Biot Savart Law

Did you wonder why electricity is essential for loudspeakers to work? We all have used loudspeakers during parties and other different occasions. However, there is mostly a moment when the light goes off, and there is silence as the loudspeakers cannot work. Sometimes, we were in the middle of our favorite song, or our favorite song was about to begin. Pretty long wait for electricity, right?

Fig. 1 - The figure shows a loudspeaker membrane that vibrates during sound generation.

Well, electricity is essential for the generation of sound from loudspeakers. A changing electric current is passed through the coil inside the loudspeaker to generate a magnetic field. This coil is passed through magnets placed inside the speaker. The magnetic field generated by the coil is opposite to the direction of a magnetic field due to these magnets. This opposing magnetic field causes the membrane's vibration, producing sound. A well-known law in physics that helps to understand this magnetic field generated due to the passing of an electric current through the coil is the Biot-Savart law. In this article, we will study the Biot-Savart law, the direction of the magnetic field induced due to a current carrying element, features of the Biot-Savart law, similarities and dissimilarities between the Biot-Savart law and Coulomb's law, and special cases of the Biot-Savart law.

Biot-Savart Law Definition

Before understanding Biot-Savart law, let's briefly explain what we mean by a magnetic field

Hans Christian Oersted discovered the magnetic effect of a current-carrying wire. Oersted's experiment confirms the formation of a magnetic field around a current-carrying wire with the deflection of a magnetic needle placed near the wire.

The Biot-Savart law helps to calculate the magnetic field around this current-carrying wire.

Biot Savart Law Derivation

Let's consider a small current element \(\mathrm{AB}\) of a conducting wire \(\mathrm{XY}\) carrying current \(I\). The length of an infinitesimally small element is \(\mathrm{d}\vec{l}\). Let \(\vec{r}\) be the position vector of an observation point P from the current element. Let \(\theta\) be an angle between \(\mathrm{d}\vec{l}\) and \(\vec{r}\).

Fig. 2 - The figure shows a small current element on a conducting wire carrying current \(I\) and an observation point P for an induced magnetic field.

Biot-Savart law states that the magnetic field induced at point P due to a current-carrying wire is

1. directly proportional to the current \(I\),

2. directly proportional to the length of a current element \(\mathrm{d}\vec{l}\),

3. directly proportional of sine of angle between \(\vec{r}\) and \(\mathrm{d}\vec{l}\),

4. inversely proportional to the square of the position vector, i.e., \(r^2\).

The mathematical form of Biot-Savart law is

\[\mathrm{d}\vec{B}\propto\frac{I\mathrm{d}\vec{l}\sin{\left(\theta\right)}}{r^2}\]

or

\[\mathrm{d}B=K\frac{I\mathrm{d}l\sin{\left(\theta\right)}}{r^2}\]

Where \(K\) is a proportionality whose value depends upon the medium between the point P and the current element \(\mathrm{d}\vec{l}\). It is defined as \[K=\frac{\mu_0}{4\pi},\]

where \(\mu_0=4\pi\times10^{-7}\,\mathrm{T\,A^{-1}\,m^{-1}}\) is the magnetic permeability of free space/vacuum.

Biot-Savart Law Formula

Let's take a look at how the Biot-Savart Law can give us a formula for the total magnetic field around a current carrying wire.

Superposition Principle

Biot-Savart law follows the superposition principle. According to this law, the net magnetic field at the point due to several current elements is the algebraic sum of a magnetic field due to each current element.

Let \(\mathrm{d}\vec{B_1},\mathrm{d}\vec{B_2},\mathrm{d}\vec{B_3},...\) and so on are the magnetic field at the point P due to the current elements \(I\mathrm{d}\vec{l_1},I\mathrm{d}\vec{l_2},I\mathrm{d}\vec{l_3},...\) and so on. Then the net magnetic field at point P is

\[\mathrm{d}\vec{B}=\mathrm{d}\vec{B_1}+\mathrm{d}\vec{B_2}+\mathrm{d}\vec{B_3}+...\]

This equation shows the superposition principle/vector addition.

