Magnetic Field on the Axis of a Circular Current Loop 🔄

The setup is a loop of radius R lying in the y-z plane with its center at the origin and current I flowing counter-clockwise (when viewed from +x). We want the field at a point P on the axis, a distance x from the center. 🎯:contentReference[oaicite:0]{index=0}

Key Geometry 📐

  • Every tiny length dl of the loop is perpendicular to the line joining it to P. So \(|\mathbf{dl}\times\mathbf{r}| = r\,dl\). :contentReference[oaicite:1]{index=1}
  • The distance from any element to P is \(r=\sqrt{x^{2}+R^{2}}\). :contentReference[oaicite:2]{index=2}
  • The angle between \(\mathbf{r}\) and the axis obeys \[\cos\theta=\frac{R}{\sqrt{x^{2}+R^{2}}}\ .\] :contentReference[oaicite:3]{index=3}

Building the Field Step-by-Step 🔧

Start with the Biot–Savart contribution from one element:

\[d\mathbf{B}= \frac{\mu_{0}}{4\pi}\,\frac{I\,\mathbf{dl}\times\mathbf{r}}{r^{3}} \tag{4.8}\]:contentReference[oaicite:4]{index=4}

Because of the right-angle, its size is

\[dB=\frac{\mu_{0} I\,dl}{4\pi\,(x^{2}+R^{2})}\ . \tag{4.9}\]:contentReference[oaicite:5]{index=5}

Only the x-component survives when all elements are added (sideways pieces cancel in pairs ✨). Hence

\[dB_{x}=dB\cos\theta=\frac{\mu_{0} I R\,dl}{4\pi\,(x^{2}+R^{2})^{3/2}}\ .\]

Adding every dl around the loop gives a total length \(2\pi R\). Substitute \(dl \to 2\pi R\) to harvest the full field:

\[ \boxed{\;B_{x}= \frac{\mu_{0} I R^{2}}{2\,(x^{2}+R^{2})^{3/2}}\;} \quad\text{(points along +x)} \tag{4.11} \]:contentReference[oaicite:6]{index=6}

Special cases make life easy 😄:

  • Center of the loop (x=0): \[ B=\frac{\mu_{0} I}{2R}. \tag{4.12} \]:contentReference[oaicite:7]{index=7}
  • N tightly-wound turns: just multiply by N → \(B=\mu_{0}NI/(2R)\).

Direction? Use the Thumb! 👍

Curl the fingers of your right hand around the loop in the current’s direction; the thumb points along the field on the axis. The upper face acts like a north pole, the lower like a south pole. 🔄:contentReference[oaicite:8]{index=8}

Quick Worked Examples 🧮

  1. Semi-circular arc (radius R)
    • Straight leads give zero field because \(\mathbf{dl}\parallel\mathbf{r}\). ✅
    • Arc’s field is half of a full loop: \(B=\mu_{0}I/(4R)\). Direction depends on which side the arc faces. :contentReference[oaicite:9]{index=9}
  2. 100-turn coil (R=0.10 m, I=1 A) \[ B=\frac{\mu_{0} N I}{2R}=6.3\times10^{-4}\ \text{T}. \]:contentReference[oaicite:10]{index=10}

Field Lines Snapshot 🌐

The lines loop through the center, exit on one face, wrap around outside, and re-enter on the other face—just like a small bar magnet. :contentReference[oaicite:11]{index=11}

High-Yield NEET Takeaways 🚀

  • Axial field of a loop: \(B_{x}= \dfrac{\mu_{0} I R^{2}}{2\,(x^{2}+R^{2})^{3/2}}\).
  • Field at the center: \(B=\dfrac{\mu_{0}I}{2R}\); becomes \(\mu_{0}NI/(2R)\) for a coil.
  • Right-hand thumb rule quickly tells the field’s direction. 👍
  • Semi-circular arc gives exactly half the field of a full loop; straight segments often add nothing. 🌓
  • Perpendicular components cancel in symmetrical pairs—only the axial part stays. 🎯