no speed on that demonstration \u2014 the closed form for it rests on a dipole model and a numerical coefficient that this cluster could not derive from first principles, and an invented figure would be worse than none.<\/p>\n\n<h3>Metal detectors and non-destructive testing<\/h3>\n\n<p>A coil carrying alternating current induces eddy currents in any conductor nearby, and those currents push a field back into the coil and change what it measures. Do that crudely and you have a metal detector. Do it at a fixed, well-chosen frequency and you have eddy-current testing, which finds cracks and corrosion in aircraft skins and heat-exchanger tubes without touching them. What is read there is the change in the loss, which is why the skin depth matters so much.<\/p>\n\n<h2>Common Misconceptions About Eddy Currents<\/h2>\n\n<h3>&#8220;Eddy currents are always a waste&#8221;<\/h3>\n\n<p>They are a waste in a transformer and the entire product in an induction furnace. Every application listed above except the first is somebody arranging for the loops on purpose \u2014 heating a pan, stopping a train, finding a crack in an aircraft skin. Treating the topic as pure loss is reading half of it.<\/p>\n\n<h3>&#8220;Forty slices must cut the loss 1600 times&#8221;<\/h3>\n\n<p>Forty slices cut it 849.79 times in the copper block measured here, not 1600. The squared rule is a fine description of thin sheets; the number it is being compared against, the loss in the solid block, is the one it gets wrong \u2014 by 88.28 per cent, because that block screens its own interior.<\/p>\n\n<h3>&#8220;Doubling the frequency always quadruples the loss&#8221;<\/h3>\n\n<p>It does while the sheet stays thin, and problem 6 below is exactly that case: a 0.500 mm copper sheet really does lose 64 times as much at 400 Hz as at 50 Hz. At two skin depths it does not, and the exact loss rises by 1.8762 instead of 4. A rule with a validity boundary is only useful if you carry the boundary around with it.<\/p>\n\n<h3>&#8220;A thicker piece of metal always burns more&#8221;<\/h3>\n\n<p>More in total, up to a point; more per unit volume, only up to 2.2542 skin depths. Beyond that the extra metal is screened, contributes almost no current, and the loss per cubic metre falls \u2014 which is the single most counter-intuitive thing in this whole topic.<\/p>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A lamination 0.500 mm thick, of resistivity 50 microhm-centimetres, sits in a 1.0 T peak field at 50 Hz. What is the eddy-current loss per cubic metre?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Use P\/V = \u03c0\u00b2B\u00b2f\u00b2d\u00b2 \/ (6\u03c1), with every quantity in SI units.\n\nStep 2: Convert and square. d = 0.500 mm = 5.00 \u00d7 10<sup>\u22124<\/sup> m, so d\u00b2 = 2.50 \u00d7 10<sup>\u22127<\/sup> m\u00b2. B\u00b2 = 1.00 T\u00b2 and f\u00b2 = 2500 Hz\u00b2. The resistivity is 50 \u00b5\u03a9 cm = 5.0 \u00d7 10<sup>\u22127<\/sup> \u03a9 m, and \u03c0\u00b2 = 9.8696.\n\nStep 3: Top: 9.8696 \u00d7 1.00 \u00d7 2500 \u00d7 2.50 \u00d7 10<sup>\u22127<\/sup> = 6.1685 \u00d7 10<sup>\u22123<\/sup>. Bottom: 6 \u00d7 5.0 \u00d7 10<sup>\u22127<\/sup> = 3.0 \u00d7 10<sup>\u22126<\/sup>. Divide: 6.1685 \u00d7 10<sup>\u22123<\/sup> \/ 3.0 \u00d7 10<sup>\u22126<\/sup> = 2056.17.\n\n<strong>Answer: P\/V = 2056.17 W\/m\u00b3<\/strong>\n\nThat metal is posited, not named: no material on this page carries a resistivity that was not sourced.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">The same lamination is rolled down to 0.250 mm. What happens to the loss per cubic metre?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Only d changes, and the formula contains d\u00b2.\n\nStep 2: d = 0.250 mm = 2.50 \u00d7 10<sup>\u22124<\/sup> m, so d\u00b2 = 6.25 \u00d7 10<sup>\u22128<\/sup> m\u00b2 \u2014 exactly one quarter of 2.50 \u00d7 10<sup>\u22127<\/sup> m\u00b2.