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Physics · Ch 10 — Magnetic Fields due to Electric Current

Ampere's Law

10.15

Ampere's Law

The Biot-Savart law (Section 10.10) is completely general -- it can, in principle, be used to find the magnetic field of ANY current distribution, however complicated -- but, as the sometimes-lengthy integrations of Sections 10.10.1 through 10.13 show, DIRECTLY carrying out that integration can be a genuinely laborious business, even for fairly simple, highly symmetric current shapes. Recall the exactly parallel situation in electrostatics: Coulomb's law is likewise completely general, but Gauss's law offers a dramatic shortcut whenever the CHARGE distribution has enough symmetry (spherical, cylindrical, or planar). Ampere's law is the magnetostatic counterpart of exactly that idea, offering the same kind of shortcut whenever the CURRENT distribution is sufficiently symmetric.

Ampere's circuital law states that

∮B⃗⋅dl⃗=μ0I,\oint \vec{B}\cdot d\vec{l} = \mu_0 I,

where the integral (denoted by the circle through the integral sign) is taken all the way around any closed path, called an "Amperian loop," and II on the right-hand side is the NET current actually encircled (enclosed) by that particular loop -- currents lying entirely OUTSIDE the chosen loop contribute nothing to the integral, since their field's tangential (along-the-loop) component integrates to exactly zero over a full loop that does not enclose them.

To apply this law correctly, a definite sign convention is needed for currents that might flow in different directions relative to the chosen loop. The convention (Fig. 10.22) uses the same right-hand curl idea seen throughout the chapter: curl the fingers of the right hand along the SENSE OF INTEGRATION chosen for traversing the Amperian loop; a current flowing along the direction the outstretched THUMB then points is counted as POSITIVE, and a current flowing in the opposite direction is counted as NEGATIVE. Currents lying outside the loop are simply not counted (their net contribution to the integral cancels regardless of sign, as noted above). With this convention, for a loop encircling currents I1I_1, I2I_2 (out of the page, hence positive by the chosen sense) and I3I_3 (into the page, hence negative), with a fourth current I4I_4 lying outside the loop altogether,

∮B⃗⋅dl⃗=μ0(I1+I2−I3).\oint \vec{B}\cdot d\vec{l} = \mu_0(I_1+I_2-I_3).

As an immediate demonstration of the power of this law, apply it to re-derive the field of a long straight wire (already found the "hard way," by direct Biot-Savart integration, in Section 10.10.1). Choose a CIRCULAR Amperian loop of radius rr, centred on and coaxial with the wire (Fig. 10.23). By the wire's cylindrical symmetry, BB has the same magnitude everywhere on this loop, and is directed exactly TANGENT to the circle at every point -- exactly parallel to dl⃗d\vec{l} everywhere along this particular choice of loop, so B⃗⋅dl⃗=B dl\vec{B}\cdot d\vec{l}=B\,dl (no cosine factor needed) at every point. Ampere's law then gives, almost by inspection,

∮B dl=B∮dl=B(2πr)=μ0I⇒B=μ0I2πr,\oint B\,dl = B\oint dl = B(2\pi r) = \mu_0 I \quad\Rightarrow\quad B = \frac{\mu_0 I}{2\pi r}, …

Figure 10.22Fig. 10.22: Amperian loop
Fig. 10.22 — Fig. 10.22: Amperian loop

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. Cross-sections of four long straight wires, carrying currents I1I_1, I2I_2, I3I_3 and I4I_4, are shown as dots/crosses perpendicular to the plane of the page (into or out of the page, per wire, as individually marked). A closed Amperian loop (drawn as an irregular closed curve, not necessarily circular) is drawn so that it encircles three of the four wires -- I1I_1, I2I_2 and I3I_3 -- while wire I4I_4 lies entirely OUTSIDE the loop. Wires I1I_1 and I2I_2 are marked as coming OUT of the page (parallel to the curled-right-hand-thumb sense chosen for the loop, hence counted positive), while I3I_3 is marked as going INTO the page (opposite sense, hence counted negative) -- so Ampere's law for this loop gives ∮B⃗⋅dl⃗=μ0(I1+I2−I3)\oint\vec{B}\cdot d\vec{l}=\mu_0(I_1+I_2-I_3), wi …

Figure 10.23Fig. 10.23: Long straight current-carrying wire (Amperian loop)
Fig. 10.23 — Fig. 10.23: Long straight current-carrying wire (Amperian loop)

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 single long straight current-carrying wire, carrying current I, is shown with a CIRCULAR Amperian loop of radius r drawn concentric with the wire, in the plane perpendicular to it. At a representative point on this circular loop, the magnetic field vector B⃗\vec{B} is drawn TANGENT to the circle (consistent with the field lines being concentric circles around a straight wire, per the very first right-hand thumb rule of Section 10.1), and the length element dl⃗d\vec{l} of the Amperian loop at that same point is drawn tangent to the circle as well, i.e. exactly PARALLEL to B⃗\vec{B} there -- so B⃗⋅dl⃗=B dl\vec{B}\cdot d\vec{l}=B\,dl everywhere on this particular loop, which is precisely what makes the Ampere's-l …

Misc Ex.10.7Magnetic field inside, between, and outside a coaxial cable

Worked out. A coaxial cable consists of a central conducting core wire of radius a, and a coaxial cylindrical outer conductor of radius b, carrying an equal current I in OPPOSITE directions (one into, one out of the plane of the paper); the magnitude and direction of the magnetic field B is required for (i) the region between the two conductors, a<r<ba<r<b, and (ii) outside both conductors, r>br>b. By the cable's cylindrical symmetry, B is tangent to any circle of radius r centred on the central conductor. For a<r<ba<r<b, an Amperian circle of radius r encircles ONLY the central core's current I (the outer conductor's current lies outside this particular loop), so Ampere's law directly gives B⋅2πr=μ0IB\cdot2\pi r=\mu_0 I, i.e. B=μ0I2πrB=\frac{\mu_0 I}{2\pi r} -- identical in form to an isolated long straight wire's field, since only the inner conductor is enclosed. For r>br>b, an Amperian circle of that radius encircles BOTH conductors' currents, which are equal in magnitude but opposite in direction, so the net enclosed current is exactly zero and B=0B=0 every …