Question

Derive an expression for the magnetic field $\vec{B}$, due to a circular coil of N turns, each of radius r carrying current I, at a distance ‘x’ from the centre along its axis.

Hint:

To instantly check if your derived formula is correct, plug in $x = 0$. The messy denominator simplifies to $(r^2)^{3/2} = r^3$, resulting cleanly in $B = \mu_0 N I / 2r$, which is the famous and heavily memorized formula for the center of a loop!

Answer 1

Concept:
• The Biot-Savart Law acts as the absolute foundational cornerstone for mathematically determining the magnetic field strictly generated by arbitrary steady current distributions.
• To successfully tackle a large circular coil, we meticulously break the continuous ring into infinitesimally tiny current elements $d\vec{l}$ and integrate their individual tiny magnetic contributions over the entire ring.
• Geometric symmetry plays a massive role here: perpendicular field components elegantly and completely cancel each other out, leaving strictly only the axial field components to sum together perfectly.

Step 1: Setup the geometry and apply Biot-Savart Law
Consider a large circular coil of radius $r$ placed stably in the y-z plane, with its center perfectly locked at the origin $(0,0,0)$.
Let a steady current $I$ flow continuously through it.
We aim to find the exact magnetic field $\vec{B}$ strictly at a point $P$ located directly on the x-axis at a distance $x$ from the absolute center.
Consider a tiny, infinitesimal current element $Id\vec{l}$ situated precisely at the topmost edge of the circular coil.
The spatial distance vector $\vec{s}$ running straight from this top element down to the observation point $P$ has a rigid magnitude derived cleanly from Pythagoras:
\[ s = \sqrt{r^2 + x^2} \]
According to the rigorous Biot-Savart law, the tiny magnitude of the magnetic field $d\vec{B}$ generated uniquely by this tiny top element at point $P$ is:
\[ dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(90^\circ)}{s^2} = \frac{\mu_0}{4\pi} \frac{I dl}{(r^2 + x^2)} \]
(The angle is definitively $90^\circ$ because the tangent vector $d\vec{l}$ and the slant vector $\vec{s}$ are fundamentally mutually perpendicular in 3D space).

Step 2: Resolve components based on structural symmetry
Using right-hand cross product rules, the generated vector $d\vec{B}$ strictly points perpendicular to the slant distance $\vec{s}$.
We meticulously resolve $d\vec{B}$ into two distinct orthogonal components: a vertical perpendicular component ($dB_\perp = dB \sin\phi$) and a horizontal axial component ($dB_\parallel = dB \cos\phi$). Let the angle between $d\vec{B}$ and the vertical axis be $\phi$, which geometrically implies the angle between $s$ and $x$ is $\phi$.
(Actually, let's use angle $\theta$ between the axis $x$ and the slant distance $s$. Then the axial component along x is $dB \sin\theta$).
Let's use standard notation: let the angle between slant $s$ and axis $x$ be $\alpha$. Then the axial component of the magnetic field is precisely $dB \sin\alpha$.
From the built right triangle, we can cleanly extract the sine function:
\[ \sin\alpha = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{r}{\sqrt{r^2 + x^2}} \]
Due to the perfect rotational symmetry of the physical coil, every single top element has a perfect diametrically opposite bottom element.
The vertical perpendicular components ($dB \cos\alpha$) from these opposite pairs perfectly and violently cancel each other out entirely ($| \Sigma dB_\perp | = 0$).
Only the horizontal axial components strictly pointing along the positive x-axis actually survive and add up.

Step 3: Integrate over the entire coil
The total net magnetic field $B$ is the mathematical integral of strictly the surviving axial components:
\[ B = \int dB \sin\alpha \]
Substitute the previously established specific expressions meticulously into the integral:
\[ B = \int \left[ \frac{\mu_0}{4\pi} \frac{I dl}{(r^2 + x^2)} \right] \left( \frac{r}{\sqrt{r^2 + x^2}} \right) \]
\[ B = \frac{\mu_0 I r}{4\pi (r^2 + x^2)^{3/2}} \int dl \]
The mathematical integral $\int dl$ simply and cleanly evaluates strictly to the total physical circumference of the large circular loop, which is exactly $2\pi r$:
\[ B = \frac{\mu_0 I r}{4\pi (r^2 + x^2)^{3/2}} (2\pi r) \]
Cancel the constants securely:
\[ B = \frac{\mu_0 I r^2}{2 (r^2 + x^2)^{3/2}} \]

Step 4: Final expression for an N-turn coil
If the physical coil actually strictly consists of $N$ tightly wound identical turns, each single turn contributes an identical amount of magnetic field.
Therefore, we simply and aggressively multiply the single-turn result directly by the integer $N$:
\[ B_{axis} = \frac{\mu_0 N I r^2}{2 (r^2 + x^2)^{3/2}} \]
The final magnetic field vector points strictly along the central axial line.