\n\nStep 3: Every other factor is untouched, so P\/V = 2056.17 \/ 4 = 514.04.\n\n<strong>Answer: P\/V = 514.04 W\/m\u00b3, exactly a quarter<\/strong>\n\nSay what this is: the formula restated, not a discovery. Halving d quarters the loss <em>because<\/em> the formula has d\u00b2 in it, and the same is true of every sheet thin compared with its skin depth.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">Find the skin depth in copper at 50 Hz and at 1000 Hz, and check the ratio between them.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Use the skin depth relation, \u03b4 = sqrt( \u03c1 \/ (\u03c0f\u03bc<sub>0<\/sub>) ), with \u03c1 = 1.68 \u00d7 10<sup>\u22128<\/sup> \u03a9 m and \u03bc<sub>0<\/sub> = 4\u03c0 \u00d7 10<sup>\u22127<\/sup> N\/A\u00b2.\n\nStep 2: At 50 Hz the denominator is \u03c0 \u00d7 50 \u00d7 4\u03c0 \u00d7 10<sup>\u22127<\/sup> = 1.97392 \u00d7 10<sup>\u22124<\/sup>. Then 1.68 \u00d7 10<sup>\u22128<\/sup> \/ 1.97392 \u00d7 10<sup>\u22124<\/sup> = 8.51098 \u00d7 10<sup>\u22125<\/sup> m\u00b2.\n\nStep 3: Take the root: sqrt(8.51098 \u00d7 10<sup>\u22125<\/sup>) = 9.2255 \u00d7 10<sup>\u22123<\/sup> m. Since \u03b4 goes as one over the square root of f, at 1000 Hz it is 9.2255 \/ 4.472136 = 2.0629 mm, and 4.472136 is the square root of 20.\n\n<strong>Answer: 9.2255 mm at 50 Hz and 2.0629 mm at 1000 Hz<\/strong>\n\nTwenty times the frequency buys a factor of only 4.472136 in depth. Frequency is an expensive way to thin a current.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A solid copper block of 100 cubic centimetres is 20.0 mm thick across a 0.50 T field at 50 Hz. What does the textbook formula predict, and what is the truth?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Work out the textbook figure first. d = 20.0 mm = 0.0200 m, so d\u00b2 = 4.00 \u00d7 10<sup>\u22124<\/sup> m\u00b2; B\u00b2 = 0.25 T\u00b2; f\u00b2 = 2500 Hz\u00b2; \u03c1 = 1.68 \u00d7 10<sup>\u22128<\/sup> \u03a9 m.\n\nStep 2: Top: 9.8696 \u00d7 0.25 \u00d7 2500 \u00d7 4.00 \u00d7 10<sup>\u22124<\/sup> = 2.4674. Bottom: 6 \u00d7 1.68 \u00d7 10<sup>\u22128<\/sup> = 1.008 \u00d7 10<sup>\u22127<\/sup>. So P\/V = 2.4674 \/ 1.008 \u00d7 10<sup>\u22127<\/sup> = 2.44782 \u00d7 10<sup>7<\/sup> W\/m\u00b3, and over 1.00 \u00d7 10<sup>\u22124<\/sup> m\u00b3 that is 2447.82 W.\n\nStep 3: Now test whether the formula is allowed here. The skin depth is 9.2255 mm, so 20.000 \/ 9.2255 = 2.1679 skin depths, well past the 1.2542 mark where the error reaches 10 per cent. The exact slab solution gives 1300.08 W, and 2447.82 \/ 1300.08 = 1.8828.\n\n<strong>Answer: the formula claims 2447.82 W; the exact solution gives 1300.08 W, so the formula reads 88.28 per cent high<\/strong>\n\nKeep this one open for problem 5, which cuts the identical block into forty slices.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">Cut that same copper block into 40 insulated laminations. By how much does the loss really fall, and why is it not 1600 times?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Each slice is 20.0 mm \/ 40 = 0.500 mm thick, which is 0.500 \/ 9.2255 = 0.0542 skin depths. That is deep inside the valid region, so here the textbook formula and the exact solution agree: both give 1.53 W.\n\nStep 2: What the squared rule predicts. Each sheet is a fortieth as thick and the loss goes as d\u00b2, so the rule promises a reduction of 40\u00b2 = 1600.\n\nStep 3: What the metal delivers is 849.79 times, from 1300.08 W down to 1.53 W. Divide the two predictions: 1600 \/ 849.79 = 1.8828, which is exactly the factor by which the formula overstated the solid block in problem 4.