1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Magnetic force problem - please help!

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data

    So I'm working through textbook problems in preparation for my physics final, and I just can't seem to get the answers to this question (I'm 99.9% certain I got it wrong as my friends got a different value, but I don't actually know the answer as it's not in the back of the textbook); any help (particularly if you can walk me through your thinking processes...physics isn't my forte) would be greatly appreciated!

    "A segment of a high-voltage power line is 250 m long and carries 110 A current. At this particular location, Earth's magnetic field is directed toward the North, at an angle of 72 degrees below horizontal (i.e., downward into the ground at an angle). The field is uniform and has a magnitude of 0.59 Gauss. If current in the power line is flowing horizontally toward the east,

    A) what is the magnitude of the magnetic force on the wire? What is the direction of the magnetic force on the wire?
    B) If the current is flowing horizontally toward the south, what is the magnitude of the magnetic force on the wire? What is the direction of the magnetic force on the wire?"

    Thank you so much in advance!

    2. Relevant equations

    Units: 1 Gauss is equal to 10-4 T.

    I think this maybe calls for F=ILBsin(theta), in which I=current, L=length, B=magnetic field, F=magnetic force, and theta=the angle between the magnetic field lines and the current (I think? We kind of skipped over most of this in class)

    3. The attempt at a solution

    For the magnitude of the magnetic force in situation A: F=110X250X(5.9X10^-5)Xsin(72 degrees)=1.54...N
  2. jcsd
  3. Jun 5, 2013 #2
    First, look carefully at the geometry, Is there a 72 deg angle with the wire?
  4. Jun 5, 2013 #3
    you should use the more general formulae for force that is in vector form and gives both magnitude and direction.

    F = I L X B
  5. Jun 5, 2013 #4
    Alright, so I think I see what barryJ is saying - it's supposed to be an 18 degree angle above the horizon, correct (because it says 72 degrees below...I missed that the first few times), so then the equation would be F=ILBsin(theta) in which theta is 18 degrees, and this would yield 0.5013800734...N, but I just checked and that's not what my friends have gotten. =/ Thoughts/suggestions?
  6. Jun 5, 2013 #5
    you are definitely doing it wrong, The reason is that you have just memorised the formulae.

    Force on a current carrying conductor placed in a magnetic field is given by

    F = I L(cross )B, which reduces to> F = I*L*B*sin(theta)

    where theeta is the angle between the direction of current and the direction of magnetic field.

    Hint: you are taking theeta wrong.
  7. Jun 5, 2013 #6
    Maybe Annette123 has not learned cross products. It doesn't matter whether the angel is 72 deg or 50 degrees the force will be the same. Draw a diagram and look at the angle between the magnetic field and the axis of the wire.
  8. Jun 5, 2013 #7
    I should have said that the magnitude of the force does not depend on the angle but the direction will.
  9. Jun 5, 2013 #8
    That is why i mentioned cross product. And the problem itself is asking the direction of force. I think Annette123 has missed the topic cross products. I would advice Annette123 to go through the text once more.
  10. Jun 5, 2013 #9
    This problem can be solved by using the right hand rule and not knowing cross products, yes? The right hand rule is taught first and when the student advances, then they learn the cross product.
  11. Jun 6, 2013 #10
    Right hand rule only useful when the the B and L are along the axes( x, y or z) NOT when the B or L vectors are oblique. To solve using right hand rule one must first break the components along the orthogonal axes which will be quite tedious and impractical. The question asks direction so one must use the CROSS product.
  12. Jun 6, 2013 #11
    I don't think so. The right hand rule is when two of the three vectors are perpendicular to each other and this is the case. The wire goes East/west and the B field is North although down from the horizontal. The B field and the wire are perpendicular so we can use the right hand rule . Granted the RHR is a simplification of the cross product but in starting physics the RHR will suffice.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted