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Find magnetic field strength

  1. Dec 20, 2012 #1
    1. The problem statement, all variables and given/known data

    magneticfieldstrength_zps065c6961.jpg

    2. Relevant equations

    force = current * length * field strength * sin (theta)

    current = 5.6 A
    mass/ length = 80g/m
    right angles = sin(90) = 1

    3. The attempt at a solution

    i noticed the mass is given in grams, 80 grams to be exact or .08 kg . how would I find the acceleration? if I can find acceleration(m/s^2) I can multiply that with kg to get the force
     
    Last edited: Dec 20, 2012
  2. jcsd
  3. Dec 20, 2012 #2

    haruspex

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    You don't need to worry about acceleration. This is equlibrium. Just balance the forces (in this case, forces per unit length).
     
  4. Dec 20, 2012 #3
    I do not understand. The question gave the units in grams and that unit threw me off. How do you mean balance the forces?
     
  5. Dec 20, 2012 #4

    haruspex

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    The magnetic force balancing the weight.
     
  6. Dec 21, 2012 #5
    the example in the book has something similar but the length is given in km and all the variables were given unlike this one. their equation is like this:

    magnetic force = current * length * magnetic field strength * sin theta

    i am going to use that with my variables

    i wind up converting grams to kilogram seeing as that is convention.

    magnetic force = 5.6 A * .08kg * magnetic field strength * (sin 90)

    the magnetic force is coming from the Earth. the textbook says the earth field is approx of a dipole in mu. units are in amperes * meters ^ 2. but it does not seem right. seeing as magnetic force is N/m

    so I was thinking what I learned many months ago in class: mass * gravity = weight. but you said I do not need to worry about acceleration
     
  7. Dec 21, 2012 #6

    haruspex

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    I was afraid you'd misinterpret that. You do have to consider gravitational force, but nothing is accelerating. The gravitational force (per unit length) is balancing the magnetic force (per unit length). Write that equation.
     
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