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How to Model a Magnet Falling Through a Solenoid

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class="blog-meta"></div><div class='entry-content-wrapper clearfix standard-content'><header class="entry-content-header" aria-label="Post: How to Model a Magnet Falling Through a Solenoid"><h1 class='post-title entry-title ' itemprop="headline" >How to Model a Magnet Falling Through a Solenoid<span class="post-format-icon minor-meta"></span></h1><span class="post-meta-infos"><time class="date-container minor-meta updated" itemprop="datePublished" datetime="2022-06-05T12:48:47-05:00" >June 5, 2022</time><span class="text-sep">/</span><span class="comment-container minor-meta"><a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comments" class="comments-link" >11 Comments</a></span><span class="text-sep">/</span><span class="blog-categories minor-meta">in <a href="https://www.physicsforums.com/insights/science-math-tutorials/physics-tutorials/" rel="tag">Physics Tutorials</a></span><span class="text-sep">/</span><span class="blog-author minor-meta">by <span class="entry-author-link" itemprop="author" ><span class="author"><span class="fn"><a href="https://www.physicsforums.com/insights/author/kuruman/" title="Posts by kuruman" rel="author">kuruman</a></span></span></span></span></span></header><div class="entry-content" itemprop="text" ><div><strong>Estimated Read Time:</strong> 9 minute(s)</div><div><strong>Common Topics:</strong> dipole, emf, solenoid, figure, magnet</div><div><div id="ez-toc-container" class="ez-toc-v2_0_72 counter-hierarchy ez-toc-counter ez-toc-custom ez-toc-container-direction"><p class="ez-toc-title" style="cursor:inherit">Table of Contents</p> <label for="ez-toc-cssicon-toggle-item-67bec1df7594b" class="ez-toc-cssicon-toggle-label"><span class="ez-toc-cssicon"><span class="eztoc-hide" style="display:none;">Toggle</span><span class="ez-toc-icon-toggle-span"><svg style="fill: #4c4c4c;color:#4c4c4c" xmlns="http://www.w3.org/2000/svg" class="list-377408" width="20px" height="20px" viewBox="0 0 24 24" fill="none"><path d="M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z" fill="currentColor"></path></svg><svg style="fill: #4c4c4c;color:#4c4c4c" class="arrow-unsorted-368013" xmlns="http://www.w3.org/2000/svg" width="10px" height="10px" viewBox="0 0 24 24" version="1.2" baseProfile="tiny"><path d="M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z"/></svg></span></span></label><input type="checkbox" id="ez-toc-cssicon-toggle-item-67bec1df7594b" aria-label="Toggle" /><nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class="ez-toc-link ez-toc-heading-1" href="#Introduction" title="Introduction">Introduction</a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class="ez-toc-link ez-toc-heading-2" href="#The-motional-emf" title="The motional emf">The motional emf</a></li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class="ez-toc-link ez-toc-heading-3" href="#A-single-ring" title="A single ring">A single ring</a></li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class="ez-toc-link ez-toc-heading-4" href="#From-ring-to-solenoid" title="From ring to solenoid">From ring to solenoid</a></li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class="ez-toc-link ez-toc-heading-5" href="#The-emf-profile" title="The emf profile">The emf profile</a></li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class="ez-toc-link ez-toc-heading-6" href="#Term-corrections" title="Term corrections">Term corrections</a></li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class="ez-toc-link ez-toc-heading-7" href="#A-relation-between-amplitudes" title="A relation between amplitudes">A relation between amplitudes</a></li></ul></li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class="ez-toc-link ez-toc-heading-8" href="#Summary" title="Summary">Summary</a></li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class="ez-toc-link ez-toc-heading-9" href="#Acknowledgments" title="Acknowledgments">Acknowledgments</a></li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class="ez-toc-link ez-toc-heading-10" href="#Appendix" title="Appendix">Appendix</a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class="ez-toc-link ez-toc-heading-11" href="#More-Related-Articles" title="More Related Articles">More Related Articles</a></li></ul></li></ul></nav></div><h2><span class="ez-toc-section" id="Introduction"></span>Introduction<span class="ez-toc-section-end"></span></h2><p>Modeling a magnet realistically is a task best done numerically.  Even the simplified model of two separated disks with uniform surface magnetization ##\pm~\sigma_M## involves elliptic integrals simplifying assumptions. As a model, the point dipole may be unrealistic to some but the math is tractable and accessible.  The usefulness of the point dipole model in electrodynamics is analogous to that of the point mass in kinematics and dynamics.  It brings forth the salient features and serves as a starting point for more involved treatments.  Indeed, analyses of the falling magnet found in the literature rely on modifications of the point dipole equations with an eye toward bringing their predictions into better agreement with experiments.