\n\n<strong>Answer: the loss falls 849.79-fold, not 1600-fold \u2014 and the squared rule was wrong about the starting block, not about the slicing<\/strong>\n\nOne caution on arithmetic. Dividing the two rounded powers above gives about 849.7, not 849.79; the published factor comes from the unrounded values, so do not check it by hand from the printed figures.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A 0.500 mm copper lamination is moved from 50 Hz to 400 Hz. By what factor does the loss change, and is the simple rule trustworthy here?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Nothing but f changes, and the formula contains f\u00b2.\n\nStep 2: The frequency ratio is 400 \/ 50 = 8, so the predicted factor is 8\u00b2 = 64.\n\nStep 3: Check that the sheet is still thin enough. At 400 Hz copper&#8217;s skin depth is 3.2617 mm, so 0.500 \/ 3.2617 = 0.1533 skin depths, far inside the 0.7051 boundary. The exact solution follows the rule: the same 100 cm\u00b3 of copper, cut into 0.500 mm sheets, loses 1.530 W at 50 Hz and 97.91 W at 400 Hz.\n\n<strong>Answer: the loss rises by a factor of 64, and here the rule can be trusted<\/strong>\n\nWorth noticing which way this cuts. The rule fails on thick metal, not on high frequency \u2014 a thin sheet obeys it happily at 400 Hz, while the solid block of problem 4 does not obey it even at 50 Hz.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">How thin must a lamination be to keep the loss to 10000 W per cubic metre at 1.0 T and 50 Hz, in a metal of resistivity 50 microhm-centimetres?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Rearrange for d. Multiplying out gives d\u00b2 = 6\u03c1(P\/V) \/ (\u03c0\u00b2B\u00b2f\u00b2), and d is the positive square root.\n\nStep 2: Top: 6 \u00d7 5.0 \u00d7 10<sup>\u22127<\/sup> \u00d7 10000 = 3.0 \u00d7 10<sup>\u22122<\/sup>. Bottom: 9.8696 \u00d7 1.00 \u00d7 2500 = 24674.\n\nStep 3: So d\u00b2 = 3.0 \u00d7 10<sup>\u22122<\/sup> \/ 24674 = 1.21585 \u00d7 10<sup>\u22126<\/sup> m\u00b2, and d = sqrt(1.21585 \u00d7 10<sup>\u22126<\/sup>) = 1.1027 \u00d7 10<sup>\u22123<\/sup> m.\n\n<strong>Answer: d = 1.1027 mm, and any thinner sheet is safely below the target<\/strong>\n\nSquaring loses a sign, so the negative root satisfies the algebra just as exactly. A thickness of minus 1.1027 mm is not a sheet, so it is discarded.\n\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">At what thickness does the textbook eddy-current formula read 10 per cent high, for copper at 50 Hz?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: The boundary is a property of the shape of the two solutions, not of any one metal, and it sits at a thickness of 1.2542 skin depths. That figure was found by scanning the exact solution against the formula in steps of 0.0001 skin depths.\n\nStep 2: Convert it into a real thickness using copper&#8217;s skin depth at 50 Hz, which problem 3 gave as 9.2255 mm.\n\nStep 3: d = 1.2542 \u00d7 9.2255 = 11.5706 mm.\n\n<strong>Answer: 11.5706 mm \u2014 above that, the formula is at least 10 per cent high for copper at 50 Hz<\/strong>\n\nPut the page&#8217;s two blocks against it. A 0.500 mm lamination sits at 0.0542 skin depths, nowhere near the line; the 20.0 mm block of problem 4 sits at 2.1679, well beyond it, which is why it read 88.28 per cent high.\n\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What are eddy currents in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\n\nEddy currents are loops of current that a changing magnetic field stirs up inside a solid piece of metal. There is no wire and no circuit anyone built: the current simply circulates in the metal itself, following whatever closed paths the material offers. They heat the metal, and they always flow the way that opposes the change that created them.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why are transformer cores laminated instead of solid?