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United StatesPhysics C: Electricity and MagnetismSyllabus dot point

What is magnetic flux, and how is it computed as a surface integral of the field?

Topic 13.1 Magnetic Flux: define magnetic flux as the surface integral of the field and compute it for uniform and changing configurations.

A calculus-based answer to AP Physics C E&M Topic 13.1, covering magnetic flux as the surface integral of B, the area vector and angle dependence, flux through a coil of N turns, and how flux changes with field, area or orientation.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Defining magnetic flux
  3. The angle and the area vector
  4. Flux linkage and the three ways flux changes
  5. Try this

What this topic is asking

The College Board (Topic 13.1) wants you to define magnetic flux as the surface integral of the field through a loop, compute it for uniform and angled configurations, and recognize the three ways it can change. Magnetic flux is the quantity whose rate of change drives induction in the rest of the unit.

Defining magnetic flux

The dot product means only the component of B\vec{B} along the surface normal threads the surface. A field skimming parallel to the surface produces no flux; a field perpendicular to it produces the maximum, BABA. When the field is non-uniform or the surface is curved, the flux is the genuine surface integral BdA\displaystyle\int\vec{B}\cdot d\vec{A}, summing BdA\vec{B}\cdot d\vec{A} over every patch. For the loops in this course the field is usually uniform across a flat area, so the simple BAcosθBA\cos\theta form applies, but the integral definition is what makes magnetic flux the same kind of quantity as the electric flux of Unit 8.

The angle and the area vector

The angle θ\theta is measured between B\vec{B} and the surface normal, not the surface itself. A common slip is to use the field's angle with the plane of the loop; the two differ by 9090^\circ. When the loop's plane faces the field (normal parallel to B\vec{B}), θ=0\theta = 0 and Φ=BA\Phi = BA; when the plane lies along the field (normal perpendicular to B\vec{B}), θ=90\theta = 90^\circ and Φ=0\Phi = 0.

Flux linkage and the three ways flux changes

For a coil of NN identical turns, the total flux linkage is NΦBN\Phi_B, which is what matters for induction: each turn contributes the same flux, so a coil of many turns multiplies the effect, which is why generators and transformers use coils of hundreds of turns. The flux can change in exactly three ways:

  • the field BB changes (for example a magnet moves closer),
  • the area AA enclosed changes (a loop is stretched or a bar slides),
  • the orientation θ\theta changes (the loop rotates, as in a generator).

Try this

Q1. A 0.500.50 T field passes perpendicularly through a 0.100.10 m squared loop. Find the flux. [1 point]

  • Cue. Φ=BAcos0=(0.50)(0.10)=0.050\Phi = BA\cos 0 = (0.50)(0.10) = 0.050 Wb.

Q2. List the three ways the flux through a loop can change. [1 point]

  • Cue. Change the field BB, the area AA, or the orientation angle θ\theta.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice). A flat coil of area AA sits in a uniform field BB with its plane perpendicular to the field. The coil is then rotated so its plane is parallel to the field. The magnetic flux changes from (A) BABA to zero (B) zero to BABA (C) BABA to BABA (D) BABA to BA-BA. Justify your reasoning.
Show worked answer →

A 1-point MCQ on the orientation dependence of flux. The answer is (A).

Flux is Φ=BAcosθ\Phi = BA\cos\theta, with θ\theta between B\vec{B} and the area vector (the normal). When the plane is perpendicular to B\vec{B}, the normal is parallel to B\vec{B} (θ=0\theta = 0, Φ=BA\Phi = BA). When the plane is parallel to B\vec{B}, the normal is perpendicular to B\vec{B} (θ=90\theta = 90^\circ, Φ=0\Phi = 0). The trap is confusing the plane's orientation with the normal's.

AP 2024 (style)4 marksSection II (FRQ, quantitative). A circular coil of 5050 turns and radius 0.0400.040 m sits in a uniform 0.300.30 T field. (a) Calculate the flux through one turn when the coil's plane is perpendicular to the field. (b) Calculate the total flux linkage through all turns. (c) State what happens to the flux if the field is halved.
Show worked answer →

A 4-point FRQ on flux through a multi-turn coil.

(a) Single-turn flux (2 points): area A=πr2=π(0.040)2=5.03×103A = \pi r^2 = \pi(0.040)^2 = 5.03\times10^{-3} m squared. With the plane perpendicular to B\vec{B}, θ=0\theta = 0: Φ=BA=(0.30)(5.03×103)=1.51×103\Phi = BA = (0.30)(5.03\times10^{-3}) = 1.51\times10^{-3} Wb.
(b) Flux linkage (1 point): NΦ=(50)(1.51×103)=7.5×102N\Phi = (50)(1.51\times10^{-3}) = 7.5\times10^{-2} Wb.
(c) Halving the field (1 point): flux is proportional to BB, so halving the field halves the flux.

Markers reward the single-turn flux, the NΦN\Phi linkage, and the proportionality to BB.

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