</p><p>The goal of this article is to gather in one place what can be said about a magnet dipole falling through a solenoid.  A second article, now in preparation, has the same goal but for a point <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-conducting-pipe/">magnetic dipole falling through a solid conducting pipe</a>.</p><h3><span class="ez-toc-section" id="The-motional-emf"></span>The motional emf<span class="ez-toc-section-end"></span></h3><p>The inverse Lorentz transformation equations for the components of the electric and magnetic fields are $$\begin{align}&amp; \mathbf{E}_{\parallel}=\mathbf{E&#8217;}_{\parallel}~;~~\mathbf{B}_{\parallel}=\mathbf{B&#8217;}_{\parallel}\nonumber \\&amp; \mathbf{E}_{\perp} =\gamma\left(\mathbf{E&#8217;}_{\perp} -\mathbf{v}\times\mathbf{B&#8217;}\right)~;~~\mathbf{B}_{\perp}=\gamma\left(\mathbf{B&#8217;}_{\perp}-\frac{1}{c^2}\mathbf{v}\times\mathbf{E&#8217;}\right).\nonumber \end{align}$$</p><p>Consider a magnetic point dipole ##\mathbf{m}=m~\hat z## falling with velocity ##\mathbf{v}=v~\hat z##.  Taking &#8220;down&#8221; as positive  and for ordinary magnet velocities (##\gamma \approx 1##), we write in cylindrical coordinates,$$\mathbf{E}_{\perp}=-\mathbf{v}\times \mathbf{B&#8217;}=-\mathbf{v}\times \mathbf{B}=-v~\hat z\times\left(B_{\rho}~\hat {\rho}+B_z~\hat {z}\right)=-vB_{\rho}~\hat {\theta}.$$The line integral ##\int_a^b \mathbf{E}_{\perp}\cdot d\mathbf{l}## is the motional emf between points ##a## and ##b##.</p><p>The relativistic treatment might be a bit too much, but it helps get the signs right.  Anyway, from here on we will discontinue the use of primes to denote coordinates in the moving frame. Instead, we will use primes for source coordinates and no primes for coordinates of points of interest. Integrals will be taken over primed coordinates.</p><h3><span class="ez-toc-section" id="A-single-ring"></span>A single ring<span class="ez-toc-section-end"></span></h3><p>At point of interest ##\mathbf{r}(\rho ,z)## when the dipole is at ##\mathbf{r&#8217;}(\rho&#8217;,z&#8217;)## the dipolar field is $$\mathbf{B}=\frac{\mu_0}{4\pi} \left[ \frac{ \left(3 \mathbf{m}\cdot (\mathbf{r}-\mathbf{r}&#8217;))(\mathbf{r}-\mathbf{r}&#8217;)-|\mathbf{r}-\mathbf{r}&#8217;|^2 \mathbf{m}\right)}{|\mathbf{r}-\mathbf{r}&#8217;|^5}  -\frac{\mathbf{m}}{ |\mathbf{r}-\mathbf{r}&#8217;|^3}  \right].$$We place the dipole on the ##z##-axis and orient it so that ##\mathbf{m}=m~\hat z##. At  ##\rho=a##, the radial component of the magnetic field is $$B_{\rho}=\frac{3\mu_0ma}{4\pi}  \frac{ z-z&#8217;} {\left[a^2+(z-z&#8217;)^2\right]^{5/2}}.$$</p><p>The motional emf in a coaxial single ring of radius ##a## at ##z&#8217;## is $$\begin{align}\text{emf}_{\text{ring}}=\int \mathbf{E}_{\perp}\cdot d\mathbf{l}=-vB_{\rho}(2\pi a) = -\frac{3\mu_0ma^2}{2}  \frac{ (z-z&#8217;)~v} {\left[a^2+(z-z&#8217;)^2\right]^{5/2}}.\end{align}$$Note that the ring could be an intangible loop, not necessarily conducting, and the emf around it would still exist.  If, however, the ring is a conducting loop, there will be an induced current in it.  A plot of the spatial dependence of the emf is shown in Figure 1 below. We note that when the dipole is farther than about 4 radii from the ring&#8217;s center, the emf is below 1% of its peak value.</p><div id="attachment_41007" style="width: 430px" class="wp-caption aligncenter"><img data-lazyloaded="1" src="data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI0MjAiIGhlaWdodD0iMjU4IiB2aWV3Qm94PSIwIDAgNDIwIDI1OCI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0iI2NmZDRkYiIvPjwvc3ZnPg==" decoding="async" aria-describedby="caption-attachment-41007" class="wp-image-41007" data-src="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring.png" alt="Single ring emf as a function of dipole position" width="420" height="258" data-srcset="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring.png 640w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring-768x471.png 768w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring-705x432.png 705w" data-sizes="(max-width: 420px) 100vw, 420px" /><noscript><img decoding="async" aria-describedby="caption-attachment-41007" class="wp-image-41007" src="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring.png" alt="Single ring emf as a function of dipole position" width="420" height="258" srcset="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring.png 640w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring-768x471.png 768w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emfring-705x432.png 705w" sizes="(max-width: 420px) 100vw, 420px" /></noscript><p id="caption-attachment-41007" class="wp-caption-text">Figure 1. Single ring emf as a function of dipole position.</p></div><p>As the dipole falls in free space,  a wave of motional emf with the waveform shown in Figure 1 is traveling with velocity ##v(t)## down an intangible tube consisting of  loops of radius ##a.##  In the next section, we will explore what to expect when a conducting finite solenoid of radius ##a## occupies part of that space.