Q.A circular loop of radius $3\ \text{cm}$ carries a current of $4\ \text{A}$. Find the magnetic field at a point on the axis of the loop, $4\ \text{cm}$ from its centre.

Each current element contributes a Biot–Savart field perpendicular to the line joining it to the axial point; by symmetry the components perpendicular to the axis cancel in pairs, and only the axial components survive, integrating to $B=\mu_0IR^2/2(R^2+x^2)^{3/2}$.

Setup

Consider a circular coil of radius $R$, carrying current $I$, lying in a plane. Let $P$ be a point on the axis of the coil, at a distance $x$ from the centre $O$.

Every current element $I\,d\vec l$ on the coil is at the same distance from $P$:

$r=\sqrt{R^2+x^2}$

Biot–Savart law for one element

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

Since $d\vec l$ (tangential to the circle) is perpendicular to $\vec r$ (the line from the element to $P$) for every element on the coil, $|d\vec l\times\hat r|=dl$, so the magnitude of each contribution is:

$dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{r^2}=\frac{\mu_0}{4\pi}\frac{I\,dl}{R^2+x^2}$

The direction of $d\vec B$ is perpendicular to the plane containing $d\vec l$ and $\vec r$.

Resolving into axial and perpendicular components

By symmetry, resolve each $d\vec B$ into a component along the axis ($dB_x$) and a component perpendicular to the axis ($dB_\perp$, lying in the plane perpendicular to the axis). For every element, there is a diametrically opposite element on the coil whose perpendicular component $dB_\perp$ points in exactly the opposite direction — so summing over the whole coil, all the perpendicular components cancel out in pairs. Only the axial components survive and add up.

The axial component of each element's contribution is:

$dB_x = dB\sin\theta,\qquad \text{where}\ \sin\theta=\frac{R}{r}=\frac{R}{\sqrt{R^2+x^2}}$

(here $\theta$ is the angle between $\vec r$ and the plane of the coil, i.e. between $d\vec B$ and the perpendicular-to-axis direction, chosen so that $\sin\theta=R/r$ picks out the axial projection).

$dB_x = \frac{\mu_0}{4\pi}\frac{I\,dl}{r^2}\cdot\frac{R}{r}=\frac{\mu_0 I R}{4\pi r^3}\,dl$

Integrating around the full loop

$B=\int dB_x = \frac{\mu_0 I R}{4\pi r^3}\oint dl = \frac{\mu_0 I R}{4\pi r^3}\times(2\pi R)$

$B = \frac{\mu_0 I R^2}{2r^3}$

Substituting $r=\sqrt{R^2+x^2}$, so $r^3=(R^2+x^2)^{3/2}$:

$\boxed{B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}}$

directed along the axis (by the right-hand rule).

Check: at the centre of the coil ($x=0$), this reduces to $B=\dfrac{\mu_0IR^2}{2R^3}=\dfrac{\mu_0I}{2R}$, matching the known formula for the field at the centre of a circular loop.

Applying the Biot-Savart law to every current element of the loop and integrating, the components perpendicular to the axis cancel by symmetry, leaving a net axial field.

Consider a circular loop of radius R carrying current I, lying in a plane, and let P be a point on its axis at a distance x from the centre O.

By the Biot-Savart law, a current element $Id\vec{l}$ at the top of the loop produces a field at P of magnitude:

$dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin 90°}{r^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{R^2+x^2}$

(since $d\vec{l}$ is always perpendicular to $\vec{r}$, the line joining the element to P, and $r = \sqrt{R^2+x^2}$), directed perpendicular to r, in the plane containing the axis and r.

This $dB$ can be resolved into a component $dB\cos\theta$ along the axis (where $\cos\theta = R/\sqrt{R^2+x^2}$) and a component $dB\sin\theta$ perpendicular to the axis. By symmetry, for every element there is a diametrically opposite element whose perpendicular component exactly cancels it, while the axial components from all elements add up.

