Question

Derive an expression for the magnetic field \(B\), 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:

At the center of the loop (\(x=0\)): \[ B = \frac{\mu_0 N I}{2r} \] Always use symmetry to avoid full integration.

Answer 1

Concept: The magnetic field at a point on the axis of a circular current loop is derived using the Biot–Savart law. Due to symmetry, only axial components survive while perpendicular components cancel.

Step 1: Biot–Savart law
The magnetic field due to a small current element is: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I\, d\vec{l} \times \hat{r}}{r^2} \] For a circular loop, we integrate around the entire coil.

Step 2: Geometry of the circular loop
Consider:
• Radius of loop = \(r\)
• Point on axis at distance = \(x\)
• Distance of any current element from point = \(\sqrt{r^2 + x^2}\)

Step 3: Symmetry argument
Each current element produces a magnetic field:
• Radial components cancel due to symmetry
• Only axial components add up So we calculate only axial component \(dB_x\).

Step 4: Expression for axial field of one turn
After integrating around the loop: \[ B_{\text{one turn}} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}} \]

Step 5: For N turns
For \(N\) identical turns, fields add linearly: \[ B = N \cdot B_{\text{one turn}} \] \[ B = \frac{\mu_0 N I r^2}{2(r^2 + x^2)^{3/2}} \] Final Answer: \[ \boxed{B = \frac{\mu_0 N I r^2}{2(r^2 + x^2)^{3/2}}} \]