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Biot-Savart Law: How to Calculate Arbitrary Magnetic Fields

Important Formula

Formula: Biot-Savart law

$$\class{violet}{\boldsymbol{B}}(\boldsymbol{r}) ~=~ \frac{\mu_0}{4\pi} \int_{V} \class{red}{\boldsymbol{j}} \, (\boldsymbol{R}) \times \frac{\boldsymbol{r}-\boldsymbol{R}}{|\boldsymbol{r}-\boldsymbol{R}|^3} ~ \text{d}v$$ $$\class{violet}{\boldsymbol{B}}(\boldsymbol{r}) ~=~ \frac{\mu_0}{4\pi} \int_{V} \class{red}{\boldsymbol{j}} \, (\boldsymbol{R}) \times \frac{\boldsymbol{r}-\boldsymbol{R}}{|\boldsymbol{r}-\boldsymbol{R}|^3} ~ \text{d}v$$

What do the formula symbols mean?

Magnetic flux density (B-field)

$$ \class{violet}{\boldsymbol{B}}(\boldsymbol{r}) $$ Unit $$ \mathrm{T} = \frac{\mathrm{kg}}{\mathrm{A} \, \mathrm{s}^2} $$

Magnetic flux density indicates how strong the magnetic field is at the location $ \boldsymbol{r} $, which is generated by a stationary current or a (not too fast) point charge.

Position vector to the field point

$$ \boldsymbol{r} $$ Unit $$ \mathrm{m} $$

The position vector to the field point at which the B field is to be calculated using the Biot-Savart law.

Position vector to the volume element

$$ \boldsymbol{R} $$ Unit $$ \mathrm{m} $$

Position vector pointing to the infinitesimal volume element $\text{d}v$.

Electric Current Density

$$ \class{red}{\boldsymbol j} $$ Unit $$ \frac{ \mathrm A }{ \mathrm{m}^2 } $$

Current per volume at the location $ \boldsymbol{R} $ of the infinitesimal volume element.

Volume

$$ V $$ Unit $$ \mathrm{m}^3 $$

The volume of the extended conductor.

Vacuum permeability

$$ \mu_0 $$ Unit $$ \frac{\mathrm{Vs}}{\mathrm{Am}} = \frac{ \mathrm{kg} \, \mathrm{m} }{ \mathrm{A}^2 \, \mathrm{s}^2 } $$

The vacuum permeability is a physical constant and has the following experimentally determined value: $$ \mu_0 ~=~ 1.256 \, 637 \, 062 \, 12 ~\cdot~ 10^{-6} \, \frac{\mathrm{Vs}}{\mathrm{Am}} $$

Table of contents

Important Formula
Exercises with Solutions

The Biot-Savart law is used to calculate the magnetic field $ \class{violet}{\boldsymbol B} $ generated by the current density $ \class{red}{\boldsymbol j} $ in a volume element $ \mathrm{d}v$:

$$ \begin{align} \class{violet}{\boldsymbol{B}}(\boldsymbol{r}) ~=~ \frac{\mu_0}{4\pi} \int_{V} \class{red}{\boldsymbol{j}} \, (\boldsymbol{R}) \times \frac{\boldsymbol{r}-\boldsymbol{R}}{|\boldsymbol{r}-\boldsymbol{R}|^3} ~ \text{d}v \end{align} $$

The Biot-Savart law is particularly useful for calculating the magnetic field around current-carrying wires or coils.

$$ \begin{align} \class{violet}{\boldsymbol{B}}(\boldsymbol{r}) ~=~ \frac{\mu_0 \, \class{red}{I}}{4\pi} \int_{S} \frac{\boldsymbol{r}-\boldsymbol{R}}{|\boldsymbol{r}-\boldsymbol{R}|^3} \times \text{d}\boldsymbol{s} \end{align} $$

Position Vectors for the Biot-Savart Law for a thin Wire

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise #1: Magnetic Field of a Semi-Circular Current-Carrying Wire

A thin current-carrying wire in the x-y plane runs along the x-axis and then bends into a semi-circular shape with radius $R$. The wire then runs along the x-axis.

