Electromagnetism

Discussion in 'Homework Help' started by user123456, Feb 26, 2014.

  1. user123456

    Thread Starter New Member

    Jan 26, 2014
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    What is the force of attraction between a north and south pole separated by air? The flux density in the air is 0.84 Tesla and the rectangular poles measure 2cm by 3 cm. (168N)

    F = B^2/2u0

    0.84^2/2 x 1.257 x 10^-6

    Not getting 168Newtons, can anyone please give me a hand here?
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,767
    4,801
    Could you show your work? With units?

    If

    F = B^2/(2μ0)

    What are the units on B?

    What are the units on μ0?

    What are the units when you divide the first by (twice) the second?
     
  3. jjw

    Member

    Dec 24, 2013
    173
    31
    The formula is missing something.
    Hint: the area of the magnetic poles has an effect to the force.
     
  4. user123456

    Thread Starter New Member

    Jan 26, 2014
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    0
    Thanks for your reply,

    I multiplied 2 x 3 = 6 cm, still not getting the right figure.
     
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    cm are not SI units, you should be using metres.

    further 2cm x 3cm do not multiply to get 6cm.
     
  6. user123456

    Thread Starter New Member

    Jan 26, 2014
    12
    0
    2.66 x 10^-10 if i used meters.

    Thanks
     
  7. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    lol

    AREA is squared.
     
    user123456 likes this.
  8. WBahn

    Moderator

    Mar 31, 2012
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    Well, since the OP can't be troubled to answer these questions, I guess I will.

    The units on B are tesla, which is (Ns)/(Cm)

    The units on μ0 are H/m or N/(A^2)

    Thus, the units of B^2/(2μ0) are [(Ns)^2/(Cm)^2]*[(A^2)/N]

    which reduce to N/(m^2)

    Thus, a simple dimensional analysis reveals not only that the equation given simply cannot yield the desired answer, but gives a really strong indication of exactly what needs to be done in order to correct it.

    But, hey, units are just a troublesome bother that only waste time.
     
  9. jjw

    Member

    Dec 24, 2013
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    Your result is about 10^-10 times too small.
    The constant 1.257 x 10^-6 should be in denominator.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    <br />
F = \frac{B^2 A}{2 \mu_0}<br />
\ <br />
F = \frac{(0.84 T)^2 (2 cm \cdot 3 cm)}{2 \( 4\pi \cdot 10^{-7} \frac{N}{A^2} \)}<br />
\ <br />
F = \frac{ \( 0.84 \frac{Ns}{Cm} \)^2 (2 cm \cdot 3 cm)}{2 \( 4\pi \cdot 10^{-7} \frac{N}{\(C/s\)^2} \)}<br />
\ <br />
F = 1684000 \ \( \frac{ N^2 C^2 s^2 cm^2 }{N C^2 s^2 m^2} \)<br />
\ <br />
F = 1684000 \ \( \frac{ N \ cm^2 }{m^2} \)<br />
\ <br />
F = 1684000 \ \( \frac{ N \ cm^2 }{m^2} \) \cdot \( \frac{1 m}{100 cm} \)^2<br />
\ <br />
F = 168.4 N<br />

    See how wasting all that time on tracking units saves a lot of time?
     
    Last edited: Feb 28, 2014
  11. WBahn

    Moderator

    Mar 31, 2012
    17,767
    4,801
    You are right. That was an editing typo when doing a final fix-up of the LaTeX script.

    Thanks for catching it!
     
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