biot-savart_law

Biot-Savart law

 Stan Zurek, Biot-Savart law, Encyclopedia Magnetica, E-Magnetica.pl

Biot-Savart law - the mathematical equation describing the relationship between magnetic field strength H (or flux density B) and the electric current generating it.

The law is consistent with both Ampère's circuital law and Gauss's law for magnetism, but it only describes magnetostatic conditions. A generalised modification which includes displacement currents is known as Jefimenko equations.

It is named after Jean-Baptiste Biot and Félix Savart, who discovered this relationship in 1820.1) In the original publication only the directions and amplitudes were specified.

 * Helpful page? Support us! → All we need is just $0.25 per month. Come on… Mathematical equations Fig. 1. Illustration of Biot-Savart law:$P$- point with magnetic field strength$\vec H$,$R$- perpendicular distance from conductor,$I$- current,$\vec J$- current density,$d \vec l$- infinitesimal length of conductor contributing to magnetic field,$r$- distance,$\vec r$- path vector,$\hat r$- unit vector along direction of$r$,$\theta$- angle between$d \vec l$and$r$The Biot-Savart law allows calculating the value of magnetostatic field at any point in space, due to electrical current. There are several ways in which the relationship can be expressed. Similarly to the Maxwell equations, they can be written in a differential or integral form. The various forms are used mostly because of the ease of analytical or numerical calculations as required for a given application. The integration is always required for calculation of the value of H or B (even if the integral is solved analytically, for simple geometries). Fig. 2. Electric current I generates magnetic field strength H whose vector is always perpendicular to the direction of I, according to the right-hand rule Table 1 shows examples of three different ways in which the equations (all are equivalent) can be expressed in a differential form, as encountered in the literature. The variables in the equation are explained in the illustration in Fig. 1. The value and direction of the vector of magnetic field strength$H$at point$P$at a distance$R$from the conductor with current$I$is contributed by each infinitesimal part$d \vec l$of the conductor. The magnetic field is always generated in a plane perpendicular to the current and circulating around the current according to the right-hand rule (Fig. 2). For a straight conductor the contribution to magnetic field is the greatest directly above the the infinitesimal part (sin(90°) is unity), and diminishes away from it proportionally to the angle (which at infinity becomes sin(0°) is zero) - as illustrated in Fig. 3. A similar distribution of magnetic field in space is around a moving electrical charge (Fig. 4). Table 1. Various versions of equations of the Biot-Savart law 2) Based on value of current$I$Based on current density$\vec J$Based on electric charge$q$and its velocity$\vec v$$$d \vec H = \frac{1}{4·π}·\frac{I·(d \vec l \times \hat r)}{r^2}$$ or $$d \vec H = \frac{1}{4·π}·\frac{I·(d \vec l \times \vec r)}{r^3}$$ where direction of$d \vec H$is perpendicular to$d \vec l$and$\hat r$(and defined by the right-hand rule) $$d \vec H = \frac{1}{4·π}·\frac{\vec J \times \vec r}{r^3}·dV$$ where $$J = \frac{I}{A} = \frac{I·dl}{A·dl} = \frac{I·dl}{dV}$$ and$A$is cross-section area of the conductor with current and$V$is its volume $$d \vec H = \frac{1}{4·π}·q·\frac{(\vec v \times \vec r)}{r^3}$$ (see also Fig. 4) Fig. 3. Normalised$H$contributed by infinitesimal part$d \vec l$at changing distance (and the corresponding angle) from it3) Fig. 4. Magnetic field around a moving electron (because of the convention the electron moves in the opposite direction to electric current)4) H and B implications In the original experiments, Biot and Savart used a magnetic setup in which there was a magnetic force (from a conductor with current I) acting on magnetic pieces. The experiment was carried out in air, whose permeability can be very closely approximated with the permeability of free space. Therefore, in air, the equations can be expressed both, for magnetic field strength as H = fH (I), or for flux density as B = fB (I) which is equivalent to B = µ0 · fH(I). In most publications the equations are specified such that the value of B 5)6)7)8)9)10)11)12) is