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Physics · Ch 10 — Magnetic Fields due to Electric Current

Axial Magnetic Field Produced by Current in a Circular Loop

10.13

Axial Magnetic Field Produced by Current in a Circular Loop

Section 10.12 found the field only at the CENTRE of a circular current loop. This section generalises that result to any point P on the loop's AXIS -- the straight line through the centre of the loop, perpendicular to the plane of the loop -- at an arbitrary distance zz from the centre (Fig. 10.19).

Set up coordinates so the circular loop, of radius RR, carrying steady current II, lies in the xx-yy plane, centred at the origin OO, with the point P on the zz axis at distance zz from OO. For a current element dl⃗d\vec{l} anywhere on the loop, the distance to P is r=R2+z2r=\sqrt{R^2+z^2} -- the SAME for every element, by the loop's circular symmetry, exactly as it was constant for every element of a circular arc in the previous section. Because dl⃗d\vec{l} (lying in the xx-yy plane, tangent to the loop) is always exactly perpendicular to r⃗\vec{r} (which points from the element out to P, lying in a yy-zz plane containing the axis), the Biot-Savart law gives a differential field of magnitude

dB=μ04πI dlr2=μ04πI dlR2+z2.dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{r^2} = \frac{\mu_0}{4\pi}\frac{I\,dl}{R^2+z^2}.

This differential field dB⃗d\vec{B} is NOT directed purely along the axis, however -- it has both an axial (zz) component dBzdB_z and a component dB⊥dB_{\perp} perpendicular to the axis. Crucially, by the loop's symmetry, the perpendicular component dB⊥dB_{\perp} due to any one current element is EXACTLY CANCELLED by the perpendicular component due to the current element diametrically OPPOSITE it on the loop -- so when the contributions of every element are summed over the full loop, all the perpendicular components cancel out completely, leaving only the AXIAL components to add up.

Using the geometry of Fig. 10.19, cos⁡α=R/r\cos\alpha=R/r (where α\alpha is the angle between r⃗\vec{r} and the plane of the loop), so dBz=dBcos⁡αdB_z=dB\cos\alpha, and integrating this axial component over the entire loop (a simple integral, since dBdB itself is the SAME for every element, by symmetry) gives

Bz=∫dBcos⁡α=μ04πI(R2+z2)⋅RR2+z2∫dl=μ04πI (2πR) R(R2+z2)3/2=μ0IR22(z2+R2)3/2B_z = \int dB\cos\alpha = \frac{\mu_0}{4\pi}\frac{I}{(R^2+z^2)}\cdot\frac{R}{\sqrt{R^2+z^2}}\int dl = \frac{\mu_0}{4\pi}\frac{I\,(2\pi R)\,R}{(R^2+z^2)^{3/2}} = \frac{\mu_0 I R^2}{2(z^2+R^2)^{3/2}}

for a single-turn loop, or, for a coil of NN turns, …

Figure 10.19Fig. 10.19: Magnetic field on the axis of a circular current loop of radius R
Fig. 10.19 — Fig. 10.19: Magnetic field on the axis of a circular current loop of radius R

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. A circular current loop of radius R lies flat in the x-y plane, centred at the origin O, carrying a steady current I. A point P is marked on the z axis (the axis passing through O, perpendicular to the loop's plane), at a distance z from the centre O. A representative current element dl⃗d\vec{l} is marked at one point on the circumference of the loop (lying in the x-y plane), joined to P by a vector r⃗\vec{r} of length r=R2+z2r=\sqrt{R^2+z^2}; the resulting differential field dB⃗d\vec{B} at P is shown resolved into its component dBzdB_z ALONG the z axis and its component dB⊥dB_{\perp} perpendicular to the z axis, with the figure geometry making clear that cos⁡α=R/r\cos\alpha=R/r where α\alpha is the angle between r⃗\vec{r} and the loop's plane -- this resolution is exactly what lets the perpendicular components cancel …

Misc Ex.10.6Magnetic field at the centre of a large 1000-turn coil

Worked out. A closely-wound coil of 1000 turns has a radius of 1 m (R = 100 cm), and a current of 10 A passes through it; the magnitude of the magnetic field at the coil's centre is required. Using the centre-of-loop formula for N turns, B=μ0NI2R=(4π×10−7)×1000×102×1B=\frac{\mu_0 N I}{2R}=\frac{(4\pi\times10^{-7})\times1000\times10}{2\times1}, this evaluates to B≈6.28×10−3B\approx6.28\times10^{-3} T -- illustrating that even a modest current, when passed through many turns of a reasonably large coil, produces a field of a few milliteslas at the centre, considerably stronger than the Earth's own field of about 0.36 …