Biot-Savart Law

Statement of the law

Consider a finite conductor carrying a steady current $I$. Take an infinitesimal element $d\mathbf{l}$ of the conductor, and a point $P$ located by the displacement vector $\mathbf{r}$ drawn from the element to $P$. Let $\theta$ be the angle between $d\mathbf{l}$ and $\mathbf{r}$. According to the Biot-Savart law, the field $d\mathbf{B}$ at $P$ is proportional to the current $I$, to the element length $|d\mathbf{l}|$, and inversely proportional to the square of the distance $r$, with its direction perpendicular to the plane containing $d\mathbf{l}$ and $\mathbf{r}$.

In the vector notation used by NCERT [Eq. 4.7(a)],

$$d\mathbf{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\mathbf{l}\times\hat{\mathbf{r}}}{r^2}$$

and, using the property of the cross product, the magnitude is [Eq. 4.7(b)]

$$dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^2}.$$

The constant $\mu_0$ is the permeability of free space, and its value in SI units is exact [Eq. 4.7(c)]:

$$\mu_0 = 4\pi\times 10^{-7}\ \text{T·m/A} \quad\Longrightarrow\quad \frac{\mu_0}{4\pi} = 10^{-7}\ \text{T·m/A}.$$

The NIOS module (Section 18.3) states the same dependence: the field due to an element $\Delta l$ depends on the current $I$, the element length $\Delta l$, the inverse square of the distance, and the angle between the element and the line joining it to the observation point. The two sources are in exact agreement; NCERT's vector form is the one you should reproduce verbatim in answers.

Direction of dB and the cross product

The single most examined feature of the law is the direction of $d\mathbf{B}$. Because $d\mathbf{B}$ is fixed by the cross product $d\mathbf{l}\times\hat{\mathbf{r}}$, it is perpendicular to the plane containing both $d\mathbf{l}$ and $\mathbf{r}$. It does not point along $\mathbf{r}$, and it does not point along $d\mathbf{l}$ — it stands at right angles to the plane that contains them.

NCERT gives the sense of $d\mathbf{l}\times\mathbf{r}$ through the right-hand screw rule: look at the plane containing $d\mathbf{l}$ and $\mathbf{r}$, imagine rotating from the first vector to the second; if the rotation is anticlockwise the result points towards you, if clockwise it points away. The NIOS right-hand grip rule (grasp the wire with the thumb along the current, fingers curl in the direction of the field) is the same statement applied to a complete wire.

The sinθ angular factor and the 1/r² dependence

Two quantitative features of $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\,\sin\theta}{r^2}$ deserve attention. First, the field falls off as $1/r^2$ — the same inverse-square geometry as Coulomb's law. Second, there is an explicit $\sin\theta$ angular factor that has no analogue in the electrostatic case.

This angular factor produces a striking result. Along the direction of $d\mathbf{l}$ itself (the dashed forward line in NCERT Fig. 4.7), $\theta = 0$, so $\sin\theta = 0$ and $|d\mathbf{B}| = 0$. A current element produces no field directly ahead of or behind itself; the field is maximum sideways, at $\theta = 90^\circ$, where $\sin\theta = 1$.

Application: field of a long straight wire

Integrating the Biot-Savart contributions of every element of an infinitely long straight wire carrying current $I$ gives the field at perpendicular distance $r$:

$$B = \frac{\mu_0 I}{2\pi r}.$$

This is the result NCERT recovers in Example 4.7 (and confirms again via Ampère's circuital law). The field lines are concentric circles around the wire, with direction set by the right-hand grip rule. The dependence is $B \propto 1/r$ — a single inverse power, not inverse-square — because the field is the cumulative effect of the whole line of elements, not of one point source.

Application: circular loop centre and arc

For a circular loop of radius $R$, the field at the centre follows directly from the Biot-Savart law because every element $d\mathbf{l}$ is perpendicular to its $\mathbf{r}$ (so $\theta = 90^\circ$, $\sin\theta = 1$), and each contribution points along the same axial direction. NCERT obtains it as the special case $x = 0$ of the on-axis formula [Eq. 4.12]:

$$B = \frac{\mu_0 I}{2R}.$$

The NIOS module derives the identical result (Eq. 18.6) by summing $\sum \Delta l = 2\pi R$ around the loop, and notes that $n$ turns multiply the field to $B = \mu_0 n I/(2R)$. A partial arc subtending angle $\theta$ (in radians) at the centre carries only the fraction $\theta/2\pi$ of the full circle, giving

$$B_{\text{arc}} = \frac{\mu_0 I \theta}{4\pi R}.$$

A semicircle ($\theta = \pi$) therefore produces $\mu_0 I/(4R)$, exactly half the full-loop value — the conclusion NCERT reaches in Example 4.5, where the straight feed segments contribute nothing because $d\mathbf{l}\times\mathbf{r} = 0$ along them.

Biot-Savart law vs Coulomb's law

NCERT spells out the parallels and the differences between the Biot-Savart law for the magnetic field and Coulomb's law for the electrostatic field. Both are long-range inverse-square laws, both are linear in their source (so the principle of superposition applies to each). But the differences are precisely what NEET probes.

The electrostatic field is produced by a scalar source — the electric charge $q$. The magnetic field is produced by a vector source — the current element $I\,d\mathbf{l}$. The electric field lies along the displacement vector joining source and field point; the magnetic field is perpendicular to the plane containing $\mathbf{r}$ and $I\,d\mathbf{l}$. And the Biot-Savart law carries an angle dependence ($\sin\theta$) that has no counterpart in Coulomb's law.