When considering the magnetic field over a length of wire as we do in the Biot-Savart law, we split up the wire into elements of infinitesimal length \(\mathrm{d}\vec{l}\). Each of these infinitesimal lengths adds a contribution to the magnetic field of:\[\mathrm{d}\vec{B}=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{\left(\theta\right)}}{r^2}\]

To find the overall magnetic field field from a wire we us the superposition principle by integrating over the wire length:

\[\vec{B}(\vec{r})=\int\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{\left(\theta\right)}}{r^2}\]

Direction of Magnetic Field

In the Biot-Savart law, the magnetic field due to a current-carrying conductor is given by

\[\mathrm{d}\vec{B}=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\times\vec{r}}{r^3}\]

This equation shows that the direction of a magnetic field is along the direction of a cross product \(I\mathrm{d}\vec{l}\times\vec{r}\). To find the direction of this cross-product, we can use the Right-Hand rule.

For the case shown in figure 2, by using the Right-Hand Rule, the curl of your fingers is along a clockwise direction on the plane containing \(\mathrm{d}\vec{l}\) and \(\vec{r}\). Then the direction of \(\mathrm{d}\vec{B}\) is perpendicular to the plane and is directed inward.

In case when \(\theta=0^\circ\)

The direction of current element \(\left(I\mathrm{d}\vec{l}\right)\) is along the direction of position vector \(\vec{r}\), then

\[\begin{align*}\mathrm{d}\vec{B}&=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{\left(0^\circ\right)}}{r^2}\\\mathrm{d}\vec{B}&=0\end{align*}\]

This gives the minimum value of a magnetic field.

In case when \(\theta=90^\circ\)

The direction of the current element \(\left(I\mathrm{d}\vec{l}\right)\) is perpendicular to the direction position vector \(\left(\vec{r}\right)\), then

\[\begin{align*}\mathrm{d}\vec{B}&=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{\left(90^\circ\right)}}{r^2}\\\mathrm{d}\vec{B}&=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}}{r^2}\end{align*}\]

This gives the maximum value of a magnetic field.

Biot-Savart Law and Coulomb's Law

According to Biot-Savart law, the magnetic field around a current-carrying element \(\left(I\mathrm{d}\vec{l}\right)\) at a distance of \(r\) is

\[\mathrm{d}\vec{B}=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}}{r^2}\]

or

\[\mathrm{d}\vec{B}=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\times\vec{r}}{r^3}\]

According to Coulomb's law, the electric field around an electric charge \(q\) at a distance of \(r\) is

\[E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\]

or

\[\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^3}\vec{r}\]

Some of the main similarities between Biot-Savart law and Coulomb's law are,

Some of the main differences between Biot-Savart law and Coulomb's law are,

Applications of Biot-Savart Law

There are several applications of Biot-Savart law but the most important is it helps in calculating the magnetic due to a current element irrespective of its configuration. In other words, Biot-Savart law is independent of the configuration of the wire-carrying current. In this article, we will calculate the induced magnetic field due to a straight wire carrying current and a current carrying in form of a circular coil using Biot-Savart law.

For a Straight Wire Carrying Current

Imagine a straight current-carrying wire XY in a plane carrying current \(I\). Let P be a point perpendicular to the wire on a plane at a distance of \(r\) from the wire and \(I\mathrm{d}\vec{l}\) be a small current element.

Let \(\vec{r}'\) be the position vector of point P from the current element.

Fig. 4 - The figure shows a straight wire carrying current and an observation point P at a distance of \(r\) from the wire.

In the diagram, we can see \(\theta\) is an angle between position vector \(\vec{r}'\) and a current element \(I\mathrm{d}\vec{l}\) and \(\phi\) is an angle between position vector \(\vec{r}'\) and \(\vec{r}\).

According to Biot-Savart law, the magnetic field \(\mathrm{d}B\) at a point P due to the current element \(I\mathrm{d}\vec{l}\) is\[\mathrm{d}\vec{B}=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\times\vec{r}'}{r'^3}\]

or\[\mathrm{d}\vec{B}=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{\theta}}{r'^2}\tag{1}\]

From a right-angle triangle CPO,

\[\begin{align*}\theta+\phi+90^{\circ} &=180^\circ\\\theta &=90^{\circ}-\phi\end{align*}\]

Substituting this value of \(\theta\) in equation (1), \[\begin{align*}\mathrm{d}\vec{B} &=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{90^{\circ}-\phi}}{r'^2}\\\mathrm{d}\vec{B} &=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\cos{\phi}}{r'^2}\tag{2}\end{align*}\]

Also in triangle CPO \[\begin{align*}\cos{\phi}&=\frac{r}{r'}\\r'&=\frac{r}{\cos{\phi}}\tag{3}\end{align*}\]

And \[\begin{align*}\tan{\phi}&=\frac{l}{r}\\l&=r\tan{\phi}\tag{4}\end{align*}\]

By differentiating equation (4) \[\mathrm{d}l=r\sec^2{\phi}d\phi\tag{5}\]

using equations (3) and (5), equation (2) becomes\[\begin{align*}\mathrm{d}B &= \frac{\mu_0}{4\pi}\frac{Ir\sec^2{\phi}\cos{\phi}}{\frac{r^2}{\cos^2{\phi}}}\mathrm{d}\phi\\\mathrm{d}B &= \frac{\mu_0}{4\pi}\frac{Ir\cos{\phi}\mathrm{d}\phi} {\cos^2{\phi}}\frac{\cos^2{\phi}}{r^2}\\\mathrm{d}B &= \frac{\mu_0}{4\pi}\frac{I}{r}\cos{\phi}\mathrm{d}\phi\tag{6}\end{align*}\]

By integrating equation (6) with the limit \(-\phi_1\) and \(+\phi_2\), the magnetic field at the point P due to a whole straight is

\[\begin{align*}\int_{-\phi_1}^{\phi_2} \mathrm{d}B &= \frac{\mu_0}{4\pi}\frac{I}{r} \int_{-\phi_1}^{\phi_2} \cos{\left(\phi\right)} \mathrm{d}\phi\\B&=\frac{\mu_0}{4\pi}\frac{I}{r}\left[\sin{\phi}\right]_{-\phi_1}^{\phi_2}\\B&=\frac{\mu_0}{4\pi}\frac{I}{r}\left(\sin{\left(\phi_2\right)}-\sin{\left(-\phi_1\right)}\right)\\B&=\frac{\mu_0}{4\pi}\frac{I}{r}\left(\sin{\left(\phi_2\right)}+\sin{\left(\phi_1\right)}\right)\end{align*}\]This equation gives the magnetic field at point P due to a finite straight wire XY carrying current.

Let's say conductor XY is of infinite length. In this case \(\phi_1=\phi_2=90^\circ\), the magnetic field at the point P will become

\[\begin{align*}B&=\frac{\mu_0}{4\pi}\frac{I}{r}\left(\sin{\left(90^\circ\right)}+\sin{\left(90^\circ\right)}\right)\\B&=\frac{\mu_0}{4\pi}\frac{I}{r}2\\B&=\frac{\mu_0}{2\pi}\frac{I}{r}\end{align*}\]

The direction of this magnetic field is perpendicular to the plane containing the wire and an observation point.

Current Carrying Circular Coil

Let's say a circular coil of radius \(r\) is placed perpendicular to the plane of the paper. \(I\) be the current flowing through the coil. Suppose an observation point P is on the paper at the distance of \(x\) from the center of the coil (let's say O).

Fig. 5 - The figure shows the magnetic field at point P due to a current-carrying circular coil.

In the diagram, we can see that the current on the wire is perpendicular to the position vector \(\vec{MP}\). Thus, the angle between \(I\mathrm{d}\vec{l}\) and \(\vec{MP}\) is \(90^\circ\).

