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Physics · Ch 10 — Magnetic Fields due to Electric Current

Magnetic Field Produced by a Current in a Circular Arc of a Wire

10.12

Magnetic Field Produced by a Current in a Circular Arc of a Wire

Having worked out the field of a straight wire (Section 10.10.1), the next natural shape to consider is a CIRCULAR arc of wire -- and, as a special case, a complete circular loop, which turns out to be considerably simpler to handle than the straight-wire case, on account of a special geometric fact unique to circles.

Consider a circular arc ABAB of a wire, of radius rr, centred at a point OO, subtending an angle θ\theta (measured in RADIANS) at that centre, and carrying a steady current II (Fig. 10.18). Take a current element dl⃗d\vec{l} anywhere along this arc. Because every point on a circle is, by the very definition of a circle, always exactly the SAME distance rr from the centre, and because the tangent direction to a circle at any point is always exactly PERPENDICULAR to the radius drawn to that point, the current element dl⃗d\vec{l} (which points along the local tangent direction of the wire) is ALWAYS perpendicular to the radius vector r⃗\vec{r} drawn from that element to the centre OO -- for EVERY element along the arc, without exception. This means the angle between dl⃗d\vec{l} and r⃗\vec{r} in the Biot-Savart law is always exactly 90∘90^\circ, so sin⁡θ=1\sin\theta=1 throughout, and moreover rr itself stays constant (equal to the arc's own radius) as the integration proceeds along the arc. Both of these simplifications together mean the general Biot-Savart integral collapses to something very simple:

B=∫dB=μ04πIr2∫dl=μ04πIr2(rθ)=μ04πIrθB = \int dB = \frac{\mu_0}{4\pi}\frac{I}{r^2}\int dl = \frac{\mu_0}{4\pi}\frac{I}{r^2}(r\theta) = \frac{\mu_0}{4\pi}\frac{I}{r}\theta

(using ∫dl=rθ\int dl = r\theta, the standard relation between arc length, radius, and the SUBTENDED ANGLE in radians), directed, throughout the arc, into (or out of) the plane of the page according to the right-hand rule applied to the current's direction of flow.

As an important special case, letting the arc become a COMPLETE circle -- that is, θ=2π\theta=2\pi radians, the full angle around a circle -- gives the field at the CENTRE of a full circular current loop of radius rr, carrying current II: …

Figure 10.18Fig. 10.18: Current-carrying wire in the shape of a circular arc
Fig. 10.18 — Fig. 10.18: Current-carrying wire in the shape of a circular arc

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 arc of wire AB, part of a circle of radius r centred at point O, carries a current I flowing along the arc from A to B. A representative differential length element dl⃗d\vec{l} is marked at some point along the arc, together with the radius vector r⃗\vec{r} drawn from that element straight to the centre O -- explicitly drawn (and labelled) as being PERPENDICULAR to dl⃗d\vec{l} at every point along the arc, which is the key geometric fact (unique to a circular shape) that makes the Biot-Savart integration for the field at the centre O particularly simple, since $\sin\theta=\ …

Misc Ex.10.5Magnetic field at the centre of a 3-segment wire: two straight parts and a semicircular arc

Worked out. A wire consists of three sections joined together: a straight section (i), a semicircular section (ii) of radius R, and another straight section (iii), forming a path that passes near a point O (the centre of the semicircle), and the field at O due to each of the three sections individually, and their total, is required. For sections (i) and (iii), the current-length element Idl⃗Id\vec{l} is found to be exactly PARALLEL (angle 180∘180^\circ) or ANTI-PARALLEL (angle 0∘0^\circ) to the radius vector R⃗\vec{R} drawn to O for every point on those straight sections, so sin⁡(180∘)=sin⁡(0∘)=0\sin(180^\circ)=\sin(0^\circ)=0 and each contributes ZERO field at O. For section (ii), the semicircular arc, dl⃗d\vec{l} is always PERPENDICULAR to R⃗\vec{R} (as in the general circular-arc case above), giving, via the arc-field formula with θ=π\theta=\pi (a semicircle, i.e. half of 2π2\pi), B=μ0I4RB=\frac{\mu_0 I}{4R}. Since sections (i) and (iii) contribute zero, the TOTAL field at O is simply this same value, $B_{total}=0+\frac{\ …