Ampere's Circuital Law

Statement

• Consider an open surface bounded by a closed curve $C$, with current passing through the surface. Break the boundary into small elements $d\vec{l}$, take the tangential component $B_t$ of the field at each, form the products $B_t\,dl = \vec{B}\cdot d\vec{l}$ and add them; in the limit the sum becomes an integral.
• Ampere's circuital law: the line integral of the magnetic field round a closed loop equals $\mu_0$ times the total current threading the surface bounded by that loop. $$\displaystyle\oint_{C} \vec{B}\cdot d\vec{l} = \mu_0 I$$
• Sign convention: curl the fingers of the right hand in the sense in which the boundary is traversed in the integral; the thumb then gives the direction in which current is counted positive.
• Validity: steady currents that do not vary with time.

The working (simplified) form

• Choose an amperian loop such that at every point either (i) $\vec{B}$ is tangential to the loop and constant in magnitude, or (ii) $\vec{B}$ is normal to the loop, or (iii) $\vec{B}$ vanishes.
• Then with $L$ the length of loop over which $\vec{B}$ is tangential and $I_e$ the enclosed current, $$B L = \mu_0 I_e$$
• This is the form actually used to get the field of a straight wire, a thick wire, and a solenoid.

Application 1: the long straight wire

• Symmetry says $\vec{B}$ has the same magnitude at all points of a circle of radius $r$ centred on the wire, and is tangential to that circle. Take the circle as the amperian loop, so $L = 2\pi r$: $$B\left(2\pi r\right) = \mu_0 I \;\Rightarrow\; B = \dfrac{\mu_0 I}{2\pi r}$$
• Points worth making in an answer: the field has cylindrical symmetry, depending only on $r$; the lines are closed concentric circles (unlike electrostatic lines, which begin and end on charges); the field is finite at any non-zero distance even though the wire is infinite; and the direction follows the right-hand rule (thumb along the current, fingers curling with $\vec{B}$).
• This result is the theoretical justification of Oersted's observation.

Application 2: a thick wire of radius $a$

• Outside, $r>a$: the loop encloses the whole current, $B(2\pi r) = \mu_0 I$, so $B = \dfrac{\mu_0 I}{2\pi r}$ and $B\propto 1/r$.
• Inside, $r<a$: the current is uniformly distributed, so the enclosed current is $I_e = I\dfrac{\pi r^{2}}{\pi a^{2}} = \dfrac{I r^{2}}{a^{2}}$, and $$B\left(2\pi r\right) = \dfrac{\mu_0 I r^{2}}{a^{2}} \;\Rightarrow\; B = \dfrac{\mu_0 I r}{2\pi a^{2}}, \qquad B\propto r$$
• Graph: $B$ rises linearly from zero at the axis to a maximum $\dfrac{\mu_0 I}{2\pi a}$ at the surface, then falls off as $1/r$ outside. Draw it as a straight line then a hyperbola, with the peak at $r=a$.

How Biot-Savart leads to Ampere's law

• The five-mark version. Take a circular amperian loop of radius $r$ round a long straight wire. From Biot-Savart's law, the field everywhere on this loop has the same magnitude $B = \dfrac{\mu_0}{4\pi}\dfrac{2I}{r}$ and is tangential, so it is parallel to $d\vec{l}$ at every point and $\vec{B}\cdot d\vec{l} = B\,dl\cos 0^\circ = B\,dl$.
• Hence $$\displaystyle\oint \vec{B}\cdot d\vec{l} = \oint B\,dl = \dfrac{\mu_0}{4\pi}\dfrac{2I}{r}\oint dl = \dfrac{\mu_0}{4\pi}\dfrac{2I}{r}\times 2\pi r = \mu_0 I$$
• The result, obtained here for a straight wire, is generalised by superposition to any closed loop and any distribution of steady current threading it. That generalisation is Ampere's circuital law.

Relation to Biot-Savart and to Gauss's law

• Ampere's law contains no new physics: it can be derived from the Biot-Savart law, exactly as Gauss's law can be derived from Coulomb's law.
• Both Ampere's and Gauss's laws relate a field quantity on the boundary to a source quantity inside — current for one, charge for the other.
• The law always holds, but it is only *useful* where the symmetry lets $B$ be taken out of the integral. It cannot, for instance, be used to extract $B = \dfrac{\mu_0 I}{2R}$ at the centre of a circular loop; there the Biot-Savart law is needed.

Ampere-Maxwell form

• Extended to time-varying situations by adding the displacement current: $$\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0\left(i_c + i_d\right)$$ where $i_c$ is the conduction current and $i_d$ the displacement current threading the loop. This is the one-mark answer when the question says "Ampere-Maxwell".

Diagram in words

• Draw the wire as a dot (current out of the page) with two concentric circles round it labelled loop 1 ($r<a$) and loop 2 ($r>a$), each with an arrow showing the sense of traversal and a tangential $\vec{B}$ arrow at one point.
• Beside it, sketch $B$ against $r$: a straight line from the origin up to $r=a$, then a $1/r$ decay.

Results to carry away

• $\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0 I$; simplified $BL = \mu_0 I_e$.
• Straight wire $\dfrac{\mu_0 I}{2\pi r}$; inside a thick wire $\dfrac{\mu_0 I r}{2\pi a^{2}}$.
• Ampere-Maxwell: $\displaystyle\oint \vec{B}\cdot d\vec{l} = \mu_0(i_c+i_d)$.
• Ampere is to Biot-Savart as Gauss is to Coulomb.