<\/summary><div class=\"pf-faq-item-answer\">\n\nBecause the eddy-current loss goes as the square of the sheet thickness, so slicing a core into thin insulated sheets cuts the waste heat dramatically without removing any metal. In the copper block measured on this page, forty slices take the loss from 1300.08 W to 1.53 W. The insulation does no physics itself; it just stops the loops closing across the whole block.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is skin depth and why does it matter for eddy currents?<\/summary><div class=\"pf-faq-item-answer\">\n\nSkin depth is the distance a changing field penetrates into a conductor before it is substantially screened out by the currents it induces. It is given by sqrt of resistivity over pi times frequency times permeability, and in copper at 50 Hz it is 9.2255 mm. It matters because the standard loss formula assumes the field reaches the middle of the sheet, which stops being true near one skin depth.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does cutting a core into 40 slices really cut the loss 1600 times?<\/summary><div class=\"pf-faq-item-answer\">\n\nNo. In the copper block measured here it cuts the loss 849.79 times, not 1600. The squared rule describes thin sheets accurately, but the solid block it is being compared against already screens its own interior, so the formula overstated that starting figure by 88.28 per cent. The shortfall is in the block, not in the lamination.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why does a magnet fall slowly down a copper pipe?<\/summary><div class=\"pf-faq-item-answer\">\n\nAs the magnet drops, the flux through each ring of pipe wall changes, inducing eddy currents that circulate round the wall. By Lenz&#8217;s law those currents oppose the change, so they pull back on the magnet and it settles to a gentle steady drift. No figure for that drift speed is given here, because the closed form for it rests on a coefficient this page could not derive or source.\n\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Do eddy currents flow with direct current?<\/summary><div class=\"pf-faq-item-answer\">\n\nOnly while something is changing. A steady direct current produces a steady magnetic field, the flux through the metal is constant, and no EMF appears anywhere inside it. Switch that current on or off, vary it, or move the metal and the magnet relative to one another, and eddy currents appear immediately for as long as the change lasts.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>An eddy current is a loop of current a changing magnetic field drives inside solid metal, with no wire to guide it. In a thin sheet the power it wastes is P\/V = pi^2 B^2 f^2 d^2 \/ (6 rho), which is why a metal core is cut into slices: the same 100 cubic centimetres of copper goes from 1300.08 W as one 20.0 mm block to 1.53 W as forty sheets. That is a cut of 849.79 times where the squared rule promises 1600, and this page measures exactly where the formula stops telling the truth. Eight worked problems, three original diagrams, an interactive lab and a calculator.<\/p>\n","protected":false},"author":1,"featured_media":2093,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-2087","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-electromagnetism"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2087","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=2087"}],"version-history":[{"count":3,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2087\/revisions"}],"predecessor-version":[{"id":2113,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/2087\/revisions\/2113"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/2093"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=2087"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=2087"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=2087"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}