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I'm trying to plot magnetic field lines for a rectangular bar magnet in 3 dimensional space. The position of the magnet is known and we wish to produce a plot such as the one shown below.

Can something like the Biot Savart law be modified for an exact or approximate solution to this to enable plottingenter image description here

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You can model a bar magnet by a rectangular box with a constant magnetization in one direction. Let's take the box $[0,a]\times[0,b]\times[0,c]$, with a constant magnetization $\mathbf M(\mathbf x) = M_0 \ \hat{\mathbf k}$, where $\hat{\mathbf k}$ is the unit vector in the $z$ direction. The bound volume and surface current densities are: $$\mathbf J_b(\mathbf x) = \boldsymbol{\nabla}\times\mathbf M(\mathbf x)$$ $$\mathbf K_b(\mathbf x) = \mathbf M(\mathbf x) \times \hat {\mathbf n}$$ The volume current density is zero because $\mathbf M$ is constant. For the surface current density, the top and bottom faces don't contribute since $M_0 \hat{\mathbf k}\times\hat {\mathbf k}=0$. For the other four faces we have: $$\mathrm{x=0 \ face:} \ \mathbf K_1 = M_0 \ \hat{\mathbf k}\times (-\hat{\mathbf i}) = -M_0 \ \hat{\mathbf j}$$ $$\mathrm{x=a \ face:} \ \mathbf K_2 = M_0 \ \hat{\mathbf k}\times \hat{\mathbf i} = M_0 \ \hat{\mathbf j}$$ $$\mathrm{y=0 \ face:} \ \mathbf K_3 = M_0 \ \hat{\mathbf k}\times (-\hat{\mathbf j}) = M_0 \ \hat{\mathbf i}$$ $$\mathrm{y=b \ face:} \ \mathbf K_4 = M_0 \ \hat{\mathbf k}\times \hat{\mathbf j} = -M_0 \ \hat{\mathbf i}$$ Now that you know the bound current distribution, you can simply use the Biot-Savart law to calculate the magnetic field: $$\mathbf B(\mathbf x) = \frac{\mu_0}{4\pi}\int_{\mathbb S}d\mathbf a' \ \mathbf K(\mathbf x') \times \frac{\mathbf{x-x'}}{|\mathbf{x-x'}|^3}$$ I believe you can take it from here.

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  • $\begingroup$ Thanks is that equations for magnetic field inside the magnet? $\endgroup$ – ffejrekaburb Jul 20 '18 at 16:59
  • $\begingroup$ I gave you equations for the bound surface currents on the magnet. You can derive the magnetic fields yourself by using Biot-Savart's law and evaluating the integral (the last equation in my answer). $\endgroup$ – Sahand Tabatabaei Jul 20 '18 at 18:25

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