A-Level Physics October/November 2024 Q1(b)(iii): Calculate the magnetic flux cut by spoke AX during one revolution of the wheel. Give a…

1 answer

• ▲0
  accepted ✓

  The magnetic flux $\Phi$ is the product of the magnetic flux density B and the area A perpendicular to the field. In one revolution, the spoke AX sweeps out the entire area of the wheel.

  $\Phi = BA$

  The area swept is the area of a circle with radius $r = 0.85$ m. $A = \pi r^2$

  Substituting the values: $\Phi = B \times (\pi r^2)$ $\Phi = 0.18 \times \pi \times (0.85)^2$ $\Phi = 0.4084...$ Wb

  Magnetic flux = 0.41 Wb (to 2 s.f.)

  ---

  How the marks are awarded

  • C1 — Correctly stating the formula for magnetic flux, $\Phi = BA$.
  • C1 — Correctly substituting the given magnetic flux density and the area of the circle ($\pi r^2$) into the formula, as in $0.18 \times \pi \times 0.85^2$.
  • A1 — Calculating the final correct answer of 0.41 Wb, including the correct unit.

  Common mistakes

  • Using the circumference ($2\pi r$) or just the radius (r) instead of the area ($pi r^2$) for the 'A' term in the flux equation.
  • Forgetting to square the radius in the area calculation, i.e., calculating $A = \pi r$ instead of $A = \pi r^2$.
  • Omitting the unit 'Wb' from the final answer, which is explicitly requested in the question.
  • Using the diameter instead of the radius for the area calculation, leading to an answer four times too large.

  Examiner tip: Recognise that the total magnetic flux cut by a rotating conductor in one full revolution is equal to the flux passing through the entire area it sweeps out.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers✓ Official

Your answer

Sign in to answer this question.