Derivation of Biot Savart’s Law

The Biot-Savart Law is a key concept in electromagnetism that describes how a magnetic field is originated by an electric current flowing through a conductor. This law is essential for calculating the magnetic field produced by various current configurations. For example, it enables the determination of the magnetic field created by an infinitely long, straight wire by integrating the contributions of current elements along its length, thus allowing for the calculation of the field at any point in space. The Biot-Savart Law is also vital for finding the magnetic field induction of a circular current loop, both at its center and along its axis. This capability is crucial for the design and working of technologies such as magnetic resonance imaging (MRI) machines and electromagnets.

1.0Statement of Biot Savart’s Law

The Biot-Savart Law states that the magnetic field B at a point in space due to a small element of current-carrying conductor is directly proportional to the current I in the conductor, the length of the element dl, and the sine of the angle θ between the current element and the line connecting the element to the observation point where the field is being calculated. The magnetic field is inversely proportional to the square of the separation r from the element to the point.

Mathematically It can be expressed as,

$dB=\frac{{\mu }_0}{4\pi }\frac{Idl\mathrm{Sin}\theta }{r^2}$

Current Element-: A very small element of length dl of a thin current carrying conductor (wire). It is represented by $I(\overrightarrow{dl}$) . It is a vector quantity and its direction is the same as that of current.

2.0Diagram of Biot Savart’s Law

3.0Derivation Of Biot Savart’s Law

Experimentally, it was found, magnetic field dB at point P due to current element varies as:

$dB\propto I$

$dB\propto dl$

$dB\propto \mathrm{Sin}\theta$

$dB\propto \frac{1}{r^2}$

Combining all the factors we get,

$dB\propto \frac{Idl\mathrm{Sin}\theta }{r^2}$

$dB=K\frac{Idl\sin \theta }{r^2}$

$K=\frac{{\mu }_0}{4\pi }=\mathrm{Tm}{A}^{-1}$ =Constant of Proportionality

$dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

Permeability-Also known as magnetic permeability. It indicates to what extent a material/medium can be magnetized when placed inside an externally applied magnetizing field.

${\mu }_0$ : Permeability of free space

$\mu$: Permeability of a medium

${\mu }_r$: Relative Permeability of a medium

${\mu }_r=\frac{\mu }{{\mu }_0}\text{ where }{\mu }_0=4\pi \times 10^{-7}{\mathrm{Tma}}^{-1}$

4.0Biot Savart’s Law in Vector and Integral Form

Experimentally, it is found that direction of magnetic field due to a current element is perpendicular to plane containing $\overrightarrow{dl}$ and $\vec{r}$.

$\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{Idl\mathrm{Sin}\theta }{r^2}\hat{n}$

($\hat{n}$ -Unit vector normal to plane containing $\overrightarrow{dl}$  and $\vec{r}$)

$\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{(Idl)r\sin \theta }{r^3}\hat{n}$

$dB=\frac{\overrightarrow{{\mu }_0}}{4\pi }\frac{I(\vec{d}l\times \vec{r})}{r^3}\Rightarrow \text{ Vector Form }$

$\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{dB}}=\frac{\overrightarrow{{\mu }_0}}{4\pi }\int \frac{\mathrm{I}(\overrightarrow{\mathrm{d}}l\times \overrightarrow{\mathrm{r}})}{{\mathrm{r}}^3}$

5.0Application of Biot Savart’s Law

1. Magnetic field surrounding a thin straight current carrying conductor

• AB is a straight conductor carrying current i from B to A. At a point P, whose perpendicular distance from AB is OP = a, the direction of field is perpendicular to the plane of paper, inwards (represented by a cross)

$l=a\mathrm{Tan}\theta \Rightarrow dl=a{\sec }^2\theta d\theta$    ..….(1)

$\alpha =90^{\circ }-\theta \text{ and }r=a\sec \theta$

By using Biot Savart’s Law

$dB=\frac{{\mu }_0}{4\pi }\frac{idl\mathrm{Sin}\alpha }{r^2}$

$\Rightarrow B=\int dB=\int \frac{{\mu }_0}{4\pi }\frac{idl\sin \alpha }{r^2}(\text{ due to wire }AB)$

