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

4.1.6 Magnetic flux

$B$ is called the magnetic flux density, and like every flux density it is integrated over surfaces. The magnetic flux through a surface $S$ is determined by

Definition · Magnetic flux

Φ = \int_{S} B \cdot d S [Wb], 1 Wb = 1 T m^{2} = 1 V s . (4.1.19)

The rain picture of Section 3.2 applies unchanged: $Φ$ counts how many lines of $B$ pass through $S$ , each crossing weighted by the cosine of the angle between $B$ and the normal. Substituting $B = \nabla \times A$ into the above integral and taking into account Stokes' theorem (Section 2.4), we can easily find

Flux and the vector potential

Φ = \int_{S} (\nabla \times A) \cdot d S = \oint_{C} A \cdot d l, (4.1.20)

where $C$ is the rim of $S$ , traversed in the sense linked to the normal of $S$ by the right-hand rule.

$The flux through a surface depends only on its rim. The flat disk S_1 and the dome S_2 share the rim C. Together they form a closed surface, through which the net flux of \vec B is zero by Eq. (4.1.16), so the same flux passes through both. This is exactly what \Phi=\oint_C\vec A\cdot d\vec l expresses: the right-hand side knows only C.$

Figure 4.1.8. The flux through a surface depends only on its rim. The flat disk

S_{1}

and the dome

S_{2}

share the rim

C

. Together they form a closed surface, through which the net flux of

B

is zero by Eq. (4.1.16), so the same flux passes through both. This is exactly what

Φ = \oint_{C} A \cdot d l

expresses: the right-hand side knows only

C

In plain words

Because field lines never end, every surface stretched over the same rim is crossed by the same lines. The flux "belongs to the rim", and $\oint_{C} A \cdot d l$ is the formula that says so. One weber is one volt second: a flux of $1$ Wb through a loop that falls to zero in $1$ s induces $1$ V in that loop (Faraday's law, in the chapter on time-varying fields).

Exercise 4.1.7 computes the flux of a long wire through a nearby rectangle both ways, with $\int B \cdot d S$ and with $\oint A \cdot d l$ , and Exercise 4.1.19 uses the flux linked by a coil to measure a current. In the next chapter the flux through a circuit will define its inductance.

Quick check

A circular loop of radius $5$ cm lies with its normal at $6 0^{\circ}$ to a uniform field $B = 0.2$ T. What is the flux through it? What is the flux through the hemisphere that has the same rim?

$Φ = B S cos 6 0^{\circ} = 0.2 \times π (0.05)^{2} \times 0.5 = 7.85 \times 1 0^{- 4}$ Wb $= 0.79$ mWb. The hemisphere has the same rim, so the same flux passes through it.