By using Biot-Savart law, the magnetic field at the point P due to a small current element at the point M in the diagram is

\[\begin{align*}\mathrm{d}\vec{B}&=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{\left(90^\circ\right)}}{\left(\sqrt{r^2+x^2}\right)^2}\\\mathrm{d}\vec{B}&=\frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}}{\left(r^2+x^2\right)}\tag{7}\end{align*}\]

Using Right Hand Rule, the direction of this magnetic field is perpendicular to the plane containing the current element \(I\mathrm{d}\vec{l}\) at M and a position vector \(\vec{MP}\).

By resolving this magnetic field along the X and Y axis as shown in the figure, we got \(\mathrm{d}B\cos{\left(\phi\right)}\) along the Y-axis and \(\mathrm{d}B\sin{\left(\phi\right)}\) along the X-axis.

Similarly, due to a current element at N, the component of a magnetic field is \(-\mathrm{d}B\cos{\left(\phi\right)}\) along the Y-axis and \(\mathrm{d}B\sin{\left(\phi\right)}\) along the X-axis.

Therefore, the net magnetic field along the Y-axis is

\[\mathrm{d}B_{\mathrm{y}}=\mathrm{d}B\cos{\left(\phi\right)}-\mathrm{d}B\cos{\left(\phi\right)}=0\]

And the magnetic field along the X-axis is

\[\mathrm{d}B_{\mathrm{x}}=\mathrm{d}B\sin{\left(\phi\right)}+\mathrm{d}B\sin{\left(\phi\right)}=2\mathrm{d}B\sin{\left(\phi\right)}\]

So, the net magnetic field at point P due to the current element at M and its alternate on the coil (i.e. at N) is

\[\begin{align*}\mathrm{d}B_{\mathrm{net}}&=\mathrm{d}B_{\mathrm{y}}+\mathrm{d}B_{\mathrm{x}}\\\mathrm{d}B_{\mathrm{net}}&=0+2\mathrm{d}B\sin{\left(\phi\right)}\\\mathrm{d}B_{\mathrm{net}}&=2\mathrm{d}B\sin{\left(\phi\right)}\tag{8}\end{align*}\]

Using equation (7), equation (8) becomes

\[\mathrm{d}B_{\mathrm{net}}=2\times \frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}\sin{\left(\phi\right)}}{\left(r^2+x^2\right)}\]

From triangle MPO, \(\sin{\left(\phi\right)}=\frac{r}{\sqrt{r^2+x^2}}\), thus, \(\mathrm{d}B_{\mathrm{net}}\) becomes

\[\mathrm{d}B_{\mathrm{net}}=2\frac{\mu_0}{4\pi}\frac{Ir\mathrm{d}\vec{l}}{\left(r^2+x^2\right)^{3/2}}\]

For the half current-carrying coil and its alternate part (another half of the coil), the net magnetic field at point P is

\[\begin{align*}\int \mathrm{d}B_{\mathrm{net}}&=2\frac{\mu_0}{4\pi}\frac{Ir}{\left(r^2+x^2\right)^{3/2}}\int \mathrm{d}\vec{l}\\B_{\mathrm{net}}&=2\frac{\mu_0}{4\pi}\frac{Ir}{\left(r^2+x^2\right)^{3/2}}\pi r\\B_{\mathrm{net}}&=\frac{\mu_0}{2}\frac{Ir^2}{\left(r^2+x^2\right)^{3/2}}\end{align*}\]

If point P is at the center of a coil, i.e., \(x=0\), the net magnetic field at point P due to a current-carrying circular coil is

\[\begin{align*}B_{\mathrm{net}}&=\frac{\mu_0}{2}\frac{Ir^2}{\left(r^2+0\right)^{3/2}}\\B_{\mathrm{net}}&=\frac{\mu_0}{2}\frac{Ir^2}{r^3}\\B_{\mathrm{net}}&=\frac{\mu_0}{2}\frac{I}{r}\end{align*}\]

For \(n\) number of turns in the coil

\[B_{\mathrm{net}}=\frac{\mu_0}{2}\frac{nI}{r}\]

This equation shows the magnetic field at the center of the current-carrying coil having \(n\) number of turns which is calculated using Biot-Savart law.In conclusion, Biot-Savart law helps in calculating the magnetic field due to a current-carrying conductor irrespective of its configuration.