$\therefore B=\frac{{\mu }_{0}i}{4\pi a}\int \mathrm{Cos}\theta d\theta$

Taking limits of Integration as $-{\varphi }_2$ to ${\varphi }_1$

$B=\frac{{\mu }_{0}i}{4\pi a}{\int }_{-{\varphi }_2}^{{\varphi }_1}\mathrm{Cos}\theta d\theta =\frac{{\mu }_{0}i}{4\pi a}[\mathrm{Sin}\theta {]}_{-{\varphi }_2}^{{\varphi }_1}=\frac{{\mu }_{0}i}{4\pi a}\left[\mathrm{Sin}{\varphi }_1+\mathrm{Sin}{\varphi }_2\right]\text{ inwards }$

2. Magnetic Field Due to ∞ wire :

$B=\frac{{\mu }_{0}i}{4\pi R}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{Cos}\theta d\theta =\frac{{\mu }_{0}i(2)}{2\pi R}\Rightarrow B=\frac{{\mu }_{0}i}{2\pi R}$

3. Magnetic field lines due to an infinite wire carrying current I 

In fact, the direction of the magnetic field due to a long straight wire can be determined by the right-hand rule (Figure). 

Figure: Direction of the magnetic field due to an infinite straight wire

4. Direction of MFI due to Straight Wire

NOTE: In this case magnetic field lines are circular :Hold the wire with your right hand with your thumb pointing along the current then direction of curling of finger will give direction of MFI.

5. Variation of Magnetic Field(B) with Distance(x)

6.0Sample Questions on Biot Savart’s Law

Q-1.Find the.

Solution:

$B_P=\frac{{\mu }_{0}I}{4\pi d}{\int }_{0^{\circ }}^{90^{\circ }}\mathrm{Cos}\theta d\theta$

$B_P=\frac{{\mu }_{0}i}{4\pi d}\left[\mathrm{Sin}90^{\circ }-\mathrm{Sin}{0}^{\circ }\right]$

$B_P=\frac{{\mu }_{0}I}{4\pi d}$

Q-2.The magnitude of magnetic field 80 cm from the axis of long straight infinite wire is 7 mT. Calculate the current in the wire?

Solution:

$B=\frac{{\mu }_0}{4\pi }\left(\frac{2I}{x}\right)$

$I=\frac{Bx}{2\times \left(\frac{{\mu }_0}{4\pi }\right)}=\frac{7\times 10^{-6}\times 0.8}{2\times 10^{-7}}=28\mathrm{A}$

Q-3.Calculate the magnetic field at point P?

Solution: Applying Biot Savart’s Law for straight current carrying wire,

$B=\frac{{\mu }_{0}I}{4\pi x}\left[\mathrm{Sin}{\theta }_1+\mathrm{Sin}{\theta }_2\right]$

$B=\frac{10^{-7}(10)}{(1)}\left[\mathrm{Sin}53^{\circ }+\mathrm{Sin}37^{\circ }\right]$

$B=\frac{10^{-7}(10)}{(1)}\left[\frac{4}{5}+\frac{3}{5}\right]=10^{-6}\times \frac{7}{5}=\frac{7}{5}\times 10^{-6}T\text{ Inwards }$

Q-4.Find the magnetic field due to moving charge?

Solution: Magnetic Field at a point due to a charge can be found in a way similar to that of a current element.

Magnetic field due to current element 

$dB=\frac{{\vec{\mu }}_0}{4\pi }\frac{I(\vec{d}l\times \vec{r})}{r^3}$

$I\overrightarrow{dl}=\left(\frac{dq}{dt}\right)\overrightarrow{dl}=(dq)\frac{\vec{d}l}{dt}=(dq)\vec{v}$

Magnetic Field due to moving charge,

$\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{(dq)}{r^3}\vec{v}\times \vec{r}$

1. For Moving Charge in Vector Form, $\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{(dq)}{r^3}\vec{v}\times \vec{r}$
2. For Moving Charge in Scalar Form, $\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{qv\mathrm{Sin}\theta }{r^2}$