Q.A thin insulated wire forms a plane spiral of $N=100$ tight turns carrying a current $I=8$ mA (milli ampere). The radii of the inside and outside turns are $a=50$ mm and $b=100$ mm respectively. The magnetic induction at the centre of the spiral is

From Coulomb’s Law to Currents: The Intuition

You already know that a stationary charge creates an electric field that falls off as $1/r^2$ and points radially away from the charge. But when that charge moves — when it becomes a current — something new appears: a magnetic field. The question is: how does a tiny piece of current produce a tiny piece of magnetic field?

Imagine a very short segment of wire carrying a steady current $I$. Let its length be $dl$ — so small that we can treat it as a point-like source. This little current element, $I\,d\mathbf{l}$, is the magnetic analogue of a point charge in electrostatics. Just as Coulomb’s law gives the electric field from a point charge, the Biot-Savart law gives the magnetic field from a current element.

But there’s a crucial difference. The electric field from a point charge points along the line joining the charge to the observation point. The magnetic field from a current element points perpendicular to both the direction of the current and the line joining the element to the point. This perpendicular nature is the heart of magnetism.

The Precise Statement

Consider a current element $I\,d\mathbf{l}$ located at some point. Let $\mathbf{r}$ be the position vector from the element to the point $P$ where we want the magnetic field. Then the infinitesimal magnetic field $d\mathbf{B}$ at $P$ due to this element is:

Here:

• $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space — a fundamental constant.
• $d\mathbf{l}$ points along the direction of the current.
• $\mathbf{\hat{r}}$ is a unit vector pointing from the current element to the observation point.
• The cross product $d\mathbf{l} \times \mathbf{\hat{r}}$ gives both the magnitude and direction.

What the Cross Product Tells You

The magnitude of the cross product is $|d\mathbf{l} \times \mathbf{\hat{r}}| = dl \cdot 1 \cdot \sin\theta$, where $\theta$ is the angle between $d\mathbf{l}$ and $\mathbf{\hat{r}}$. So the magnitude of $d\mathbf{B}$ is:

This is exactly the form you mentioned: proportional to $I\,dl \sin\theta / r^2$. The $\sin\theta$ factor means:

• When the current element points directly toward or away from $P$ ($\theta = 0$ or $\pi$), $\sin\theta = 0$ — no magnetic field is produced along that line.
• When the current element is perpendicular to the line joining it to $P$ ($\theta = 90^\circ$), the field is maximum.

The direction of $d\mathbf{B}$ is given by the right-hand rule: curl the fingers of your right hand from $d\mathbf{l}$ toward $\mathbf{\hat{r}}$, and your thumb points in the direction of $d\mathbf{B}$. This direction is always perpendicular to the plane containing $d\mathbf{l}$ and $\mathbf{r}$.

Why the $1/r^2$ Dependence?

Just like Coulomb’s law, the Biot-Savart law has an inverse-square dependence on distance. This is not a coincidence — both laws emerge from the same underlying structure of electromagnetism. Unlike Coulomb's law, this $1/4\pi$ prefactor is not because the field spreads uniformly over a sphere -- the $\sin\theta$ factor above already shows the elemental field is NOT isotropic, it circulates around the current direction instead. The $1/(4\pi)$ here is simply a consequence of the SI 'rationalized' unit convention, chosen so that $\mu_0$ appears without a $4\pi$ in Ampere's circuital law, $\oint \vec B \cdot d\vec l = \mu_0 I_{enc}$.

The Total Field: Integration

The Biot-Savart law gives you the field from a single infinitesimal current element. To find the total magnetic field from a complete circuit (a wire of any shape), you must integrate over the entire path: …