9478 / 2027
Electromagnetic Induction overview

Lesson 1 of 6

Flux through a coil

  • I can define magnetic flux using magnetic flux density and the cross-sectional area perpendicular to it.Syllabus

    Syllabus 9478, 18(a). Define magnetic flux using magnetic flux density and the cross-sectional area perpendicular to it.

  • I can explain and use magnetic flux linkage.Syllabus

    Syllabus 9478, 18(b). Explain and use magnetic flux linkage.

  • I can recall and use Phi = BA and N Phi = NBA for the stated perpendicular area and equally linked turns.Syllabus

    Syllabus 9478, 18(c). Recall and use Phi = BA and N Phi = NBA for the stated perpendicular area and equally linked turns.

Make a guess. You are not marked.

A flat coil sits in a uniform magnetic field with its plane perpendicular to the field. It is turned through 60 degrees about an axis in its own plane. What fraction of the original flux now passes through it?

  1. One half, because the flux depends on cos 60 degrees.
  2. Two thirds, because 60 degrees is two thirds of 90 degrees.
  3. About 0.87, because the flux depends on sin 60 degrees.
Show the answer

One half, because the flux depends on cos 60 degrees.

Flux is BA cos theta, where theta is the angle between B and the normal to the coil. The normal has turned through 60 degrees, and cos 60 degrees = 0.5.

Magnetic flux combines a field with the area it passes through. Flux linkage adds the contributions through a coil's turns. Define the area and its orientation before calculating either quantity.

For a uniform magnetic flux density B and an area A perpendicular to the field, magnetic flux is:

Φ = BA

B describes the local field and is measured in teslas, T. Flux Φ describes the field through the area and is measured in webers, Wb. A field-line drawing helps represent direction and distribution; counting its drawn lines does not determine a numerical flux.

Use the angle to the area normal

A flat surface's normal is a direction perpendicular to its plane. If θ is the angle between B and that normal, the perpendicular projected area is A cos θ:

Φ = BA cos θ

When the plane is perpendicular to B, its normal is parallel to B and flux magnitude is greatest. When the plane is parallel to B, its normal is perpendicular to B and flux is zero. If an acute angle is given to the plane instead, use the complementary angle or the corresponding sine projection.

0° between the field and the area normal

0° between the field and the area normalThe same physical flat area is viewed edge-on. The teal field points right. The selected grey normal makes 0 degrees to the field, on equal horizontal and vertical direction scales. The thick line representing the plane is perpendicular to that normal; the rest of the physical area extends out of this view. With B 0.200 tesla, area 0.00500 square metre and forty equal linked turns, the flux is 0.00100 weber and linkage is 0.0400 weber-turn. Screen line lengths do not measure the area projection.Same B, area A and turn count NBNormal nθ = 0°PlaneΦ = 0.00100 WbΛ = 0.0400 Wb-turn

The plane is perpendicular to B. Its full area is perpendicular to the field, so the flux magnitude is greatest.

60° between the field and the area normal

60° between the field and the area normalThe same physical flat area is viewed edge-on. The teal field points right. The selected grey normal makes 60 degrees to the field, on equal horizontal and vertical direction scales. The thick line representing the plane is perpendicular to that normal; the rest of the physical area extends out of this view. With B 0.200 tesla, area 0.00500 square metre and forty equal linked turns, the flux is 0.000500 weber and linkage is 0.0200 weber-turn. Screen line lengths do not measure the area projection.Same B, area A and turn count NBNormal n60°Plane30°Φ = 0.000500 WbΛ = 0.0200 Wb-turn

The angle to the normal is 60 degrees; the acute angle to the plane is 30 degrees. Use A cos(60 degrees), not A cos(30 degrees).

90° between the field and the area normal

90° between the field and the area normalThe same physical flat area is viewed edge-on. The teal field points right. The selected grey normal makes 90 degrees to the field, on equal horizontal and vertical direction scales. The thick line representing the plane is perpendicular to that normal; the rest of the physical area extends out of this view. With B 0.200 tesla, area 0.00500 square metre and forty equal linked turns, the flux is zero weber and linkage is zero weber-turn. Screen line lengths do not measure the area projection.Same B, area A and turn count NBNormal n90°PlaneΦ = 0 WbΛ = 0 Wb-turn

The plane is parallel to B. Its perpendicular projected area, flux and linkage are zero.

