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Homework Help: Electromagnetism - Cross Product of length and Magnetic Field

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data
    A stiff wire 42.0cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y=2x in the xy plane. A current of 15.0 A flows in the wire-down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field given by B = (0.318i) T.

    2. Relevant equations
    F = I * l x B

    3. The attempt at a solution

    I am given all the elements, Current, length of wire, and Magnetic Field. I know that I have to figure out the cross product of the length and the magnetic field vectors, but this is where I have the most trouble.
    If I am looking at my notebook I am using an xy plane where the y axis is up/down and x is horizontal, same as we all learned in calculus. The z axis is the depth. Since the current is flowing down the z axis it is flowing into the page, so to speak.My graph diagram might look like this for the xy axis
    ...........| / y = 2x
    _______|/___________ x
    ........./ |
    with the z axis as depth (in and out of the screen)
    The length of each half of the wire is 0.21m.

    To find the angle I use old trusty TOA (tanθ = o/a) to give me an angle of 63.4°
    To find my x and y magnitudes I use SOH and CAH
    x = cos(63.4°)*0.21m = 0.09m
    y = sin(63.4°)*0.21m = 0.19m
    and since Z is flowing into the page does that mean that my z vector is 0.21m?

    Any help at this point is appreciated.

    I am going to continue trying to work this out while I wait for help.
    Last edited: Apr 11, 2012
  2. jcsd
  3. Apr 11, 2012 #2
    For the current flowing down the z axis, direction of current is of course in the [itex]\hat{-k}[/itex] direction. B is in [itex]\hat{i}[/itex] direction. So the force on this segment of the wire would be in [itex]\hat{-k}\times \hat{i}= \hat{-j}[/itex] direction.

    For the segment of the wire in the xy plane, I is flowing in the first quadrant I suppose. So
    the x and y components of vector I are positive. B is in [itex]\hat{i}[/itex] direction.
    So the force on this segment , which is cross product of vector I and vector B, would be
    in [itex]\hat{-k}[/itex] direction.
  4. Apr 11, 2012 #3
    I thouight that L is a vector as well? At least thats what my book indicates. So is the cross product between L and B or between I and B?
  5. Apr 11, 2012 #4
    well you can think of L as a vector or you can think of I as vector. The direction of both of them is same at any point....
  6. Apr 12, 2012 #5
    Ok so I got components of "L" as 0.09i, 0.19j, 0.21k in meters. I took the cross product of those with the only component of B which is 0.318i in Tesla. I got a result of
    -0.06678j and -0.06042k in Tm which gave me an angle of 47.9° below horizontal.
    Now I was assuming that the magnitude of the force would be easy to find using pythagorean theorem of 0.06678^2 + 0.06042^2 = 0.00811 then multiplying that by the current of 15.0A to get: 0.121N but that doesn't make sense...

    It should be at least greater than one newton. Unfortunately I was using a rework of the problem on MasteringPhysics so I can't check my work. But I used the same process with different data and still ended up with a result of less than a newton.
  7. Apr 13, 2012 #6
    why would get components of L as 0.09i, 0.19j, 0.21k ? One segment is flowing down the
    z axis and other segment is flowing in the xy plane. Work differently with these two segments, since the direction of force acting on these segments is different.
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