Integrating the axial component around the full loop (circumference $2\pi R$):

$B = \oint dB\cos\theta = \dfrac{\mu_0 I}{4\pi(R^2+x^2)} \times \dfrac{R}{\sqrt{R^2+x^2}} \times 2\pi R$

$B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$

directed along the axis (given by the right-hand rule). At the centre of the loop (x=0), this reduces to the familiar $B = \mu_0 I / 2R$.

(A diagram should show the circular loop carrying current I, its centre O, a point P on the axis at distance x, a current element Idl at the top of the loop, the field dB it produces at P (perpendicular to r), and its resolved components along and perpendicular to the axis.)

Apply the Biot–Savart law to a current element, sum the axial components (the perpendicular components cancel by symmetry) around the whole loop.

Consider a circular coil of radius $a$ carrying current $I$, and a point $P$ on its axis at distance $x$ from the centre $O$.

Take a small current element $I\,d\vec l$ on the loop. The distance from this element to $P$ is

$r = \sqrt{a^2 + x^2}$

By the Biot–Savart law, the magnetic field due to this element at $P$ has magnitude

$dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\,\sin 90^\circ}{r^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{a^2+x^2}$

(since $d\vec l$ is tangent to the loop and $\vec r$ from the element to $P$ is always perpendicular to $d\vec l$), and $d\vec B$ is perpendicular to $r$, i.e. it lies in the plane containing the axis and makes an angle $\theta$ with the axis, where $\cos\theta = \dfrac{x}{r}$... actually $\sin\theta = a/r$ (angle between $d\vec B$ and the axis relates to the geometry): resolve $d\vec B$ into two components —

• Axial component $dB\cos\phi$ (along $OP$), where $\cos\phi = a/r$ — wait, precisely, $dB$ makes an angle $\phi$ with the axis such that $\sin\phi = a/r$ is the component perpendicular to axis and $\cos\phi = x/r$ is along axis... The standard result: the axial component is $dB\,\dfrac{a}{r}$ and the perpendicular (radial) component is $dB\,\dfrac{x}{r}$; by symmetry, contributions from diametrically opposite elements cancel the perpendicular components, while axial components from all elements add up.

Summing (integrating) the axial components around the full loop (circumference $2\pi a$):

$B = \oint dB\,\dfrac{a}{r} = \dfrac{\mu_0 I}{4\pi(a^2+x^2)}\times\dfrac{a}{\sqrt{a^2+x^2}} \times (2\pi a)$

$B = \dfrac{\mu_0 I a^2}{2(a^2+x^2)^{3/2}}$

This field points along the axis (direction given by the right-hand rule based on the current's sense). At the centre of the coil ($x=0$), this reduces to the familiar $B = \dfrac{\mu_0 I}{2a}$.

The net magnetic field at the center of loop A is the vector sum of the field produced by loop A itself and the field produced by loop B. Since the currents are in opposite directions, these fields oppose each other. The net magnetic field is $\boxed{\frac{214\pi}{125} \times 10^{-7}\ \text{T}}$ directed away from loop B.

The Biot–Savart law gives the magnetic field due to a small current element; integrating it around a circular loop gives the axial field $B = \mu_0 I R^2 / [2(R^2+x^2)^{3/2}]$, which reduces to $\mu_0 I/2R$ at the centre.