calculated, but the calculation of H 13)14) is more fundamental, because it is independent of the permeability of the medium. If the conductor with current is placed in a medium of different permeability (e.g. relative permeability of air is 1.00000037 and not exactly unity, as for vacuum) then this would have to be taken into account, additionally. In the literature, if the for calculating B are given with the implicit, but usually not stated assumptions that the equations apply strictly only in vacuum (free space). However, in any other uniform and isotropic medium (diamagnetic, paramagnetic, antiferromagnetic, ferrimagnetic or ferromagnetic) the equation must be adjusted with the relative permeability of that medium. This problem does not arise for the equations with H, because they are correct for any medium (as long as the medium is uniform and isotropic). Table 2. Examples of differential and integral equations used in literature Differential form for$H$15) Integral form for$H$16) Differential form for$B$17) Integral form for$B$18) $$d \vec H = \frac{I}{4·π}·\frac{d \vec l \times \hat r}{r^2}$$ $$\vec H = \frac{I}{4·π}·\int \frac{d \vec l \times \hat r}{r^2}$$ $$d \vec B = \frac{\mu_0·I}{4·π}·\frac{d \vec l \times \hat r}{r^2}$$ $$\vec B = \frac{\mu_0·I}{4·π}·\int \frac{d \vec l \times \hat r}{r^2}$$ Note: the symbols in these equations were modified to conform to those used in Fig. 1. Examples of calculations Straight conductor Fig. 5. Vector$\vec H$at the point$P$, generated by current$I$in infinitely long straight wire Starting with the integral form: 19) $$\vec H = \frac{I}{4·\pi} \int_{l={-\infty}}^{+\infty} \frac{d \vec l \times \hat r}{r^2}$$ Using the symmetry assumption, the integral can be evaluated only from 0 to +∞, and only for the amplitude (because the direction and sense of the vector is known from the right-hand rule). Also, the variables can be expressed in a different way, to allow easy calculation of the integral. An angle$\theta$can be defined as the angle between$r$and the direction of the constant$I$(Fig. 5). The length$l$is measured along the axis of$I$. $$\frac{H}{2} = \frac{I}{4·\pi} \int_{l=0}^{+\infty} \frac{dl · \sin \theta}{r^2}$$ Further,$\sin \theta = \frac{R}{r}$which can be rearranged as$r = \frac{R}{\sin \theta}$. Additionally, it can be written that$\tan \theta = \frac{R}{l}$, so$l = \frac{R}{\tan \theta} = R · \cot \theta$which can be differentiated, so$dl = R· csec^2 \theta d \theta = \frac{-R}{\sin^2 \theta} d \theta$. These can be substituted into the main equation and simplified so: $$\frac{H}{2} = \frac{I}{4·\pi} \int_{\theta=\frac{\pi}{2}}^{0} \frac{-R}{\sin^2 \theta}·d \theta · \sin \theta · \frac{\sin^2 \theta}{R^2} = \frac{I}{4·\pi} \int_{\theta=\frac{\pi}{2}}^{0} \frac{-\sin \theta}{R} d \theta$$ The integral of$-\sin \theta$is$\cos \theta$and$R$is constant, hence: $$\frac{H}{2} = \frac{I}{4·\pi·R} · \left( \cos 0 - \cos \frac{\pi}{2} \right) = \frac{I}{4·\pi·R} · \left( 1 - 0 \right) = \frac{I}{4·\pi·R}$$ The final value needs to be doubled (due to earlier assumption of symmetry), so:  $$H = \frac{I}{2·\pi·R}$$ (A/m) Therefore, the magnetic field at a point P at a distance of R from a straight conductor is directly proportional to the amplitude of current and inversely proportional to the magnetic path length (i.e. circumference of the circle with radius R). Loop of wire Fig. 6. Vector of$\vec H$at the point$P$at the centre of a loop with current$I$The calculation of H at the centre of a loop of wire with current is a basis for calculating the magnetic field of a solenoid. The integral has to be calculated over the whole circumference of the loop of wire (Fig. 6), and it is somewhat simpler than for the case of the straight wire:20) $$\vec H = \frac{I}{4·\pi · r^2} · \int_{loop} d \vec l \times \hat r$$ But$d \vec l \times \hat r = dl · \sin \theta $. However, because of the circular shape, and the point being placed exactly at the centre, then for every$d \vec l$the angle$\theta$is always 90°, so sin(90°)=1, hence$d \vec l \times \hat r = dl$, which leads to: $$H = \frac{I}{4·\pi·r^2} · \int_{loop} dl$$ The integral of$dl$around a circle is just the circumference of the circle so: $$H = \frac{I}{2·\pi·r^2} · ( 2 · \pi · r)$$ And noting that$r = R\$ and applying simplifications, the final value is:

 $$H = \frac{I}{2·R}$$ (A/m)