Physics · Ch 4 — Moving Charges and Magnetism

Magnetic Field at the Centre and on the Axis of a Circular Current Loop

4.5

Field at the centre of a circular loop. Consider a circular loop of wire of radius $R$

carrying current $I$ , and the field point $P$ taken at the loop's own centre. Every current element

$I\,d\vec{l}$ around the loop lies at exactly the SAME distance $R$ from the centre, and, because the

loop is circular, every element's direction $d\vec{l}$ is exactly PERPENDICULAR to the line joining

it to the centre (i.e. $\theta = 90^\circ$ for every element, so $\sin\theta = 1$ throughout).

Applying the Biot-Savart law (Section 4.3) to one element,

dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2}

and, crucially, every element's contribution $d\vec{B}$ points in the SAME direction at the centre

(straight along the loop's axis, by the right-hand rule, since every element is tangent to the same

circle) -- so, unlike a general Biot-Savart problem, the individual contributions can simply be added

as plain numbers rather than needing vector resolution. Integrating $dl$ around the full

circumference $2\pi R$ ,

B_{\text{centre}} = \frac{\mu_0}{4\pi}\frac{I}{R^2}\int_0^{2\pi R} dl = \frac{\mu_0}{4\pi}\frac{I(2\pi R)}{R^2} = \frac{\mu_0 I}{2R}

For a coil of $N$ turns wound tightly together (all effectively at the same radius $R$ ), each turn

contributes the same field, so the total is simply $N$ times as large:

B_{\text{centre}} = \frac{\mu_0 N I}{2R}

Field on the axis of the loop, at a general point. Now consider a field point $P$ on the loop's

axis, at distance $x$ from the centre (rather than at the centre itself). Each current element still

subtends $\theta=90^\circ$ with the line joining it to $P$ (the element is still tangent to the

circle, and this line to an AXIAL point is still perpendicular to that tangent, by the same symmetry

argument), but the distance from each element to $P$ is now $\sqrt{R^2+x^2}$ rather than simply $R$ .

By symmetry, as the contributions $d\vec{B}$ from diametrically opposite elements are added, their

components PERPENDICULAR to the axis cancel in pairs (by the loop's rotational symmetry about the

axis), leaving only the components ALONG the axis to survive; each element's axial component is

$dB\cos\alpha$ , where $\alpha$ is the angle between $d\vec{B}$ and the axis, with

$\cos\alpha = R/\sqrt{R^2+x^2}$ . Carrying out the resulting integral around the full loop gives

B_{\text{axis}}(x) = \frac{\mu_0 I R^2}{2\,(R^2+x^2)^{3/2}}

for a single turn, or with an extra factor of $N$ for an $N$ -turn coil. This formula correctly …