Magnetic Field at the Center of a Circular Coil

December 29, 2025

Problem

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Explanation

We have a circular coil of wire with:

  • Number of turns: N=100N = 100
  • Radius of each turn: R=8.0 cm=0.080 mR = 8.0\ \text{cm} = 0.080\ \text{m}
  • Current in the coil: I=0.40 AI = 0.40\ \text{A}

We want the magnetic field magnitude BB at the center of the coil.


1. Formula for a Single Circular Loop

For a single circular loop of radius RR carrying current II, the magnetic field at the center is:

B1=μ0I2RB_1 = \frac{\mu_0 I}{2R}

where μ0=4π×107 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A} is the permeability of free space.


2. Coil with NN Turns

If there are NN closely spaced turns, each carrying the same current II and having the same radius RR, the total field at the center is the sum of the contributions from all turns:

B=NB1=Nμ0I2RB = N \cdot B_1 = N \cdot \frac{\mu_0 I}{2R}

Substitute the known values:

  • N=100N = 100
  • I=0.40 AI = 0.40\ \text{A}
  • R=0.080 mR = 0.080\ \text{m}
  • μ0=4π×107 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}

So:

B=100(4π×107)(0.40)2×0.080B = 100 \cdot \frac{(4\pi \times 10^{-7}) (0.40)}{2 \times 0.080}

Simplify step by step:

  1. Denominator: 2R=2×0.080=0.16 m2R = 2 \times 0.080 = 0.16\ \text{m}
  2. Numerator inside: (4π×107)×0.40=1.6π×107(4\pi \times 10^{-7}) \times 0.40 = 1.6\pi \times 10^{-7}
  3. Fraction for one turn: B1=1.6π×1070.16=10π×107 T=π×106 TB_1 = \frac{1.6\pi \times 10^{-7}}{0.16} = 10\pi \times 10^{-7}\ \text{T} = \pi \times 10^{-6}\ \text{T}
  4. Multiply by N=100N = 100: B=100(π×106)=100π×106 T=π×104 TB = 100 \cdot (\pi \times 10^{-6}) = 100\pi \times 10^{-6}\ \text{T} = \pi \times 10^{-4}\ \text{T}

Using π3.14\pi \approx 3.14:

B3.14×104 TB \approx 3.14 \times 10^{-4}\ \text{T}

So, the magnitude of the magnetic field at the center of the coil is:

B3.1×104 T\boxed{B \approx 3.1 \times 10^{-4}\ \text{T}}

3. What the Visualization Shows

The canvas visualization treats the coil as a glowing circular loop (with many turns implied by line thickness and color), centered on the screen. You can interactively change:

  • The radius of the coil (in centimeters)
  • The current through the coil (in amperes)
  • The number of turns

The computed magnetic field at the center is displayed as text and also encoded visually:

  • A vertical magnetic-field vector at the center, whose length scales with B|B|.
  • A halo or glow around the center, whose intensity and size grow with B|B|.

Mathematically, the field used in the visualization is:

B=Nμ0I2R,R=(radius in cm)/100B = N\,\frac{\mu_0 I}{2R}, \quad R = (\text{radius in cm})/100

You can see how increasing NN or II boosts the field, while increasing RR (making the coil larger) weakens it. The animation also gently rotates phase coloring around the loop to suggest the circulation of current, while the central arrow stays fixed to represent the static direction of B\vec{B} (out of or into the screen, here represented in 2D as an upward arrow).

Interactive Visualization

Parameters

100.00
8.00
0.40
Magnetic Field at the Center of a Circular Coil