Magnetic flux and flux density: HSC Physics Module 6

Area of the loop:

$A = \pi r^2 = \pi (0.10)^2 = 3.14 \times 10^{-2}$ m$^2$.

(a) Plane perpendicular to the field means the field passes through the loop along its normal, so $\theta = 0°$ between $\vec{B}$ and the area vector:

$\Phi = B A \cos 0° = 0.50 \times 3.14 \times 10^{-2} \times 1 = 1.57 \times 10^{-2}$ Wb.

(b) "Plane at 30° to the field" means the field makes 30° with the plane, so the normal makes 60° with the field:

$\Phi = B A \cos 60° = 0.50 \times 3.14 \times 10^{-2} \times 0.5 = 7.85 \times 10^{-3}$ Wb.

Markers reward correct interpretation of "plane at X degrees to field" versus "normal at X degrees to field," correct area calculation, and units in webers.

Magnetic flux density $B$ is a vector quantity describing the strength and direction of the magnetic field at a point. Its SI unit is the tesla (T), where 1 T = 1 Wb/m$^2$ = 1 N/(A m). It tells you the force per unit current per unit length on a conductor placed at that point, or the force per unit charge per unit velocity on a moving charge.

Magnetic flux $\Phi$ is a scalar quantity describing the total magnetic field passing through a surface of area $A$. Its SI unit is the weber (Wb), where 1 Wb = 1 T m$^2$. It is given by $\Phi = B A \cos \theta$, where $\theta$ is the angle between the field direction and the normal to the surface.

Markers reward correct units for both, the area dependence of flux, and a clear statement that $B$ is per unit area and $\Phi$ is over the whole area.

Flux density $B$ is measured in tesla (T); flux $\Phi$ is measured in weber (Wb), where $1\ \text{Wb} = 1\ \text{T m}^2$.

$\Phi = BA\cos\theta$.

Marks: one for both correct units (T and Wb), one for the correct equation with $\theta$ identified as the angle to the normal.

Plane perpendicular to $\vec{B}$ means the normal is parallel to $\vec{B}$, so $\theta = 0^{\circ}$ and $\cos\theta = 1$.

$\Phi = BA\cos\theta = 0.40 \times 2.5 \times 10^{-2} \times 1 = 1.0 \times 10^{-2}\ \text{Wb}$.

Marks: one for correctly identifying $\theta = 0^{\circ}$ from "plane perpendicular to the field", one for the substitution, one for $\Phi = 1.0 \times 10^{-2}\ \text{Wb}$ with the correct unit.

The flux $\Phi = BA\cos\theta$ starts at its maximum value $\Phi = BA$ when $\theta = 0^{\circ}$ (all field lines thread the coil) and decreases smoothly, following a cosine curve, to zero at $\theta = 90^{\circ}$ (the field lies entirely in the plane of the coil, so none of it "passes through" along the normal).

Marks: one for stating the flux decreases from $BA$ to zero, one for linking this to the $\cos\theta$ dependence rather than a linear decrease.

(a) $\Phi = BA\cos\theta$, and for a fixed coil in a fixed field, $B$ and $A$ are constants, so $\Phi$ is directly proportional to $\cos\theta$: a straight line through the origin with gradient $BA$.

(b) Gradient $= \dfrac{\Delta \Phi}{\Delta(\cos\theta)} = \dfrac{(8.0 - 2.0) \times 10^{-3}\ \text{Wb}}{0.80 - 0.20} = \dfrac{6.0 \times 10^{-3}}{0.60} = 1.0 \times 10^{-2}\ \text{Wb}$.

(c) The gradient equals $BA$, so $B = \dfrac{\text{gradient}}{A} = \dfrac{1.0 \times 10^{-2}}{2.5 \times 10^{-2}} = 0.40\ \text{T}$.

Marks: one for the direct-proportionality explanation with $\Phi = BA\cos\theta$, one for a correctly calculated gradient with unit weber, one for identifying gradient $= BA$, one for $B = 0.40\ \text{T}$.

Area: $A = 0.15 \times 0.080 = 1.2 \times 10^{-2}\ \text{m}^2$.

"Plane at $50^{\circ}$ to the field" means the normal is at $90^{\circ} - 50^{\circ} = 40^{\circ}$ to the field.

$\Phi = BA\cos\theta = 0.60 \times 1.2 \times 10^{-2} \times \cos 40^{\circ} = 0.60 \times 1.2 \times 10^{-2} \times 0.766 = 5.5 \times 10^{-3}\ \text{Wb}$.

Marks: one for the correct area, one for correctly converting "plane at $50^{\circ}$" to "normal at $40^{\circ}$", one for $\Phi = 5.5 \times 10^{-3}\ \text{Wb}$ with the unit.

