3.7.5.3
The angle θ is between the field and the normal — not the coil face
Magnetic Flux & Flux Linkage — AQA A-Level Physics
- When the field hits a coil at an angle, only the component of the field perpendicular to the coil face contributes to the flux. The component parallel to the face slides along without passing through.
- The angle θ in the flux equation is measured between the magnetic field direction and the normal to the plane of the coil. The normal is a line sticking straight out from the coil face, perpendicular to it.
$$\Phi = BA \cos \theta$$
- $Φ$: magnetic fluxThe product of magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³. and the area perpendicular to the field. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area. (Wb). through the coil (Wb)
- $B$: magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³.The strength of a magnetic field. The force per unit length per unit currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). on a currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).-carrying conductor perpendicular to the field. Measured in teslaThe SI unit of magnetic flux density. One tesla is the flux density when a force of 1 N acts on a 1 m conductor carrying 1 A perpendicular to the field. (T). (T)
- $A$: area of the coil (m²)
- $θ$: angle between the field direction and the normal to the coil (°)
The two key cases
- Field perpendicular to the coil face: the normal is parallel to the field, so θ = 0°. cos 0° = 1. Flux is MAXIMUM: $\Phi = BA. Every field line passes straight through$.
- Field parallel to the coil face: the normal is perpendicular to the field, so θ = 90°. cos 90° = 0. Flux is ZERO. The field slides along the surface without passing through.
Angle θ between field and normal
Two diagrams side by side. Left: coil face perpendicular to field, normal parallel to B, θ = 0°, Φ = BA (maximum). Right: coil face parallel to field, normal perpendicular to B, θ = 90°, Φ = 0. The normal vector is drawn as a dashed arrow from the centre of the coil.
Common Mistake
MEDIUM
Students often: Measuring θ from the coil face (the plane of the coil) instead of from the normal to the coil.
Instead: θ is the angle between B and the NORMAL to the coil face. If you measure from the plane instead, you'll get the complementary angle and use sin instead of cos. When the field is perpendicular to the coil (maximum flux), θ = 0°, not 90°.
Instead: θ is the angle between B and the NORMAL to the coil face. If you measure from the plane instead, you'll get the complementary angle and use sin instead of cos. When the field is perpendicular to the coil (maximum flux), θ = 0°, not 90°.
Common Mistake
MEDIUM
Students often: Using sin θ instead of cos θ in the flux equation.
Instead: Φ = BA cos θ, not BA sin θ. The cosine is needed because θ is measured from the normal. When θ = 0° (field along the normal), cos 0° = 1 and flux is maximum. This is the correct physics — the field passes straight through.
Instead: Φ = BA cos θ, not BA sin θ. The cosine is needed because θ is measured from the normal. When θ = 0° (field along the normal), cos 0° = 1 and flux is maximum. This is the correct physics — the field passes straight through.
Worked Example
A circular coil of area 3.0 × 10⁻² m² is placed in a uniform magnetic field of flux densityMass per unit volume of a material. Measured in kg m⁻³. 0.15 T. The normal to the coil makes an angle of 40° with the field direction. Calculate the magnetic fluxThe product of magnetic flux density and the area perpendicular to the field. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area. (Wb). through the coil.
Show Solution
1
List known values
- Flux density: $B = 0.15 \text{ T}$
- Area: $A = 3.0 \times 10^{-2} \text{ m}^2$
- Angle between field and normal: $\theta = 40°$
2
Select the equation
The coil is at an angle to the field, so use:
$$\Phi = BA \cos \theta$$3
Substitute values
$$\Phi = 0.15 \times 3.0 \times 10^{-2} \times \cos 40°$$
4
Evaluate
$$\cos 40° = 0.766$$
$$\Phi = 0.15 \times 3.0 \times 10^{-2} \times 0.766$$
$$= 3.4 \times 10^{-3} \text{ Wb}$$
$$= 3.4 \text{ mWb}$$Answer
$\Phi = 3.4 \times 10^{-3}$ Wb (3.4 mWb)
Examiner Tips and Tricks
- If a question says 'the coil is tilted at 40° to the field', you need to work out what θ is.
- Is 40° measured from the field to the coil face, or from the field to the normal?
- Draw a quick sketch.
- If 40° is the angle between the field and the coil face, then θ = 90° − 40° = 50° (because the normal is perpendicular to the face).