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Magnetic force on sides of square solenoid

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A square 5-turn coil with sides .03 m long and carrying 2 A of current is placed in a uniform magnetic field of .07 T, as shown.

    What is the magnitude and direction of the force on each of the four sides?

    What is the magnitude and direction of the torque [tex]\tau[/tex] on the coil about its center, if any? Use the fact from mechanics that [tex]\tau[/tex] = r x F due to a force F applied with lever arm r.

    What is the magnetic moment [tex]\mu[/tex] of this coil? What is the torque on the coil evaluated using the relation [tex]\tau[/tex] = [tex]\mu[/tex] x B?


    2. Relevant equations
    Force on a current-carrying wire
    F=I[tex]\hat{l}[/tex]x[tex]\vec{B}[/tex]

    Field of ideal solenoid
    B=[tex]\mu[/tex][tex]_{0}[/tex]NI[tex]\overline{l}[/tex]

    3. The attempt at a solution
    I wasn't sure how to translate F=I[tex]\hat{l}[/tex]x[tex]\vec{B}[/tex] into a coil with 5 loops, so I just multiplied IN[tex]\hat{l}[/tex]x[tex]\vec{B}[/tex]=(2A)(5 loops)(.03m)(.07T)=.0021 N for sides ab and cd. By right hand rule, I said ab is in the +[tex]\hat{k}[/tex] direction and cd is in the -[tex]\hat{k}[/tex] direction. I said bc and da were parallel to B so there was no force on them.

    My biggest question: When you calculate force of a coil in a uniform magnetic field, do you have to take into account the force that comes from the current itself?
    The uniform magnetic field seems to be from an external source, so to find total force do I also have to use the solenoid to find field and force from that, then add that force to the uniform field's force?

    Thanks very much for any help; I'm rather stumped by all this.
     

    Attached Files:

  2. jcsd
  3. Mar 22, 2010 #2
    Nevermind, I figured it out.
     
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