1. Biot–Savart law: The magnetic field $d\vec{B}$ at a point P due to a small current element $I\,d\vec{l}$ is: $d\vec{B} = \dfrac{\mu_0}{4\pi}\,\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}$ where $r$ is the distance from the current element to P, and $\hat{r}$ is the unit vector from the element to P. In magnitude, $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$, where $\theta$ is the angle between $d\vec{l}$ and $\hat{r}$.
2. Field on the axis of a circular current loop: Consider a circular loop of radius R carrying current I, and a point P on its axis at a distance x from the centre O. For every current element $d\vec{l}$ on the loop, the distance to P is $r=\sqrt{R^2+x^2}$, and $d\vec{l}$ is always perpendicular to $\vec{r}$ (so $\sin\theta = 1$). Hence: $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{R^2+x^2}$ This $d\vec{B}$ has a component along the axis ($dB\cos\phi$, where $\cos\phi = R/\sqrt{R^2+x^2}$) and a component perpendicular to the axis. By symmetry, the perpendicular components from diametrically opposite elements cancel, leaving only the axial components to add up: $B = \oint dB\cos\phi = \dfrac{\mu_0 I}{4\pi(R^2+x^2)}\cdot\dfrac{R}{\sqrt{R^2+x^2}}\oint dl$ Since $\oint dl = 2\pi R$ (circumference of the loop): $B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ For N turns: $B = \dfrac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}$. At the centre of the loop ($x=0$), this reduces to the familiar $B = \dfrac{\mu_0 I}{2R}$ (or $\dfrac{\mu_0 NI}{2R}$ for N turns).

Integrating the Biot-Savart contributions of all current elements around the loop, only the axial components survive, giving $B=\dfrac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}$.

Consider a circular loop of radius $R$ carrying current $I$, and a point P on its axis at distance $x$ from the centre O. Consider a current element $Id\vec l$ on the loop; its distance from P is $s=\sqrt{R^2+x^2}$, and it is always perpendicular to the line joining the element to P.

By the Biot-Savart law, the field due to this element at P has magnitude:

$dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{s^2} = \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2+x^2}$

directed perpendicular to $s$, i.e. it has a component along the axis and a component perpendicular to the axis.

By symmetry, as we sum contributions from all elements around the loop, the perpendicular (radial) components cancel out in pairs from diametrically opposite elements, while the axial components all add up. The axial component of each $dB$ is:

$dB_x = dB\sin\theta = dB\cdot\frac{R}{s} = dB\cdot\frac{R}{\sqrt{R^2+x^2}}$

Integrating around the full loop ($\oint dl = 2\pi R$):

$B = \oint dB_x = \frac{\mu_0 I}{4\pi(R^2+x^2)}\cdot\frac{R}{\sqrt{R^2+x^2}}\oint dl = \frac{\mu_0 I}{4\pi(R^2+x^2)^{3/2}}\cdot R\cdot 2\pi R$

$B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$

At the centre of the loop ($x=0$), this reduces to $B=\dfrac{\mu_0I}{2R}$, the well-known field at the centre.

Biot–Savart: $d\vec B=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}$; axial field of a coil $=\dfrac{\mu_0 I r^2}{2(r^2+x^2)^{3/2}}$.

1. Statement. The magnetic field $d\vec B$ due to a current element $I\,d\vec l$ at a point P at position $\vec r$ from the element has magnitude $dB=\dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$ ($\theta$ = angle between $d\vec l$ and $\vec r$), directed perpendicular to the plane of $d\vec l$ and $\vec r$. In vector form: $d\vec B=\frac{\mu_0}{4\pi}\,\frac{I\,d\vec l\times\hat r}{r^2}.$
2. Field on the axis of a circular coil. Consider a coil of radius $r$ carrying current $I$; P is on the axis at distance $x$ from the centre. Each element is at distance $\sqrt{r^2+x^2}$ from P, and $d\vec l\perp\vec r$, so $dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{(r^2+x^2)}.$ By symmetry, the components perpendicular to the axis cancel around the loop; only the axial components survive. The axial component is $dB\cos\alpha$ where $\cos\alpha=\dfrac{r}{\sqrt{r^2+x^2}}$: $B=\oint dB\cos\alpha=\frac{\mu_0 I}{4\pi(r^2+x^2)}\cdot\frac{r}{\sqrt{r^2+x^2}}\oint dl.$ With $\oint dl=2\pi r$, $\boxed{B=\frac{\mu_0 I r^2}{2(r^2+x^2)^{3/2}}},$ directed along the axis. (At the centre, $x=0$: $B=\dfrac{\mu_0 I}{2r}$.)