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!

Force due changing magnetic field

  1. May 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Uniform magnetic field ##= Bt## exist in cylindrical region of radius ##R## is pointing into the plane of figure (as shown in figure). A frictionless groove of length ##L## is fixed symmetrically from the centre O at a distance of ##\frac{L}{2}## .A charged particle of mass m and charge q is at rest at end B at ##t =0##. Neglect gravity and assume the particle just fits in the groove. Velocity of the particle at the end A is
    phy forums.JPG


    2. Relevant equations
    --

    3. The attempt at a solution
    Time varying magnetic field will generate an induced emf and hence an electric field. A component of this field will accelerate the particle. Now I have the potential as ##\pi r^2B## where ##r## is the distance from the origin. Now I am having difficulties calculating the electric field. What would be it's direction? Would it be in a concentric circle like induced current? If so, would differentiating w.r.t. r be correct?(since r would not change as potential changes).
     
  2. jcsd
  3. May 28, 2017 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Sounds right to me. Can you figure out the potential between the two ends of the groove?
     
  4. May 28, 2017 #3
    I don't really understand between which points exactly is the potential difference ##-\frac{d\phi}{dt}##. I can only think of it as the potential difference between the point and it's rotation by ##2\pi## radians, which seems nonsensical. But thinking about it this way I find that since the two points are at ##\frac{1}{4}## th rotation of a circle, the potential difference is ##\frac{1}{4} \pi Br^2## where ##r## is ##\frac{L}{\sqrt{2}}##.
     
  5. May 28, 2017 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What do you mean by this statement? The induced electric field is non-conservative, so you don't have a potential so to speak. What you can calculate is the potential difference between two points along some path ##C##:
    $$\Delta V = -\int_C \vec{E}\cdot d\vec{s}$$
     
  6. May 28, 2017 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    yes, that dawned on me in the middle of the night. The only way is to integrate the force / field along the groove.
     
  7. May 31, 2017 #6
    So ##\pi r^2 B ## is the potential difference when ##E## is integrated along a circular path. Assuming magnitude of E to be constant along the circular path ##2\pi r## times ##E## is ##\pi r^2 B##. So ##E## is ##\frac{r}{2}B##. Since you mentioned that the induced field is non-conservative, I guess the path does matter? So the potential difference between the two ends of the groove is
    $$\Delta V = -\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} B\frac{L}{4} \sec(\phi) \cos(\phi) d\phi $$
    Here I have written r in terms of ##\phi## which is the angle between the perpendicular bisector of the groove and the position vector. Only the ##\cos(\phi)## component of the Field is taken.
    phy forums upload.png

    From this ##\Delta V## is ##-\frac{\pi}{8}BL## which is very similar to the pot. difference I got in post#3 but seems dimensionally incorrect :frown:


    Also sorry for the late reply
     
  8. May 31, 2017 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I did not follow how you set up that integral.
    anyway, I think there is an easier way. Consider the triangular loop consisting of the groove, then the radius to the centre, then the radius back to the start of the groove.
    What is ∫E.dl along the radii? What is the area enclosed? What does that give for ∫E.dl along the groove?
     
  9. May 31, 2017 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You integrated ##E\cos\phi\,d\phi##, but you want to integrate ##E \cos\phi\,dx##. You need to write ##dx## in terms of ##d\phi##.
     
  10. May 31, 2017 #9
    ∫E.dl along the radii is zero. The area enclosed is ##\frac{L^2}{4}##. I am guessing this gives ∫E.dl along the groove, although I don't see how.
    Anyway I too, noticed something. ##E \cos(\phi)##, where ##\phi## is the angle I marked in my Paint diagram/scribble but forgot to label. So there is a constant field acting on the particle, i.e. a constant force and hence constant acceleration. Using ##v^2 = 2as## I got the correct velocity. What I get when using the area as potential difference is very similar, leading me to believe that the potential is somehow "B" times the area?
     
  11. May 31, 2017 #10
    Ahh thanks ! I knew something was wrong with my integral. But using ##x = \frac{L}{2} \sec\phi## and thus ##dx = \frac{L}{2}\sec \phi \tan \phi## would lead the integral being equal to ##-B\frac{L}{4} \sec \phi##. Putting in the limits I would get zero.....
     
  12. May 31, 2017 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You mixed up ##r = \frac L2 \sec\theta## and ##x = \frac L2 \tan\theta##. But your observation earlier that the component of the electric field along that path is constant lets you say
    $$\Delta V = -\int \vec{E}\cdot d\vec{r} = E_x \int dx = E_x L.$$
    (Too lazy to check the signs.)
     
    Last edited: May 31, 2017
  13. May 31, 2017 #12
    Thanks again! One last question though. Between which two points exactly is the potential difference ##-\frac{d\phi}{dt}## in cases such as these where the magnetic field varies with time. ?
     
  14. May 31, 2017 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, that's what ##\int_C\vec E.\vec {ds}=-\frac{\partial \phi}{\partial t}## says. φ Is the total flux through the area inside the loop C. Bt is the flux density at time t.
     
  15. May 31, 2017 #14
    Thank you! This makes much more sense now
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted