Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 2 emf problems

  1. Mar 3, 2006 #1
    1] A square loop of wire, with sides of length 'a' lies in the first quadrant of the xy-plane, with one corner at the origin. In this region there is a non-uniform time-dependent magnetic field [itex]\vec B (y,t) = ky^3t^2\hat z[/itex]. Find the induced emf in the loop.

    I applied the flux rule here.

    [tex]\varepsilon = -{d\Phi \over dt} = -{d(\vec B \cdot \vec A) \over dt} = {d(Ba^2) \over dt} = -a^2\left({d(ky^3t^2)\over dt}\right)[/tex]

    Am I going right here?

    2] A perfectly conducting spherical shell of radius 'a' rotates about the z-axis with angular velocity [itex]\omega[/itex] in a uniform magnetic field [itex]\vec B = B_0\hat z[/itex]. Calculate the emf developed between the "north pole" and the equator.

    I evaluted the flux here:
    [tex]\Phi = \vec B \cdot \vec A = B_0\left(4\pi a^2\right)[/tex]

    But the answer given here is: [tex]{1\over 2}B_0\omega a^2[/tex]

    How do I incorporate [itex]\omega[/itex] in my flux?
    Last edited: Mar 3, 2006
  2. jcsd
  3. Mar 3, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    1) What is [itex]\vec A[/itex]? From your equation, I think you mean it to be the vector area of the surface. You have to be careful, since the field is not uniform, it depends strongly on y. So the flux through the square is not simply the product of B and the area A. You have to integrate over the surface to find the flux at a certain instant of time:

    [tex]\Phi(t)=\int \limits_{\square}\vec B \cdot d\vec a[/itex].
    Then find [itex]\varepsilon[/itex] by taking the time-derivative.

    ONLY when [itex]\vec B[/itex] does not depend on position can you remove it from the integral and then:
    [tex]\Phi=\int \vec B \cdot d\vec a=\vec B \cdot \int d\vec a=\vec B \cdot \vec A[/tex]

    2) This problem is a bit different. You have a conducting surface rotating a magnetic field, so there is a magnetic force acting on each surface element, given by the Lorentz force law. The integral of this force per unit charge gives the emf.
  4. Mar 4, 2006 #3
    Thank you very much for your reply :smile:.
    Yes, I should have been more careful (since [itex]\vec A[/itex] could also mean the magnetic potential.)
    OK, so the flux will be:
    [tex]\Phi = k\int_0^a \int_0^a y^3t^2 dxdy[/tex]

    [tex]\Phi (t) = k{a^5\over 4}t^2[/tex]

    So, the emf will be:
    [tex]\varepsilon = -2k{a^5\over 4}t[/tex]

    Is my evaluation correct?
  5. Mar 4, 2006 #4
    Lorentz force? You mean:
    [tex]\vec F_{mag} = \int(\vec v \times \vec B)\sigma d\vec a[/tex]

    So, I take the force per unit charge as [itex]\vec f[/itex] & [itex]\vec v = \omega r[/itex]
    So the emf will be:
    [tex]\varepsilon = -{d\Phi \over dt} = \oint \vec f \cdot d\vec l = \int_0^a B_0\omega rdr = B_0\omega {a^2\over 2}[/tex]

    Yippee, I hope it is correct.
    Last edited: Mar 4, 2006
  6. Mar 4, 2006 #5


    User Avatar
    Science Advisor
    Homework Helper

    1) Looks good.
    2) [itex]\vec f \times \vec v=\omega rB_0 \hat r[/itex] is correct, but note that the direction is pointing away from the axis of rotation and that r is the distance from this axis (cylindrical coordinates as opposed to spherical).

    Anyways, both answers look correct :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook