Theory

Sources of the magnetic field

How currents produce the magnetic field, from the Biot-Savart law to the fields of wires, loops and arcs, the force between parallel wires and the definition of the ampere, up to Ampère's law with its applications to a wire, a coaxial cable, a solenoid and a toroid, the flux of the magnetic field and a brief look at magnetic materials.

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Biot-Savart law
Field of a straight wire

Biot-Savart law

Throughout this topic the currents are steady, the conductors are in vacuum, except where magnetic materials are treated, and the wires are thin, that is with a cross section negligible compared with the distances involved.

Consider a thin circuit carrying the steady current $I$ . An infinitesimal segment of the wire, represented by the vector $d l$ tangent to the wire and oriented in the sense of the current, produces at the point $P$ the magnetic field

d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}},

where $r$ is the vector that goes from the segment to the point $P$ , $r$ is its magnitude and $\overset{r}{^}$ its unit vector. The field of the whole circuit is the sum of all the contributions:

B (P) = \frac{μ _{0} I}{4 π} \oint \frac{d l \times r ^}{r ^{2}} .

The constant $μ_{0} = 4 π \cdot 1 0^{- 7} T \cdot m/A$ is the magnetic permeability of vacuum. The Biot-Savart law is an experimental law, obtained in 1820 from measurements of the field produced by wires of various shapes, and it gives the magnetic field of currents as Coulomb’s law gives the electric field of charges.

The magnitude of the contribution is $d B = \frac{μ _{0}}{4 π} \frac{I d l sin φ}{r ^{2}}$ , with $φ$ the angle between $d l$ and $r$ , and it is zero at the points aligned with the segment. For example, a segment $1.0 mm$ long of a wire carrying $10 A$ produces, at a point $5.0 cm$ away in the direction perpendicular to the wire, the contribution $d B = 1 0^{- 7} \cdot \frac{10 \cdot 1.0 \cdot 1 0 ^{- 3}}{( 5.0 \cdot 1 0 ^{- 2} ) ^{2}} T = 4.0 \cdot 1 0^{- 7} T$ .

The segment

d l

of the wire produces at

P

the contribution

d B

, perpendicular both to

d l

and to

r

: here it points into the page

Direction and sense of the field

The contribution $d B$ is perpendicular to the plane containing $d l$ and $r$ , and its sense is found with the right-hand rule: with the thumb in the sense of the current, the curled fingers indicate the sense in which the field circulates around the wire. In the figures the symbol $⊙$ denotes a vector out of the page and the symbol $\otimes$ a vector into the page.

A steady current flows in a closed circuit, so only the integral extended to the whole circuit has a measurable meaning; the contribution of a part, for example of a straight segment, must be added to that of the rest of the circuit, or neglected if the rest is far away.

Field of a straight wire

1. Field of a straight segment

Consider a straight segment of a circuit, carrying the current $I$ , and a point $P$ at distance $d$ from the line of the segment. Let $H$ be the foot of the perpendicular drawn from $P$ to the line, and let $s$ be the coordinate along the line with origin at $H$ , oriented in the sense of the current; the segment goes from $s_{1}$ to $s_{2}$ , with $s_{1} < s_{2}$ . The contribution of the segment to the field at $P$ is perpendicular to the plane containing the wire and $P$ , and has magnitude

B = \frac{μ _{0} I}{4 π d} (\frac{s _{2}}{s _{2}^{2} + d ^{2}} - \frac{s _{1}}{s _{1}^{2} + d ^{2}}) = \frac{μ _{0} I}{4 π d} (sin θ_{2} - sin θ_{1}),

where $θ_{1}$ and $θ_{2}$ are the angles between $P H$ and the lines joining $P$ to the endpoints, taken positive on the side toward which the current flows and negative on the opposite side.

The formula is used for circuits made of straight segments, such as a square or rectangular loop: the field at a point is the vector sum of the contributions of the sides. For a segment $30 cm$ long carrying $5.0 A$ , at a point at $d = 10 cm$ from its midpoint we have $s_{2} = - s_{1} = 0.15 m$ , so $B = \frac{1 0 ^{- 7} \cdot 5.0}{0.10} \cdot 2 \cdot \frac{0.15}{0.1 5 ^{2} + 0.1 0 ^{2}} T \approx 8.3 \cdot 1 0^{- 6} T$ .