What is the magnitude of the magnetic field $B$ at the coordinate origin $x=0$?

Solution to Exercise #1

The magnetic field from an arbitrarily shaped thin wire can be calculated using the Biot-Savart Law: 1 \[ \boldsymbol{B}(\boldsymbol{r}) ~=~ \frac{\mu_0 \, I}{4\pi} \int_{S} \frac{\boldsymbol{r}-\boldsymbol{R}}{|\boldsymbol{r}-\boldsymbol{R}|^3} \times \text{d}\boldsymbol{s} \]

Since we need to calculate the magnetic field at the origin, the position vector $ \boldsymbol{r} = 0 $ to the field point where the magnetic field is to be calculated is zero. For shorter notation, the magnitude $R := |\boldsymbol{R}| $ is defined: 2 \[ \boldsymbol{B}(0) ~=~ \frac{\mu_0 \, I}{4\pi} \int_{S} \frac{-\boldsymbol{R}}{R^3} \times \text{d}\boldsymbol{s} \]

In cylindrical coordinates, the position vector $ \boldsymbol{R} $, pointing to the infinitesimal wire element $ \text{d}\boldsymbol{s} $, can be written using the basis vector $\hat{\boldsymbol{r}}_{\perp}$ (which points radially outward from the z-axis) and the magnitude $R$: 3 \[ \boldsymbol{B}(0) ~=~ \frac{\mu_0 \, I}{4\pi} \int_{S} \frac{-R \, \hat{\boldsymbol{r}}_{\perp}}{R^3} \times \text{d}\boldsymbol{s} \]

The integral, with the distance $R$ held constant, only goes along the $\varphi$ coordinate, allowing the magnitude $R$ to be pulled out of the integral: 4 \[ \boldsymbol{B}(0) ~=~ -\frac{\mu_0 \, I}{4\pi \, R^2} \int_{S} \hat{\boldsymbol{r}}_{\perp} \times \text{d}\boldsymbol{s} \]

The infinitesimal element is $ \text{d}\boldsymbol{s} = R\, \text{d}\varphi \, \hat{\boldsymbol{\varphi}} $, (where $R$ before $\text{d}\varphi$ is the Jacobian determinant). Here, $\hat{\boldsymbol{\varphi}}$ is the basis vector in cylindrical coordinates along the $\varphi$ coordinate: 5 \[ \boldsymbol{B}(0) ~=~ -\frac{\mu_0 \, I}{4\pi \, R^2} \int_{S} \hat{\boldsymbol{r}}_{\perp} \times \hat{\boldsymbol{\varphi}} \, R\, \text{d}\varphi \]

The cross product $ \hat{\boldsymbol{r}}_{\perp} \times \hat{\boldsymbol{\varphi}} $ of two basis vectors is orthogonal to both and thus corresponds to the basis vector in the z-direction $ \hat{\boldsymbol{z}} $ in cylindrical coordinates: 6 \[ \boldsymbol{B}(0) ~=~ -\frac{\mu_0 \, I}{4\pi \, R} \int_{S} \hat{\boldsymbol{z}} \, \text{d}\varphi \]

The basis vector $ \hat{\boldsymbol{z}} = (0,0,1) $ is independent of $\varphi$ and is pulled out of the integral. The integral over the $\varphi$ coordinate runs on a semi-circle, from 0 to $\pi$: 7 \[ \boldsymbol{B}(0) ~=~ -\frac{\mu_0 \, I}{4\pi \, R} \, \hat{\boldsymbol{z}} \int^{\pi}_{0} \, \text{d}\varphi \]

The integral is equal to $\pi$. Thus, the magnetic field at the coordinate origin is: 8 \[ \boldsymbol{B}(0) ~=~ -\frac{\mu_0 \, I}{4 \, R} \, \hat{\boldsymbol{z}} \]