Each panel keeps the same B and physical area while changing the angle between B and the chosen normal. The edge-on plane is perpendicular to that normal; its full area extends out of the view. Use the supplied dimensions and projection, not a screen measurement.

Choosing a positive normal makes flux signed. A field in the opposite direction gives negative flux, and reversing the chosen normal reverses the assigned sign. The physical field does not reverse merely because a reference changes. For a nonuniform field, a single point's B cannot generally represent the whole area without an appropriate approximation.

Sum flux through the turns

For N turns, each linking the same flux Φ, define flux linkage Λ by:

Λ = NΦ = NBA   for perpendicular area
Λ = NBA cos θ   for the stated normal angle

The equal-flux condition matters. If different turns link different fluxes, add their individual fluxes. Increasing N increases linkage for the same imposed field and per-turn area; it does not itself increase that external B.

Worked orientation comparison

Separate one-turn flux from total linkage

A rough field of 0.2 T through about 5 × 10-3 m2 perpendicular to it gives flux of order 10-3 Wb. A few tens of equally linked turns then give a few hundredths of a weber-turn.

Use B = 0.200 T, A = 5.00 × 10-3 m2 per turn and N = 40. At a 60° angle between B and the chosen normal:

Φ = (0.200)(5.00 × 10-3) cos 60°
= 5.00 × 10-4 Wb
Λ = 40Φ = 2.00 × 10-2 Wb-turn

The angle between the plane and B is 30°, not 60°. With the normal instead parallel to B, one-turn flux is 1.00 × 10-3 Wb and linkage is 0.0400 Wb-turn. At a 90° normal angle, both are zero.

Check units and notation

1 Wb = 1 T m2 = 1 V s
= 1 kg m2 s-2 A-1

The turn count N is dimensionless. Writing linkage in Wb-turn makes the counting explicit, but a turn introduces no new physical dimension. Here Λ denotes linkage; it is distinct from Φ for flux through one turn and from symbols used for wavelength or decay elsewhere.

Try it step by step

Work it out in steps: Flux through one turn, BA cos theta, in Wb, then Flux linkage N x flux in Wb.

  1. Problem 1. A coil of 200 turns and area 4.0 x 10-3 m2 has its normal parallel to a uniform field of 0.050 T. Find the flux through one turn and the flux linkage.
  2. Problem 2. A coil of 150 turns and area 6.0 x 10-3 m2 is in a uniform field of 0.080 T. Its normal is at 30 degrees to the field. Find the flux linkage.
  3. Problem 3. A coil of 500 turns and area 2.0 x 10-3 m2 is in a uniform field of 0.12 T. Its normal is at 60 degrees to the field. Find the flux through one turn and the flux linkage.
Show the answer

Problem 1.

Flux through one turn, BA cos theta, in Wb: Flux = BA cos 0 = 0.050 x 4.0 x 10-3 x 1 = 2.0 x 10-4 Wb

Flux linkage N x flux in Wb: Flux linkage = 200 x 2.0 x 10-4 = 0.040 Wb

Problem 2.

Flux through one turn, BA cos theta, in Wb: Flux = 0.080 x 6.0 x 10-3 x cos 30 = 4.16 x 10-4 Wb

Flux linkage N x flux in Wb: Flux linkage = 150 x 4.16 x 10-4 = 0.0624 Wb

Problem 3.

Flux through one turn, BA cos theta, in Wb: Flux = 0.12 x 2.0 x 10-3 x cos 60 = 1.2 x 10-4 Wb

Flux linkage N x flux in Wb: Flux linkage = 500 x 1.2 x 10-4 = 0.060 Wb

Check your understanding

A 40-turn coil has area 5.00 x 10-3 m2 per turn in a uniform 0.200 T field. Every turn links the same flux and its positive normal is at 60 degrees to B. What is the flux linkage?

  1. 0.0400 Wb-turn.
  2. 0.0346 Wb-turn.
  3. 0.0200 Wb-turn.
Show the answer

0.0200 Wb-turn.

Each turn has Phi = BA cos(60 degrees) = 0.000500 Wb. Multiplying by 40 equally linked turns gives 0.0200 Wb-turn. The angle to the plane would be 30 degrees.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Define magnetic flux linkage. A coil of 200 turns and area 4.0 x 10-4 m2 is at right angles to a uniform magnetic field of 0.050 T. The field is reduced steadily to zero in 0.020 s. Calculate the initial flux linkage and the average induced e.m.f. [4]

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