Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

The force between two current loops

We shall analyse the force which two current loops apply to each other (Figure 4.2.9). The current $I_{1}$ flowing along the loop $C_{1}$ causes a magnetic flux density $B_{1}$ everywhere in space, in particular at the points of the second loop $C_{2}$ . Due to the Lorentz law (Section 4.1) the force on a charge $ρ d V$ moving with the velocity $v_{2}$ in this field is $ρ d V v_{2} \times B_{1}$ , and the force on the whole loop $C_{2}$ is

F_{12} = \int ρ d V v_{2} \times B_{1} .

Since $ρ v_{2} = J_{2}$ is the current density in the second loop, and for a thin wire $J_{2} d V \to I_{2} d l$ is concentrated along the wire,

Force of one current loop on another

F_{12} = \int_{C_{2}} (J_{2} \times B_{1}) d l = \oint_{C_{2}} I_{2} d l \times B_{1} [N], (4.2.14)

and in the same way the force of loop 2 on loop 1 is $F_{21} = \int_{C_{1}} (J_{1} \times B_{2}) d l$ .

In plain words

Each loop makes a field, and the other loop's current feels it through the Lorentz force. Nothing new is needed: Biot–Savart for the field, $I d l \times B$ for the force. Problem 4.2.1 applies this to two parallel wires and finds the force per unit length $μ_{0} I_{1} I_{2} / (2 π d)$ , attractive for parallel currents, which is the definition of the ampere that served until 2019.

Two current loops. The current I_1 in C_1 produces the field \vec B_1 (one of its lines is drawn), which exerts a force on C_2 and sends a flux \Phi_{12} through the surface S_2 bounded by C_2. — **Figure 4.2.9.** Two current loops. The current $I_{1}$ in $C_{1}$ produces the field $B_{1}$ (one of its lines is drawn), which exerts a force on $C_{2}$ and sends a flux $Φ_{12}$ through the surface $S_{2}$ bounded by $C_{2}$ .

Two long loops seen end-on: drag them and reverse a current to see attraction turn into repulsion.

InteractiveLoading simulation…

Mutual inductance

In the same system we observe a magnetic flux due to $B_{1}$ passing through the surface $S_{2}$ that is bounded by $C_{2}$ . We represent this flux by $Φ_{12}$ ,

Φ_{12} = \int_{S_{2}} B_{1} \cdot d S [Wb] . (4.2.15)

The Biot–Savart law indicates that $B_{1}$ is directly proportional to $I_{1}$ : doubling the current doubles the field at every point. Hence $Φ_{12}$ must be proportional to $I_{1}$ as well, and the ratio of the two depends only on the shapes and positions of the loops and on the medium. We define this ratio as the mutual inductance.

Definition · Mutual inductance

L_{12} = \frac{Φ _{12}}{I _{1}} [Wb/A = H] . (4.2.16)

In plain words

$L_{12}$ is "webers through loop 2 per ampere in loop 1". It measures how strongly the two circuits are linked magnetically. Its unit, the henry, is named after J. Henry, who discovered self induction in 1832.

Flux linkage

A coil usually has many turns. If the number of turns in the second loop is $N_{2}$ , the flux $Φ_{12}$ threads each of them, and what matters for the circuit (as Faraday's law will show in Chapter 5) is the flux counted once for every turn. We define the flux linkage

Λ_{12} = N_{2} Φ_{12} [Wb], (4.2.17)

and the mutual inductance then becomes

L_{12} = \frac{Λ _{12}}{I _{1}} = \frac{N _{2} Φ _{12}}{I _{1}} . (4.2.18)

It can be shown that $L_{12} = L_{21}$ always: the flux that loop 1 sends through loop 2 per ampere equals the flux that loop 2 sends through loop 1 per ampere, however different the two loops are. This reciprocity is very useful in practice, because one of the two calculations is nearly always much easier than the other (Worked example 4.2.2, Exercise 4.2.7).

Self inductance

Besides linking $C_{2}$ , the field $B_{1}$ of course threads the loop $C_{1}$ that produces it. Part of the flux of $B_{1}$ links $C_{1}$ itself,

Φ_{11} = \int_{S_{1}} B_{1} \cdot d S, (4.2.19)

and considering the number of turns $N_{1}$ in loop $C_{1}$ we write $Λ_{11} = N_{1} Φ_{11}$ . Finally we define the self inductance.

