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Bar Magnet Torque problem

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A bar magnet experiences a torque of magnitude 0.075 Nm when it is perpendicular to a 0.50 T external magnetic field. What is the strength of the bar magnet's on-axis magnetic field at a point 20 cm from the center of the magnet?

    2. Relevant equations

    [tex]\tau = \vec{\mu} B sin\vartheta[/tex]

    [tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3}[/tex]

    3. The attempt at a solution

    since its perpendicular, sin90 = 1 so:

    [tex]\tau = \vec{\mu} B \Rightarrow \vec{\mu} = \frac{\tau}{B} = \frac{0.075}{0.5} = 0.15[/tex]

    [tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3} = 10^-7 \frac{2(0.15)}{0.2^3} = 37.5 T[/tex]

    Any help would be appreciated. Thank You.
     
  2. jcsd
  3. Mar 25, 2008 #2
    i don't get it.. what is ur problem as such? as far as i see it, you have already solved it...
     
  4. Mar 25, 2008 #3
    I inputted 37.5 and it was wrong.
     
    Last edited: Mar 25, 2008
  5. Mar 26, 2008 #4
    Considering that 37.5 T is 750,000 times greater than that on the surface of the earth, you might want to recalculate that value.
     
  6. Mar 26, 2008 #5
    I see what I did wrong. I forgot to multiply it by 10^-7 so the answer should be 3.75 x 10^-6, and it is.
     
    Last edited: Mar 26, 2008
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