</p><h3><span class="ez-toc-section" id="From-ring-to-solenoid"></span>From ring to solenoid<span class="ez-toc-section-end"></span></h3><p>We imagine the solenoid as a stack of  ##N## turns. The beginning of the ##k##th turn is the end of the ##(k-1)##th and the end of the ##k##th turn is the beginning of the ##(k+1)##th.  Thus, the turns are a collection of rings connected in series.  We are seeking an expression for the emf across the free ends of a solenoid of ##N## turns, radius ##a##, and length ##L##. It will be an integral of  ##\text{emf}(z)## over the length of the solenoid.</p><p>The number of rings in a stack of height ##dz&#8217;## is ##dn=\dfrac{N}{L}dz&#8217;##. Integrating over the length, $$\begin{align}\text{emf}(z)=\frac{N}{L}\int_{\frac{-L}{2}}^{\frac{L}{2}}\text{emf}_{\text{ring}}dz&#8217;=\frac{\mu _0 N m a^2}{2 L} \left\{ \frac{v}{\left[a^2+\left(z-\frac{L}{2}\right)^2\right]^{3/2}}-\frac{v}{\left[a^2+\left(z+\frac{L}{2}\right)^2\right]^{3/2}}\right\}.\end{align}$$</p><p>We assume that the dipole has mass and is in free fall.  If the solenoid ends are open, there will be an emf across them but no current.  Even when a voltmeter is connected to measure the emf, the current through it is expected to be negligible . In either case, eddy current braking is negligible and we will ignore it together with air resistance.  Under these circumstances, the standard kinematic equations under constant acceleration apply.  Taking the origin at the midpoint of the solenoid, we write an expression for the position of the dipole. We assume that it starts from rest at height ##h## above the top of the solenoid (##z_0=-h-\frac{L}{2}##.) Substituting $$z=(-h-\frac{L}{2}+\frac{1}{2}g~t^2)~\text{ and }~v=g~t$$ in equation (2), we obtain the emf across the solenoid as a function of time. $$\begin{align}\text{emf}(t)=\frac{\mu _0 N m a^2}{2 L} \left\{ \frac{g~t}{\left[a^2+\left(-h-L+\frac{1}{2}g~t^2\right)^2\right]^{3/2}}-\frac{g~t}{\left[a^2+\left(-h+\frac{1}{2}g~t^2\right)^2\right]^{3/2}}\right\}\end{align}$$</p><h3><span class="ez-toc-section" id="The-emf-profile"></span>The emf profile<span class="ez-toc-section-end"></span></h3><p>Figure 2 below is a plot of emf(t) vs ##t## without the overall multiplicative constant.  For the plot, we used  ##g=980## for the acceleration, ##a=1## for the radius, ##h=6a## for the release height above the top and ##L=10 a## for the length. Thus,  if the units for the radius ##a## are centimeters, the numbers on the abscissa will be seconds.  The solid vertical grid lines mark the times when the dipole is at the top, midpoint, and bottom of the solenoid. The pairs of red dashed lines mark the times when the dipole is within one solenoid diameter (##\pm~2a##) from the top (left) and the bottom (right).</p><div id="attachment_41149" style="width: 430px" class="wp-caption aligncenter"><img data-lazyloaded="1" src="data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI0MjAiIGhlaWdodD0iMjQyIiB2aWV3Qm94PSIwIDAgNDIwIDI0MiI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0iI2NmZDRkYiIvPjwvc3ZnPg==" decoding="async" aria-describedby="caption-attachment-41149" class="wp-image-41149" data-src="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3.png" alt="Solenoid emf as a function of time." width="420" height="242" data-srcset="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3.png 640w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3-768x442.png 768w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3-705x405.png 705w" data-sizes="(max-width: 420px) 100vw, 420px" /><noscript><img decoding="async" aria-describedby="caption-attachment-41149" class="wp-image-41149" src="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3.png" alt="Solenoid emf as a function of time." width="420" height="242" srcset="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3.png 640w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3-768x442.png 768w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/emft3-705x405.png 705w" sizes="(max-width: 420px) 100vw, 420px" /></noscript><p id="caption-attachment-41149" class="wp-caption-text">Figure 2. Solenoid emf as a function of time.</p></div><p>Alternatively, we can obtain an expression for the emf as a function of position ##z## by substituting ##v=\sqrt{2g[z-(-h-\frac{L}{2})]}## in equation (2) to get $$\begin{align}\text{emf}(z)=\frac{\mu _0 N m a^2}{2 L} \left\{\frac{\sqrt{2 g \left(z+h+\frac{L}{2}\right)}}{\left[a^2+\left(z-\frac{L}{2}\right)^2\right]^{3/2}}-\frac{\sqrt{2 g \left(z+h+\frac{L}{2}\right)}}{\left[a^2+\left(z+\frac{L}{2}\right)^2\right]^{3/2}}\right\}.\end{align}$$A plot of the emf(##z##) vs. ##z## is not shown because it is identical to the one in figure 2 except for the labeling of the abscissa. Both plots feature extrema, one negative and one positive, with a region between when/where the emf is near zero. This is qualitatively consistent with Faraday&#8217;s law. The magnetic flux changes most rapidly when the dipole is near the ends and hardly at all when it is near the middle of the solenoid.