Definition · Self inductance

L_{11} = \frac{Λ _{11}}{I _{1}} = \frac{N _{1} Φ _{11}}{I _{1}} [H] . (4.2.20)

When only one circuit is considered we write simply $L$ .

In plain words

Self inductance is "flux linkage of a circuit per ampere of its own current". It is a property of the circuit's geometry and of the medium around it, not of the current. A coil with many turns on an iron core has a large $L$ because each ampere makes a big flux and that flux is counted $N$ times.

Inductance and geometry: how L is actually computed

The definitions contain the current, but the current always cancels. The procedure is the same every time (Figure 4.2.10).

Recipe · Computing an inductance

Assume a current $I$ in the circuit and find $B$ (Ampère's law, with $μ$ of the core).
Compute the flux through one turn, $Φ = \int_{S} B \cdot d S$ .
Multiply by the number of turns linked, $Λ = N Φ$ .
Divide by the current, $L = Λ/ I$ . The current cancels, so $L$ depends on geometry and $μ$ only. For a mutual inductance, use the current of one circuit and the turns of the other.

For a long solenoid with $n = N / l$ turns per metre and cross-section $S$ on a core of permeability $μ$ , Ampère's law gives $H = n I$ inside (Problem 4.2.3), so that

⟹ B = μ n I, Λ = (n l) B S L = μ n^{2} S l = \frac{μ N ^{2} S}{l} [H] . (4.2.21)

For a toroid of $N$ turns, mean radius $R$ and cross-section $S ≪ R^{2}$ (Home exercise 4.2.3),

L = \frac{μ N ^{2} S}{2 π R} . (4.2.22)

In both, $L \propto N^{2}$ : doubling the turns doubles $B$ and doubles the number of linkages.

Flux linkage of a solenoid. The current I makes B=\mu NI/l, each turn links the flux \Phi=BS, the N turns link \Lambda=N\Phi in all, and L=\Lambda/I=\mu N^2S/l. — **Figure 4.2.10.** Flux linkage of a solenoid. The current $I$ makes $B = μ N I / l$ , each turn links the flux $Φ = B S$ , the $N$ turns link $Λ = N Φ$ in all, and $L = Λ/ I = μ N^{2} S / l$ .

A core with an air gap shows what the boundary conditions do to an inductance. Cut a gap of length $l_{g}$ into a toroid of mean length $l_{c}$ (Figure 4.2.11a). The flux density is continuous across the pole faces ( $B_{1 n} = B_{2 n}$ ), so the same $B$ crosses the iron, where $H = B / (μ_{r} μ_{0})$ , and the gap, where $H = B / μ_{0}$ . Ampère's law round the mean path gives

N I ⟹ B = \oint H \cdot d l = \frac{B}{μ _{0} μ _{r}} l_{c} + \frac{B}{μ _{0}} l_{g} = \frac{μ _{0} N I}{l _{g} + l _{c} / μ _{r}}, (4.2.23)

and with $Λ = N B S$ ,

L = \frac{μ _{0} N ^{2} S}{l _{g} + l _{c} / μ _{r}} [H] . (4.2.24)

The denominator is the length of air that would oppose the flux as much as the whole magnetic path does: the iron path counts only as $l_{c} / μ_{r}$ metres of air ("reluctances in series").

Open the air gap and watch $L$ fall and stop depending on $μ_{r}$ , while the allowed current grows.

InteractiveLoading simulation…

(b) L versus the gap length, Eq. (4.2.24). — (a) A toroid with an air gap.

Quick check

A coil of $N = 100$ turns links the flux $Φ = 2 mWb$ when it carries $0.5 A$ . What is its inductance, and what does it become if the turns are doubled on the same core?

$L = N Φ/ I = 100 \times 2 \times 1 0^{- 3} /0.5 = 0.4 H$ . With $200$ turns the same current makes twice the flux and it is linked twice as often: $L \propto N^{2}$ , so $L = 1.6 H$ .

Common mistake · Forgetting the turns, once or twice

The flux linkage is $N Φ$ , not $Φ$ : a student who computes $B$ correctly for an $N$ -turn coil (where $N$ enters once, through $N I$ ) and then writes $L = Φ/ I$ is short by a factor $N$ . The inductance of a coil goes as $N^{2}$ , never as $N$ .

4.2.4 Mutual effect of current loops and inductance

The force between two current loops

Mutual inductance

Flux linkage

Self inductance

Inductance and geometry: how L is actually computed