1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Magnetic Field in a Charging Capacitor

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A parallel-plate capacitor of capacitance C with circular plates is charged by a constant current I. The radius a of the plates is much larger than the distance d between them, so fringing effects are negligible. Calculate B(r), the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates.

    2. Relevant equations

    When a capacitor is charged, the electric field E, and hence the electric flux Φ, between the plates changes. This change in flux induces a magnetic field, according to Ampère's law as extended by Maxwell:
    ∮B⃗ ⋅dl⃗ =μ0(I+ϵ0dΦdt).
    You will calculate this magnetic field in the space between capacitor plates, where the electric flux changes but the conduction current I is zero.

    3. The attempt at a solution
    Since the I on the left is zero, I just say the answer is μϵ0dΦdt. This becomes μI which is incorrect.
  2. jcsd
  3. Dec 6, 2013 #2


    User Avatar

    Staff: Mentor

    Interesting problem -- I would approach it with Biot–Savart's law myself, looking at the currents flowing out to each pie-shaped piece of the capacitor plates, and letting the size of the pie pieces shrink to zero. There may be an easier way to do it with the hint that you posted, but I'm not seeing that easier way offhand.

    Can you post a try at this using Biot–Savart's law?
  4. Dec 7, 2013 #3


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    What you have here is a time-dependent but (to a good approximation) spatially homogeneous electric field. You are asked for the magnetic field as function of distance from the axis.

    I'd rather solve the local Maxwell equations than to try to use the integral form. In this case, between the plates you have [itex]\vec{j}=0[/itex] and thus (in SI units)
    [tex]\vec{\nabla} \cdot \vec{B}=0, \quad\vec{\nabla} \times \vec{B}=\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}.[/tex]
    I'd use the Coulomb gauge and introduce the vector potential as
    [tex]\vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{\nabla} \cdot \vec{A}=0.[/tex]
    Then you get
    [tex]\Delta \vec{A}=-\mu_0 \epsilon_0 \partial_t \vec{E}.[/tex]
    NB: Be careful with the proper definition of the Laplace operator acting on vector fields in curvilinear coordinates (here cylinder coordinates!).
  5. Dec 7, 2013 #4


    User Avatar
    Science Advisor

    At any given instant of time you have a homogenous electric field in between the capacitor plates; you can easily calculate this electric field using either Gauss's law or just using the standard formula for the electric field in between infinite (and in this case circular) parallel plate capacitors. After that you can calculate the electric flux in between the plates and use Ampere's law to get the magnetic field. Use the symmetries of the system to argue the symmetry of the magnetic field in order to make the line integral of the magnetic field trivial in Ampere's law.
  6. Dec 7, 2013 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You should reread the problem statement and ask yourself if you're answering the question that was asked.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: The Magnetic Field in a Charging Capacitor