</p><h3><span class="ez-toc-section" id="Term-corrections"></span>Term corrections<span class="ez-toc-section-end"></span></h3><p>Examination of figure 2 reveals that the extrema occur when the dipole is near the ends of the solenoid. At ##z=+L/2## the first term in equation (4) has a maximum while the second term is small; the reverse is true at ##z=-L/2##. An exact determination of the extrema would involve solving ##\frac{d}{dz}\text{emf}(z)=0##, a daunting task. Considering, however, that the true extrema are very close to ##z=\pm \frac{L}{2}##, it is expedient to do series expansions of equation (4) about ##z=\pm \frac{L}{2}## to second-order and find corrections to the zeroth-order terms.  Here, we summarize the results and relegate some of the details to the Appendix.</p><p>The fractional corrections to the position and value of the minimum are $$\frac{\delta z}{(-L/2)}=-5.5\times 10^{-3}~;~~\frac{\delta(\text{emf})}{\text{emf}(-L/2)}=1.1\times 10^{-3}.$$The corresponding numbers for the maximum are $$\frac{\delta z}{(+L/2)}=2.1\times 10^{-3}~;~~\frac{\delta( \text{emf})}{\text{emf}(L/2)}=1.6\times 10^{-4}.$$The numbers show that the corrections have a small effect on the zeroth order terms. Thus, we will ignore them and make the algebra easier.</p><h3><span class="ez-toc-section" id="A-relation-between-amplitudes"></span>A relation between amplitudes<span class="ez-toc-section-end"></span></h3><p>We define as amplitudes the magnitudes of the extrema of ##\text{emf}(t)## (Figure 2). In the approximation that the extrema occur at ##z=\pm ~\frac{1}{2}##,  equation (4) gives, the amplitudes $$\begin{align} &amp; A_1=\left|\text{emf}(-L/2)\right|=\sqrt{2 g h}\left[\frac{1}{a^3}-\frac{1}{\left(a^2+L^2\right)^{3/2}}\right]\nonumber \\ &amp;  A_2=\text{emf}(+L/2)|=\sqrt{2 g (h+L)}\left[\frac{1}{a^3}-\frac{1}{\left(a^2+L^2\right)^{3/2}}\right].\nonumber \end{align}$$</p><p>We write the difference between the two equations as $$\begin{align} &amp;  A_2-A_1=\text{const.} \left(\sqrt{\frac{2 (h+L)}{g}}-\sqrt{\frac{2 h}{g}}\right) \nonumber \\ &amp; \text{where  }\text{const.}=\frac{\mu _0 N m a^2}{2 g L}\left[\frac{1}{a^3}-\frac{1}{\left(a^2+L^2\right)^{3/2}}\right].\nonumber \end{align}$$We recognize the expression in parentheses multiplying the constant as the transit time ##T## of the dipole from one end of the solenoid to the other. This simplifies the equation even more to $$\begin{align}A_2-A_1=(\text{const.})\times T.\end{align}$$</p><p>It is easy to imagine an experimental protocol to test equation (5) with equipment that samples and records the emf across the solenoid as a function of time. However, one should ensure that starting heights ##h## are greater than the effective dipole range of 4 solenoid radii. This will minimize experimental uncertainties that become significant at small values of ##h##. A PF thread posted by @billy_t <a href="https://www.physicsforums.com/threads/drop-height-of-a-magnet-vs-induced-emf-in-a-solenoid.1014918/#post-6629806">here</a> describes such an experiment. However, the analysis focused on the amplitudes themselves and not on their relation to the time of transit.</p><h2><span class="ez-toc-section" id="Summary"></span>Summary<span class="ez-toc-section-end"></span></h2><p>Modeling a magnet as a point dipole falling through a solenoid is sufficient to provide a sensible description of the shape and general features of the emf(##t##) curve that is in agreement with what one would expect qualitatively.  The flux varies most rapidly when the dipole is entering and exiting the solenoid and least rapidly when it is near the solenoid&#8217;s middle. The amplitude of the right peak is greater than the left. This happens because its width is narrower since the dipole is moving faster, yet the area under the curve is the same. It is unlikely that a more realistic model of the magnet will result in a different interpretation of these features. A sophisticated model needs sophisticated data analysis to match it, perhaps one involving a point-by-point simulation of the entire curve.</p><h2><span class="ez-toc-section" id="Acknowledgments"></span>Acknowledgments<span class="ez-toc-section-end"></span></h2><p>I thank @billyT_ for posting the thread quoted above; it seeded this insight. Also, the participation of @haruspex and @Charles Link honed my thinking and is greatly appreciated.</p><h2><span class="ez-toc-section" id="Appendix"></span>Appendix<span class="ez-toc-section-end"></span></h2><p>The work shown here was Mathematica-assisted.  In equation (4) we first drop ##g## and substitute reduced lengths ##\lambda=L/a,~\eta=h/a##. Then expand the equation for small values of ##\epsilon = z\pm \lambda/2## to second order.  Each expansion yielded a quadratic expression ##A\epsilon^2+B\epsilon+C.##  Setting the derivative equal to zero and solving for ##\epsilon## gives ##\epsilon=\frac{-B}{2A}##.  