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Magnetic Force

  1. Jul 30, 2012 #1
    1. The problem statement, all variables and given/known data
    the current is i, the circuit is in a uniform magnetic field B normal to the plane of the circuit as indicated
    Attached is the circuit

    2. Relevant equations
    Get the forces acting on each of their sides. Direction and data module for the following:
    L = 10cm
    R = 10cm
    I = 5A
    B = 3.4mT

    3. The attempt at a solution
    The force for the semi circunference is easy F=I.l.B = IπRB
    But I don't agree with a teammate, about the rest.
    The bottom of the rectangle should be F=IlB = I2RB
    The sides of the rectangle oppose each other so the force should be 0?
     

    Attached Files:

  2. jcsd
  3. Jul 30, 2012 #2

    TSny

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    Hi, Radicaled

    It's not quite that easy :frown: Don't forget that the force on each infinitesimal section of current is a vector (with direction!). So, it's a vector addition problem.

    ok with the bottom, except you should probably include the direction of the force. For the two vertical sides, the problem seems to want you to state the force on each side separately.
     
  4. Jul 30, 2012 #3
    Ok, so I did this:
    For the semi circumference
    [itex]F = I \bar{I}x\bar{B}[/itex] in the -j direction
    So [itex]F = 5A \pi0.1m 3.4x10^{-3}T = 5.3x10^{-3}N[/itex]

    Now for the sides of the rectangle:
    [itex]F = I \bar{I}x\bar{B}[/itex] in the -i and in +i direction. So they cancel each other?
    [itex]F = 5A 0.1m 3.4x10^{-3}T = 1.7x10^{-3}N[/itex]

    The base of the rectangle should be like this
    [itex]F = I \bar{I}x\bar{B}[/itex] in the -j direction
    [itex]F = 5A 2x0.1m 3.4x10^{-3}T = 3.4x10^{-3}N[/itex]

    If the sides of the rectangle are 0, I would be able to sum only the base of the rectangle and the semi circumference.
    So the total force should be [itex]F = 3.4x10^{-3}N + 5.3x10^{-3}N[/itex]
     
  5. Jul 30, 2012 #4

    TSny

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    You haven't yet taken into account the vector nature of the force on each infinitesimal section of the semicircle. What would be the direction of the force on a small section of the semicircle at the far left side where the section is essentially vertical? What would be the direction of the force on a section at the far right side of the semicircle? How about for a section at the top of the semicircle? Think about some other sections. (It would help if you would indicate which direction you are choosing for the current in the circuit and which direction you are choosing for B.)

    That looks good.
    OK, but is the direction of the force on the bottom going to be the same direction as the force on the semicircle?
     
  6. Jul 30, 2012 #5
    Sorry i think I forgot to add that.
    The direction of the current I took was clockwise.
    The magnetic field, comes from the exercise description, is perpendicular to the plane of the circuit.
     
  7. Jul 30, 2012 #6

    TSny

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    Ok, good. But you'll still need to choose B either "out of the page" or "into the page".
     
  8. Jul 31, 2012 #7
    We consider the B into the page
     
  9. Jul 31, 2012 #8

    TSny

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    OK, then the direction of the force on the bottom side of the loop is -j as you stated. Good.

    So, the only thing left is to get the correct force on the semicircle. The usual way to do this is by integration of the force over all the infinitesimal length elements of the semicircle.
     
    Last edited: Jul 31, 2012
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