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Related to Emf

  1. Mar 2, 2005 #1
    Hi,,

    A small circular loop of radius a and Resistance R is kept Coplanar and concentric with a large circular loop of radius b,a<<b.The large loop is kept fixed in space and a constant current I is sent through it.Now the small loop is rotated with a constant velocity w about a diameter.Neglecting the effect of back emf on the current in the small loop find the external torque required on the small loop to keep it in rotation...

    Pls help me to how to go for this Problem...I will be grateful for urs Help..
    Regards
    Heman
     
  2. jcsd
  3. Mar 4, 2005 #2

    Pls help me out...
     
  4. Mar 4, 2005 #3
    Hey Dexter...I know u can help me....Pls suggest me something how to go abt. it..
     
  5. Mar 4, 2005 #4
    Please be patient, I'm sure somebody will help you. :)
     
  6. Mar 4, 2005 #5
    Gentleman ...3 days have passed...i am thinking is that Problem really tough..
     
  7. Mar 4, 2005 #6
    Torque seems to be a function of time. After some calculations (lengthy), I am getting that torque is propotional to [tex]sin^2 \theta[/tex] where [tex]\w=d\theta/dt[/tex]. Are you supplied with the answer? Please post if you have.
     
  8. Mar 7, 2005 #7
    Thx for it...Sorry i dont have solution..bu can u tell me how did u reached here..i just need to know how to start in it and what is this back emf pointed out here
     
  9. Mar 7, 2005 #8
    Common Guys...u can tell me how to do it....Pls somehow collect some energy and Pls tell me how to go about it....Pls exercise urs brain a little bit for me..i will be highly thankful.
     
  10. Mar 7, 2005 #9
    Let B be the induced magnetic field at the small loop (since a<<b one can assume that B is uniform inside the small circle)
    [tex] B = \mu_0 I/2b [/tex]

    Let the loops be in the xy plane at the beginning. So, B is in z direction. Let the diameter about which the loop spins is along the y axis. At any time t, let [tex]\theta[/tex] be the angle between the plane of the loop and the xy plane. I have to add a diagram I guess.

    Flux [tex] \phi = B \pi a^2 cos \theta [/tex]

    Induced emf = [tex]-d\phi/dt = -B \pi a^2 w sin \theta [/tex]
    where i used [tex] w=\frac{d\theta} {dt}[/tex]

    Current I = emf/R

    [tex] I = \frac{B\pi a^2 w sin \theta}{R}[/tex]

    What is the force on the small loop due to the magnetic field B? Its easier to consider the component of B parallel and perpendicular to the loop's plane.

    [tex] B_ \bot = B cos \theta [/tex] does not contribute to the torque since the forces are directed towards the center (clockwise current)
    [tex] B_ \| = B sin \theta [/tex] is in the plane of the loop which contribute to the torque.


    To find the torque you need to find the torque due to each current element Idl and integrate. To do this consider a small current element at an angle alpha. see figure.
    [tex] \Gamma = \int_{-\pi}^{+\pi} \mu_0 I dl *cos \alpha *B_\| ( 2a *cos \alpha) [/tex]


    where [tex] dl = a d \alpha [/tex]

    [tex] \Gamma= \frac{\mu_0 w}{R}(2a^2 B \pi)^2 sin^2 (wt)[/tex]


    Comments welcome.
     
  11. Mar 7, 2005 #10
    Thanx Gamma....Thankx very very much for urs enthusiasm....
    I was unclear about the point how to take the Field....
    But still 1 doubt Gamma...
    How did u write the expression of Torque ..Pls Elaborate.
     
  12. Mar 7, 2005 #11
    Here is the diagram.

    [tex] dF= \mu_0 I dl x B_ \|= \mu_0Idl B_\| cos \alpha [/tex]

    [tex]d\Gamma = dF * 2a cos \alpha[/tex]
     

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