Omitting the gory details, the corrections for the minimum on the left and the maximum on the right are:</p><p>$$\begin{align} &amp; \epsilon_{\text{min}}  =\frac{2 \eta \left(\lambda ^2+1\right) \left[\left(\lambda ^2+1\right)^{5/2}-\left(6 \eta \lambda +\lambda ^2+1\right)\right]}{\left(12 \eta ^2+1\right) \left(\lambda ^2+1\right)^{7/2}+12 \eta ^2 \left(4 \lambda ^2-1\right)+12 \eta \lambda \left(\lambda ^2+1\right)-\left(\lambda ^2+1\right)^2}\nonumber \\ &amp; \epsilon_{\text{max}}  =\frac{ 2 \left(\lambda ^2+1\right) (\eta +\lambda ) \left[6 \eta \lambda +\left(\lambda ^2+1\right)^{5/2}+5 \lambda ^2-1\right]}    {12 \eta ^2 \left(4 \lambda ^2-1\right)+12 \eta \lambda\left(7 \lambda ^2-3\right)  +\left(\lambda ^2+1\right) \left[12 \left(\lambda ^2+1\right)^{5/2} (\eta +\lambda )^2+1\right]+\left(35 \lambda ^4-26 \lambda ^2-1\right)}\nonumber \end{align}$$Figure 3 shows a detail of figure 2 in the vicinity of the maximum.  The vertical line is at the corrected value of ##z/a##.  The zeroth order term of the expansion is to its left at ##z/a=5.00##.</p><div id="attachment_41022" style="width: 320px" class="wp-caption aligncenter"><img data-lazyloaded="1" src="data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIzMTAiIGhlaWdodD0iMTg0IiB2aWV3Qm94PSIwIDAgMzEwIDE4NCI+PHJlY3Qgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0iI2NmZDRkYiIvPjwvc3ZnPg==" decoding="async" aria-describedby="caption-attachment-41022" class="wp-image-41022" data-src="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction.png" alt="" width="310" height="184" data-srcset="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction.png 640w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction-768x456.png 768w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction-705x418.png 705w" data-sizes="(max-width: 310px) 100vw, 310px" /><noscript><img decoding="async" aria-describedby="caption-attachment-41022" class="wp-image-41022" src="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction.png" alt="" width="310" height="184" srcset="https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction.png 640w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction-768x456.png 768w, https://www.physicsforums.com/insights/wp-content/uploads/2022/05/Correction-705x418.png 705w" sizes="(max-width: 310px) 100vw, 310px" /></noscript><p id="caption-attachment-41022" class="wp-caption-text">Figure 3. Detail of the correction to the zeroth order term.</p></div></div><div class="saboxplugin-wrap" itemtype="http://schema.org/Person" itemscope itemprop="author"><div class="saboxplugin-tab"><div class="saboxplugin-gravatar"></div><div class="saboxplugin-authorname"><a href="https://www.physicsforums.com/insights/author/kuruman/" class="vcard author" rel="author"><span class="fn">kuruman</span></a></div><div class="saboxplugin-desc"><div itemprop="description"><p>I am a retired university physics professor. I have done research in biological physics, mostly studying the magnetic and electronic properties at the active sites of biomolecules and their model complexes. I have also dabbled in Physics Education research.</p></div></div><div class="clearfix"></div></div></div><style>.lwrp.link-whisper-related-posts{ margin-top: 25px; margin-bottom: 25px; } .lwrp .lwrp-title{ }.lwrp .lwrp-description{ } .lwrp .lwrp-list-container{ } .lwrp .lwrp-list-multi-container{ display: flex; } .lwrp .lwrp-list-double{ width: 48%; } .lwrp .lwrp-list-triple{ width: 32%; } .lwrp .lwrp-list-row-container{ display: flex; justify-content: space-between; } .lwrp .lwrp-list-row-container .lwrp-list-item{ width: calc(33% - 20px); } .lwrp .lwrp-list-item:not(.lwrp-no-posts-message-item){ margin-bottom: 25px; } .lwrp .lwrp-list-item img{ max-width: 100%; height: auto; object-fit: cover; aspect-ratio: 1 / 1; } .lwrp .lwrp-list-item.lwrp-empty-list-item{ background: initial !important; } .lwrp .lwrp-list-item .lwrp-list-link .lwrp-list-link-title-text, .lwrp .lwrp-list-item .lwrp-list-no-posts-message{ }@media screen and (max-width: 480px) { .lwrp.link-whisper-related-posts{ } 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itemprop="author" itemscope="itemscope" itemtype="https://schema.org/Person" ><span itemprop='name'>kuruman</span></span><span class='av-structured-data' itemprop="datePublished" datetime="2022-06-05T12:48:47-05:00" >2022-06-05 12:48:47</span><span class='av-structured-data' itemprop="dateModified" itemtype="https://schema.org/dateModified" >2022-06-11 07:07:53</span><span class='av-structured-data' itemprop="mainEntityOfPage" itemtype="https://schema.org/mainEntityOfPage" ><span itemprop='name'>How to Model a Magnet Falling Through a Solenoid</span></span></span></article><div class='single-big'></div><div class='related_posts clearfix av-related-style-full'><h5 class="related_title">You might also like</h5><div class="related_entries_container"><div class='av_one_half no_margin alpha relThumb relThumb1 relThumbOdd post-format-standard related_column'><a href='https://www.physicsforums.com/insights/symmetry-arguments-and-the-infinite-wire-with-a-current/' class='relThumWrap 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data-av_iconfont='entypo-fontello'></span></span></span><strong class="av-related-title">An Alternative Approach to Solving Collision Problems</strong></a></div></div></div><div class='comment-entry post-entry'><div class='comment_meta_container'><div class='side-container-comment'><div class='side-container-comment-inner'> <span class='comment-count'>11</span> <span class='comment-text'>replies</span> <span class='center-border center-border-left'></span> <span class='center-border center-border-right'></span></div></div></div><div class='comment_container'><ol class="commentlist" id="comments"><li class="comment byuser comment-author-hornbein even thread-even depth-1" id="li-comment-12938"><div id="comment-12938"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">Hornbein</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12938"> <time > August 21, 2022 at 1:21 am </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">I don&#039;t watch any narrative fiction. I edit other people&#039;s videos to my own satisfaction instead. This morning I paired a vid of Koreans making steel out of scrap metal with the heavy metal song Loving You Is A Dirty Job.</div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-vanhees71 odd alt thread-odd thread-alt depth-1" id="li-comment-12872"><div id="comment-12872"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">vanhees71</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12872"> <time > June 7, 2022 at 12:59 pm </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">But you can make your Insight easily complete by using your point-dipole model by calculating the equation of motion for that point dipole, falling through the pipe. See the quoted papers above.</div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-charles-link even thread-even depth-1" id="li-comment-12871"><div id="comment-12871"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">Charles Link</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12871"> <time > June 7, 2022 at 12:39 pm </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">See <a href="https://www.physicsforums.com/threads/drop-height-of-a-magnet-vs-induced-emf-in-a-solenoid.1014918/" class="link link--internal">https://www.physicsforums.com/threads/drop-height-of-a-magnet-vs-induced-emf-in-a-solenoid.1014918/</a> (thread by <a href="https://www.physicsforums.com/members/696247/" class="username" data-xf-init="member-tooltip" data-user-id="696247" data-username="@billyt_">@billyt_</a> )<br /> posts 25 and 39. I think the transition from point magnetic dipole to an extended permanent magnet is fairly straightforward, because we are taking ## d \Phi /dt ## to compute the EMF, and thereby the integral (## d \Phi/dt ## rather than ## \Phi ##) over the length of the solenoid becomes routine. The magnetic dipole is also a very good calculation, but if we can do a finite length permanent magnet in closed form, that is even somewhat better. It&#039;s a little bit of extra work, but writing out the terms for the case of the extended permanent magnet would make the reference more complete.</p><p>Edit: On second thought, it is perhaps worth mentioning, but really not worth the trouble of displaying all the terms, since it is too lengthy.</p></div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-kuruman bypostauthor odd alt thread-odd thread-alt depth-1" id="li-comment-12870"><div id="comment-12870"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">kuruman</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12870"> <time > June 7, 2022 at 12:08 pm </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">&quot;<br /> He is using a different model for the magnet, i.e., a cylinder of homogeneous magnetization, which has been demonstrated to be a quantitatively better model for the experiment with a real bar magnet than the dipole approximation:</p><p><a href="https://doi.org/10.1119/1.4864278" target="_blank" class="link link--external" rel="nofollow ugc noopener">https://doi.org/10.1119/1.4864278</a><br /> &quot;<br /> Yes, I agree that the permanently magnetized disk model produces a better quantitative model if one has the data on one hand and the sophisticated analysis on the other. Like I said, the point dipole as a model is as useful and as accurate as the point mass is in mechanics.</p><p>My goal with these articles is modest: to show, at the intermediate undergraduate level, that the point dipole model goes a long way towards understanding what is physically going on with a magnet falling through a solenoid and a pipe. It&#039;s not perfect, but it does a descent job and any refinements to the model for a magnet are not going to change the basic features of the point dipole model. A secondary goal is to gather these ideas all in one place as an easily accessible reference for PF users who have questions related to this material.</p></div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-charles-link even thread-even depth-1" id="li-comment-12869"><div id="comment-12869"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">Charles Link</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12869"> <time > June 7, 2022 at 11:53 am </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">With this topic, (of the permanent magnet), perhaps it would even be useful for some readers to read come of the fundamentals. See<br /> <a href="https://www.physicsforums.com/threads/permanent-magnets-described-by-magnetic-surface-currents-comments.900528/" class="link link--internal"></a><a href="https://www.physicsforums.com/threa" rel="ugc">https://www.physicsforums.com/threa</a>&#8230;by-magnetic-surface-currents-comments.900528/<br /> and<br /> <a href="https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/" class="link link--internal">https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/</a></p><p>When I was in college, (many years ago), we were taught the pole method, but it is important to connect that method to the magnetic surface current method, because otherwise, it is a lot of incomplete handwaving, where you have static fictitious charges creating a magnetic field. Fortunately, as it turns out, the pole method follows from the magnetic surface current method, and the two get the exact same answer for the magnetic field.</p><p>( I don&#039;t want to steer the reader away from the problem of the permanent magnet falling through the solenoid, but it is kind of important to have complete physics for the permanent magnet, in order to work this problem).</p></div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-vanhees71 odd alt thread-odd thread-alt depth-1" id="li-comment-12868"><div id="comment-12868"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">vanhees71</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12868"> <time > June 7, 2022 at 8:22 am </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">You can also complete the article by actually describe the motion of the magnet itself. It&#039;s nicely described, e.g., in</p><p><a href="https://doi.org/10.1119/1.2203645" target="_blank" class="link link--external" rel="nofollow ugc noopener">https://doi.org/10.1119/1.2203645</a></p><p>He is using a different model for the magnet, i.e., a cylinder of homogeneous magnetization, which has been demonstrated to be a quantitatively better model for the experiment with a real bar magnet than the dipole approximation:</p><p><a href="https://doi.org/10.1119/1.4864278" target="_blank" class="link link--external" rel="nofollow ugc noopener">https://doi.org/10.1119/1.4864278</a></div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-kuruman bypostauthor even thread-even depth-1" id="li-comment-12867"><div id="comment-12867"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">kuruman</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12867"> <time > June 6, 2022 at 12:54 pm </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">&quot;<br /> Very good article <a href="https://www.physicsforums.com/members/192687/" class="username" data-xf-init="member-tooltip" data-user-id="192687" data-username="@kuruman">@kuruman</a> :). Besides the thread by <a href="https://www.physicsforums.com/members/696247/" class="username" data-xf-init="member-tooltip" data-user-id="696247" data-username="@billyt_">@billyt_</a> mentioned in the above article, it may also be of interest to see posts 122-123 and 138-139 of <a href="https://www.physicsforums.com/threads/calculating-magnetic-field-strength-of-a-magnet.1005917/page-4" class="link link--internal">https://www.physicsforums.com/threads/calculating-magnetic-field-strength-of-a-magnet.1005917/page-4</a><br /> <a href="https://www.physicsforums.com/members/635497/" class="username" data-xf-init="member-tooltip" data-user-id="635497" data-username="@hutchphd">@hutchphd</a> and <a href="https://www.physicsforums.com/members/575631/" class="username" data-xf-init="member-tooltip" data-user-id="575631" data-username="@bob012345">@bob012345</a> had very good inputs in helping to solve the problem of the EMF of a magnet moving through a ring.<br /> &quot;<br /> Thank you for liking the article and for pointing out this other thread. Part of my motivation for this article was to consolidate in one place what can be said about magnets falling through a single ring and a solenoid assuming the point dipole approximation. My hope is that it can serve as a reference for future questions and as a starting point for more complicated magnet points.</p><p>I should mention (advertise?) that I am now putting the finishing touches on a companion article that is intended to serve the same purpose but for a magnet falling through a solid conducting pipe. It should be ready in a few days.</p></div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-charles-link odd alt thread-odd thread-alt depth-1" id="li-comment-12866"><div id="comment-12866"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">Charles Link</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12866"> <time > June 6, 2022 at 12:24 pm </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">Very good article <a href="https://www.physicsforums.com/members/192687/" class="username" data-xf-init="member-tooltip" data-user-id="192687" data-username="@kuruman">@kuruman</a> :). Besides the thread by <a href="https://www.physicsforums.com/members/696247/" class="username" data-xf-init="member-tooltip" data-user-id="696247" data-username="@billyt_">@billyt_</a> mentioned in the above article, it may also be of interest to see posts 122-123 and 138-139 of <a href="https://www.physicsforums.com/threads/calculating-magnetic-field-strength-of-a-magnet.1005917/page-4" class="link link--internal">https://www.physicsforums.com/threads/calculating-magnetic-field-strength-of-a-magnet.1005917/page-4</a><br /> <a href="https://www.physicsforums.com/members/635497/" class="username" data-xf-init="member-tooltip" data-user-id="635497" data-username="@hutchphd">@hutchphd</a> and <a href="https://www.physicsforums.com/members/575631/" class="username" data-xf-init="member-tooltip" data-user-id="575631" data-username="@bob012345">@bob012345</a> had very good inputs in helping to solve the problem of the EMF of a magnet moving through a ring.