Band-6 plan. (1) State the working definitions of $B$ and $\Phi = NBA\cos\theta$ for the multi-turn coil. (2) Explain the direction-finding method (rotate the coil, flux linkage is maximum when the normal is along $\vec{B}$). (3) Explain the magnitude method (measure maximum flux linkage, then $B = \Phi_{\max}/(NA)$, or use an induced-EMF search-coil technique). (4) Evaluate limitations (uniformity assumption, coil size relative to field region, alignment precision, whether a static or changing field is used). Finish with a judgement on reliability.

Model answer. The flux linked by the $N$-turn coil is $\Psi = N\Phi = NBA\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and the coil's normal. Because $B$ and $A$ are fixed for a given position, $\Psi$ depends only on $\cos\theta$, reaching its single maximum value $\Psi_{\max} = NBA$ when the normal is exactly parallel to $\vec{B}$ ($\theta = 0^{\circ}$).

To find the direction of $\vec{B}$, the student rotates the coil about a fixed point between the poles while monitoring the flux linkage (in practice, via the induced EMF as the coil is rotated or briefly withdrawn, since a static flux alone cannot be read directly without a sensor). The orientation that gives the maximum reading identifies the normal direction, and hence the field direction, at that point.

To find the magnitude, the student uses the coil in that maximum orientation. If the flux linkage $\Psi_{\max}$ can be measured (for example, by suddenly removing the coil from the field and integrating the induced EMF, or with a calibrated Hall probe as a cross-check), then $B = \dfrac{\Psi_{\max}}{NA}$, using the known $N$ and $A$ of the coil.

Limitations. The method assumes the field is uniform over the area of the coil; if the coil is large compared with the region between the poles, it averages the field rather than measuring it at a point, underestimating peak $B$ near the pole faces. Precisely aligning the normal with $\vec{B}$ by eye introduces angular uncertainty, and since $\Phi \propto \cos\theta$ is fairly flat near $\theta = 0^{\circ}$, small misalignments barely change the reading, making the maximum hard to pinpoint exactly. The method also only gives $B$ at the coil's location, not the full field map.

Marker's note: the top band gives both the direction method (rotate to find the flux-linkage maximum) and the magnitude method (measured $\Psi_{\max}$ then $B = \Psi_{\max}/(NA)$), explicitly uses $\Psi = NBA\cos\theta$, and evaluates at least two genuine limitations (field non-uniformity over the coil area, angular alignment uncertainty near the flat top of the cosine curve). A response that only states the formula without the direction-finding reasoning caps in the middle band.

Band-6 plan. (1) State the claim is false and why: give the precise definitions of $B$ (field per unit area, a vector, at a point) and $\Phi$ (total field through a whole surface, a scalar). (2) Show the unit mismatch is not just a scaling issue (T vs Wb are dimensionally different, differing by $\text{m}^2$). (3) Give a concrete scenario (two coils of different area, or one coil at different angles) where equal $B$ gives different $\Phi$, or equal $\Phi$ arises from different $B$. (4) Conclude with a precise, judged statement.

Model answer. The claim is incorrect. Magnetic flux density $\vec{B}$ is a vector field quantity describing the strength and direction of the magnetic field at a single point in space, measured in tesla. Magnetic flux $\Phi$ is a scalar describing the total amount of field passing through a whole surface of finite area, measured in weber, and is calculated from $B$ by $\Phi = BA\cos\theta$. These are not the same physical quantity measured in different units (like metres and centimetres); they differ by a factor of area and orientation, so their units are dimensionally different ($1\ \text{Wb} = 1\ \text{T m}^2$, not a pure numerical conversion).

A concrete case shows why the distinction matters. Two coils sit in the same uniform field $B = 0.50\ \text{T}$, both with their normal aligned with the field ($\theta = 0^{\circ}$): a small coil of area $1.0 \times 10^{-3}\ \text{m}^2$ and a large coil of area $1.0 \times 10^{-1}\ \text{m}^2$. Both experience the identical flux density $B = 0.50\ \text{T}$, yet their fluxes are $\Phi_1 = 0.50 \times 1.0 \times 10^{-3} = 5.0 \times 10^{-4}\ \text{Wb}$ and $\Phi_2 = 0.50 \times 1.0 \times 10^{-1} = 5.0 \times 10^{-2}\ \text{Wb}$, a hundred-fold difference. If a student incorrectly treated $B$ and $\Phi$ as interchangeable, they would wrongly conclude the two coils experience "the same field effect," when in fact any induced EMF (which depends on the rate of change of $\Phi$, not $B$) would be a hundred times larger in the big coil for the same rate of rotation.

Treating $B$ and $\Phi$ as interchangeable would therefore lead directly to an incorrect prediction of induced EMF in a generator or search coil, since Faraday's law is stated in terms of $\Phi$ (or $N\Phi$), not $B$ alone.

Marker's note: the top band gives both correct definitions with correct units, explicitly notes the dimensional difference (area factor), and uses a worked numerical scenario (not just an assertion) to show the claim fails, linking the consequence to Faraday's law. A response that only recites "B is tesla, Phi is weber" without the worked contrast or the EMF consequence caps in the middle band.