Straight segment from

s_{1}

s_{2}

and point

P

at distance

d

; with the current to the right the field at

P

points out of the page

Proof

We choose the $x$ axis along the wire, with origin at $H$ and oriented like the current, and the $y$ axis toward $P$ , so that $P = (0, d)$ . The infinitesimal segment at position $s$ is $d l = d s \overset{}{^}$ , and the vector going from the segment to $P$ is $r = - s \overset{}{^} + d \overset{}{^}$ , of magnitude $s^{2} + d^{2}$ . Since $\overset{}{^} \times \overset{}{^} = 0$ and $\overset{}{^} \times \overset{}{^} = \hat{k}$ ,

d l \times r = d d s \hat{k}, d B = \frac{μ _{0} I}{4 π} \frac{d l \times r}{r ^{3}} = \frac{μ _{0} I}{4 π} \frac{d d s}{( s ^{2} + d ^{2} ) ^{\frac{3}{2}}} \hat{k} .

All the contributions therefore have the direction and sense of $\hat{k}$ , and their magnitudes add up. With the substitution $s = d tan u$ we have $d s = \frac{d}{cos ^{2} u} d u$ and $(s^{2} + d^{2})^{\frac{3}{2}} = \frac{d ^{3}}{cos ^{3} u}$ , therefore

\int \frac{d d s}{( s ^{2} + d ^{2} ) ^{\frac{3}{2}}} = \frac{1}{d} \int cos u d u = \frac{sin u}{d} = \frac{s}{d s ^{2} + d ^{2}} + c .

Evaluating the antiderivative between $s_{1}$ and $s_{2}$ we obtain the claim; finally $\frac{s _{i}}{s _{i}^{2} + d ^{2}} = sin θ_{i}$ , because $s_{i}$ is the leg opposite $θ_{i}$ in the right triangle with leg $P H = d$ .

2. Field of an infinite straight wire

An infinite straight wire carrying the current $I$ produces, at distance $d$ from the wire, a field of magnitude

B = \frac{μ _{0} I}{2 π d} .

The field lines are circles centered on the wire, in planes perpendicular to the wire, traversed in the sense given by the right-hand rule. A semi-infinite wire ending at the foot $H$ of the perpendicular drawn from $P$ produces at $P$ a field of magnitude $\frac{μ _{0} I}{4 π d}$ , half of the previous one.

The infinite wire is the model of a long straight wire, at points whose distance from the wire is small compared with the distance from the ends and from the rest of the circuit. A wire carrying $10 A$ produces at $2.0 cm$ the field $B = \frac{4 π \cdot 1 0 ^{- 7} \cdot 10}{2 π \cdot 0.020} T = 1.0 \cdot 1 0^{- 4} T$ , of the same order of magnitude as the Earth’s magnetic field, which is a few times $1 0^{- 5} T$ .

Field lines of an infinite wire carrying the current

I

out of the page: concentric circles traversed counterclockwise, with

B = \frac{μ _{0} I}{2 π d}

Proof

We apply the theorem on the field of a straight segment. For the infinite wire $s_{1} \to - \infty$ and $s_{2} \to + \infty$ , so $sin θ_{1} \to - 1$ and $sin θ_{2} \to 1$ , and

B = \frac{μ _{0} I}{4 π d} (1 - (- 1)) = \frac{μ _{0} I}{2 π d} .

For the semi-infinite wire ending at $H$ we have $s_{1} = 0$ and $s_{2} \to + \infty$ , or $s_{1} \to - \infty$ and $s_{2} = 0$ ; in both cases the difference of the sines equals $1$ and $B = \frac{μ _{0} I}{4 π d}$ .

The theory continues with 7 more sections (field of a loop and of an arc, force between parallel wires and definition of the ampere, ampères law…), plus all worked examples, the solutions and the PDF.

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