</div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-delta2 even thread-even depth-1" id="li-comment-12865"><div id="comment-12865"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">Delta2</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12865"> <time > June 6, 2022 at 8:05 am </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">&quot;<br /> When you say &quot;not so simple&quot;, what specifically are you referring to? The math or the physics? Both, I think, are appropriate to the undergraduate intermediate level.<br /> &quot;<br /> Yes I agree they are appropriate for undergraduate level, it&#039;s just that they are not so simple as for e.g. the emf of a rotating ring in a uniform B. The expressions look a bit complex, main responsible for this is the form of the B- field from a point dipole.</div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-kuruman bypostauthor odd alt thread-odd thread-alt depth-1" id="li-comment-12864"><div id="comment-12864"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">kuruman</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12864"> <time > June 6, 2022 at 7:50 am </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">&quot;<br /> Great article for an experiment that is relatively simple to setup and perform, however it has a not so simple detailed explanation using the laws of classical electromagnetism.<br /> &quot;<br /> When you say &quot;not so simple&quot;, what specifically are you referring to? The math or the physics? Both, I think, are appropriate to the undergraduate intermediate level.</div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li><li class="comment byuser comment-author-delta2 even thread-even depth-1" id="li-comment-12863"><div id="comment-12863"><article><div class="gravatar"></div><div class='comment_content'><header class="comment-header" aria-label="Comment Info" > <cite class="author_name heading"><cite class="comment_author_name">Delta2</cite></cite> <span class="says">says:</span><div class="comment-meta commentmetadata"> <a href="https://www.physicsforums.com/insights/how-to-model-a-magnet-falling-through-a-solenoid/#comment-12863"> <time > June 5, 2022 at 10:48 pm </time> </a></div></header><div class='comment_text entry-content-wrapper clearfix' ><div class="bbWrapper">Great article for an experiment that is relatively simple to setup and perform, however it has a not so simple detailed explanation using the laws of classical electromagnetism.</div> <a rel="nofollow" class="comment-reply-login" href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">Log in to Reply</a></div></div></article></div></li></ol></div><div class="comment_container"><h3 class='miniheading '>Leave a Reply</h3><span class="minitext">Want to join the discussion?<br />Feel free to contribute!</span><div id="respond" class="comment-respond"><h3 id="reply-title" class="comment-reply-title">Leave a Reply <small><a rel="nofollow" id="cancel-comment-reply-link" href="/insights/how-to-model-a-magnet-falling-through-a-solenoid/#respond" style="display:none;">Cancel reply</a></small></h3><p class="must-log-in">You must be <a href="https://www.physicsforums.com/insights/wp-login.php?redirect_to=https%3A%2F%2Fwww.physicsforums.com%2Finsights%2Fhow-to-model-a-magnet-falling-through-a-solenoid%2F">logged in</a> to post a comment.</p></div></div></div></main><aside class='sidebar sidebar_right smartphones_sidebar_active alpha units' aria-label="Sidebar" role="complementary" itemscope="itemscope" itemtype="https://schema.org/WPSideBar" ><div class="inner_sidebar extralight-border"><section id="custom_html-2" class="widget_text widget clearfix widget_custom_html"><div class="textwidget custom-html-widget"><h2 class="widgettitle">Trending Articles</h2><ul class="wpp-list wpp-sunset"><li class="" style="--item-position: 1;--total-items: 10"><div class="item-position"></div><div class="item-data"><a href="https://www.physicsforums.com/insights/can-see-atom/" class="wpp-post-title" target="_self">Can We See an Atom?</a></div></li><li class="" style="--item-position: 2;--total-items: 10"><div class="item-position"></div><div class="item-data"><a href="https://www.physicsforums.com/insights/hand-wavy-discussion-planck-length/" class="wpp-post-title" target="_self">What Planck Length Is and It’s Common Misconceptions</a></div></li><li class="" style="--item-position: 3;--total-items: 10"><div class="item-position"></div><div class="item-data"><a href="https://www.physicsforums.com/insights/what-is-a-tensor/" class="wpp-post-title" target="_self">What Is a Tensor? 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