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Find magnitude of current of 2 wires separated by an angle

  1. Jul 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A telegraph consisted of 2 long straight wires carrying currents equal in magnitude but oppositely directed. Each was suspended by a pair of light-weight and inelastic threads that were about 6cm long. When current flowed through the wires, the magnetic force exerted between them caused them to move apart.

    Given that the linear mass density of the wire is 40 g/m, find the magnitude of the current in either wire when the angle separating the threads of the device is 16 degrees.

    2. Relevant equations
    Here's a pic of the object: http://i.imgur.com/zwZLzPX.png

    F = qv x B = qvBsin(theta)

    B = U/2∏ * I/R (Due to a long straight wire) (U is the permeability of empty space.)

    3. The attempt at a solution

    I tried solving for B from the first equation then plugging it into the 2nd then solving for I.

    I = FR2pi/qvsin(theta)

    But I'm pretty much stuck. I think I'm missing something really simple but not sure. I know I'm going to need to write an expression of the mag field due to the current in a long straight wire and force on a long straight wire carrying some current though.

    Any hints would be great :)
     
  2. jcsd
  3. Jul 21, 2014 #2

    rude man

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    Draw a free-body diagram of one of the wires, showing the magnitudes and directions of the forces involved.

    Also, you need to dig up the formula for the force between two parallel wires carrying current. (Or you can derive it from F= qv x B and Ampere's law. You've already stated that law).
     
    Last edited: Jul 21, 2014
  4. Jul 22, 2014 #3
    Yea the right hand rule is a bit confusing but I think I got it.

    The force you are talking about is F1 = B2I1L where L is the length (6cm).

    And since the current is equal and with this geometry, the forces are equal. As such, both their magnetic fields should be equal.

    Also since it's going about an angle as shown, I can use qB = mv/r I think.

    All my work has q in it from F=qvBsin(theta)... would I use a charge of a proton for q?
     
  5. Jul 22, 2014 #4

    BiGyElLoWhAt

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    I don't think 6 cm is the length of the wire. They are long straight wires, and the cables that hold them up are 6 cm (it's a radius).

    You want forces, but ultimitely you'll have forces acting on an object suspended by a string which can rotate from pi -> 2pi.
     
  6. Jul 22, 2014 #5

    rude man

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    Right. Except L is not 6 cm, as post 2 indicates. L is arbitrary.
    The forces are always equal, even if the two fields were not.

    This formula is totally irrelevant. It describes a charge moving circularly in a B field. No motion here.
     
  7. Jul 22, 2014 #6
    Oh right, duh.

    Alright so I was probably right in thinking it requires the F = qvBsin(theta). Where theta will probably be 16 degrees.

    However, that q in that equation confuses me.

    So this is what I got now:

    F = BIL = mv^2/r = qvBsin(theta)

    I = mv^2/(BLR) = qvsin(theta)/L

    So I have a velocity in one and that q and velocity in the other... both of which I don't know. I don't think substituting for one variable and inputting to the other would help at all.
     
  8. Jul 22, 2014 #7

    BiGyElLoWhAt

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    I think you're missing the point here. The theta in that equation comes from the fact that it's actually ##qv \times B = F##
    Its a cross product between the velocity of the CHARGE and the magnetic field its traveling through. You need to do a newtonian analysis of the situation, but you dont want F=qv×B. You can do that, but you have to sum up the force on every single electron. Theres a better equation that was mentioned in this thread.

    What happens when the 6cm wires make a 16 degree angle from the vertical?

    what would happen if you spontaneously removed one of the wires from this equilibrium position? What motion would the remaining wire exibit?
     
  9. Jul 24, 2014 #8
    did you ever find your answer to this problem??
     
  10. Jul 24, 2014 #9

    BiGyElLoWhAt

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    Not sure, they just kinda stopped posting.
     
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