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A problem in magnetics.

  1. Feb 8, 2012 #1
    a problem in magnetics.........

    i've been stuck here for long.................plz help
    ;
    a square coil of side A ,carrying a current I ,is placed on a table and a magnetic field B parallel to one of the edges is switched on. the mass of coil is M
    now the question says-"find min magnetic field for which the coil will start tipping over"
    .
    this is how i went
    net magnetic force = zero
    net torque due to magnetic force= BI(L^2)
    net torque due to wieght=0
    .
    have seen its solution which says
    torque due to mag force=torque due to wieght
    nd equates Mg(L/2)=BI(L^2)
    which goes over my head.
    plzzzzzzzzz helpppppppppppp
     
  2. jcsd
  3. Feb 9, 2012 #2
    Re: a problem in magnetics.........

    Why do you think the net torque due to weight is 0? You've basically solved the problem once you realize there is a torque from the weight of the coil. If the two torques are competing, then while the loop is resting on the table the torque from gravity is either stronger or equal to the magnetic torque on the loop. However, if the torque from the coil is stronger than the torque from gravity the loops tips over.
     
  4. Feb 10, 2012 #3
    Re: a problem in magnetics.........

    i cant understand it.......
    there are 2 arms of loop that experience magnetic force....consider each arm - one would have torque due to magnetic force BI(L^2)/2 and a opposite torque due to weight Mg/4 * L/2 (shouldn't mass of only the arm be considered?) and on second one both torques in same direction
    the torque due to weight on each arm is in opposite directions and cancels out
     
  5. Feb 11, 2012 #4
    Re: a problem in magnetics.........

    plzzzzzzzzzzzzzz replyyyyyyyyyyyy
     
  6. Feb 11, 2012 #5
    Re: a problem in magnetics.........

    You're viewing the loop as floating. The loop is on the table so only the end that has torque in the direction of free space will move.
     
    Last edited: Feb 11, 2012
  7. Feb 11, 2012 #6
    Re: a problem in magnetics.........

    and so?????????????????
     
  8. Feb 11, 2012 #7
    Re: a problem in magnetics.........

    So ignore the torque at the other end because that end cannot move.
     
  9. Feb 11, 2012 #8
    Re: a problem in magnetics.........

    ok then.....i get Mg/4 *L\2 = BI(L^2)/2 ......which is still wrong!
     
  10. Feb 11, 2012 #9
    Re: a problem in magnetics.........

    No, take your pivot point to be the far leg (because this leg will always be pushed into the table by both gravity and the magnetic field it will actually be anchored pretty well), and now the torque due to gravity acts on the center of mass.
     
  11. Feb 12, 2012 #10
    Re: a problem in magnetics.........

    kkkk......got it.........thanks
     
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