Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 4.1.2

Interactive: Biot–Savart Field Lines

Problem 4.1.2

Circular loop: the field on the axis

A constant current of $I$ [A] is flowing along a circular loop with a radius of $b$ located on the $x y$ -plane with its centre at the origin. Write the density of current and find the explicit expression of the magnetic field at a point $P (0, 0, h)$ on the $z$ -axis.

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 6-6.

Try it yourself first.

Strategy

Biot–Savart's law round the loop. All elements are at the same distance $R = b^{2} + h^{2}$ from P, so only the direction of $d l^{'} \times R$ varies. Its radial part cancels between diametrically opposite elements and its axial part adds, exactly as for the charged ring of Problem 3.1.2.

1. Symmetry

The loop is unchanged by any rotation about the $z$ -axis, and P lies on that axis. A field component perpendicular to the axis would single out a direction in the $x y$ -plane, which the symmetry forbids, so only $B_{z}$ can survive at P.

2. Current density

In cylindrical coordinates the loop is $ρ^{'} = b$ , $z^{'} = 0$ , and the current flows along $a_{ϕ^{'}}$ :

⟹ J = I δ (ρ^{'} - b) δ (z^{'}) a_{ϕ^{'}} J d v^{'} \to I d l^{'} = I b d ϕ^{'} a_{ϕ^{'}} .

3. Geometry: the points and R

From the source point Q $(b cos ϕ^{'}, b sin ϕ^{'}, 0)$ to P $(0, 0, h)$ ,

R R = - b a_{ρ^{'}} + h a_{z}, = b^{2} + h^{2} (the same for every element),

and with $a_{ϕ^{'}} \times a_{ρ^{'}} = - a_{z}$ and $a_{ϕ^{'}} \times a_{z} = a_{ρ^{'}}$ ,

a_{ϕ^{'}} \times R = - b (a_{ϕ^{'}} \times a_{ρ^{'}}) + h (a_{ϕ^{'}} \times a_{z}) = b a_{z} + h a_{ρ^{'}} .

4. Integrate

Since $R$ is constant it comes out of the integral:

B = \frac{μ _{0}}{4 π} \int_{0}^{2 π} \frac{I b d ϕ ^{'} ( a _{ϕ^{'}} \times R )}{R ^{3}} = \frac{μ _{0} I b}{4 π ( b ^{2} + h ^{2} ) ^{3/2}} \int_{0}^{2 π} (b a_{z} + h a_{ρ^{'}}) d ϕ^{'} .

The unit vector $a_{ρ^{'}} = cos ϕ^{'} a_{x} + sin ϕ^{'} a_{y}$ integrates to zero over a full turn, $\int_{0}^{2 π} a_{ρ^{'}} d ϕ^{'} = 0$ : the radial contributions of diametrically opposite elements cancel in pairs (Figure 4.1.19a). Only the axial part survives, $\int_{0}^{2 π} b d ϕ^{'} = 2 π b$ :

Answer

B (0, 0, h) = \frac{μ _{0} I b ^{2}}{2 ( b ^{2} + h ^{2} ) ^{3/2}} a_{z} [T]

(b) B_z on the axis. Problem 4.1.2. (a) Diametrically opposite elements of the loop give d\vec B vectors tilted symmetrically about the axis: their radial parts cancel and their axial parts add. (b) The axial field falls from \mu_0I/2b at the centre to half of it at h=0.77b and follows the dipole law \propto(b/h)^3 beyond a few radii. — **Figure 4.1.19.** (b) $B_{z}$ on the axis. Problem 4.1.2. (a) Diametrically opposite elements of the loop give $d B$ vectors tilted symmetrically about the axis: their radial parts cancel and their axial parts add. (b) The axial field falls from $μ_{0} I /2 b$ at the centre to half of it at $h = 0.77 b$ and follows the dipole law $\propto (b / h)^{3}$ beyond a few radii.

On the axis the computed $B_{z}$ follows $μ_{0} I b^{2} /2 (b^{2} + h^{2})^{3/2}$ : 12.6 μT at the centre for these values.

InteractiveLoading simulation…

Check

Units: $μ_{0} I b^{2} / R^{3}$ is (H/m)(A)/m $=$ T. Two limits (Figure 4.1.19b). At the centre, $h = 0$ : $B = μ_{0} I /2 b$ . Far away, $h ≫ b$ : $B \to μ_{0} I b^{2} /2 h^{3} = μ_{0} m /2 π h^{3}$ with the magnetic moment $m = I π b^{2} a_{z}$ , the field of a magnetic dipole, the counterpart of the electric dipole of Home exercise 3.1.4 (whose axial field is $p / (2 π ε_{0} r^{3})$ ). Direction: the right hand with the fingers curled along the current gives the thumb along $+ a_{z}$ . Off the axis the field has no closed form in elementary functions (it needs elliptic integrals), but the on-axis result is all one needs for coils: Home exercise 4.1.2 stacks these loops into a solenoid, and Problems 4.1.3 and 4.1.8 nest them into disks. For $I = 1$ A and $b = 5$ cm the centre field is $4 π \times 1 0^{- 7} /0.1 = 12.6$ μT.

Problem 4.1.2: Circular loop: the field on the axis