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Magnetic moment and magnetic field and dot product

  • #1
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Homework Statement


a magnetic moment of U = 1 (i) + 2 (k) , surrounded by a magnetic uniform field of B= 3 (i) + 4 (j) - 1 (k)


find the potential energy in mJ ?

Homework Equations


dot product of 2 vectors
( ui*bi)+(uj*bj)+(uk*bk) =

or finding the module of both vector and doing AB cos theta
but we would need to find the angle between the 2 vectors with the formula u*B / uB so dot product over product of both modules
3. The attempt at a


dot product of u and b ,so the answer should be 11.4 or 5,39 (539) ,
 

Answers and Replies

  • #2
kuruman
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What are the units of the numbers associated with the magnetic moment and the magnetic field? Without units any numerical answer is meaningless.
 
  • #3
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my apologies , yeah i know the importance of dimensions behind the units , and am quite interested in the meaning behind those value , which lead me to the heaveside and hill units , ergs units and cgs units , which , refering to your other thread , contains the abcoulomb :) and dimensional analysis

U dimensions of u are Amps per square meters so (M^1/2 L^3/2 T^2) for amps if im not mistaken
B is in millitesla mT so 10^-3 ( tesla in SI unit dimensional analysis is kg*sec^-2*Amp^-1 or kg/sec^2 *A )
 
  • #4
kuruman
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Homework Statement


a magnetic moment of U = 1 (i) + 2 (k) , surrounded by a magnetic uniform field of B= 3 (i) + 4 (j) - 1 (k)
find the potential energy in mJ ?

Homework Equations


dot product of 2 vectors
( ui*bi)+(uj*bj)+(uk*bk) =
or finding the module of both vector and doing AB cos theta
but we would need to find the angle between the 2 vectors with the formula u*B / uB so dot product over product of both modules
3. The attempt at a
dot product of u and b ,so the answer should be 11.4 or 5,39 (539) ,
It looks like neither of the answers you provided is correct. Forget the cosine of the angle and use the equation for dot product in component form because that's how the vectors are expressed. Be sure to use the correct units. If you choose to post an answer, please show your work in detail and not just a number. Also, it would help if you used LaTeX to write your equations. Finally, please add units where units belong when you post numbers.
 
  • #5
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LaTeX???

well dot product results in (1*3)i + (4*0)j + (2*-1)k = 3i -2k
module of vector is square root of 13 , equals = , so answer is 3.6 ???

that cant be right

3606 mJ?
 
Last edited:
  • #7
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ohhh its a programming language , i recently bought a mathlab book, and installed the full program. ive been using mathematica alpha, and kinda like it . anyway i dled Texlive and gonna check it out after the exam electricity and magnetism next week .
 
  • #8
Orodruin
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ohhh its a programming language
It is not. It is a text typesetting language.
 
  • #9
kuruman
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well dot product results in (1*3)i + (4*0)j + (2*-1)k = 3i -2k
module of vector is square root of 13 , equals = , so answer is 3.6 ???
Incorrect. ##(1\hat i+2\hat k)\cdot (3\hat i+4\hat j-1\hat k)=(1)(3)+(0)(4)+(2)(-1)##. The dot product is a scalar. There is no module to take. Also, the B field components are given in milliTesla. Where did you take that into account?
 
  • #10
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since B field is in milliTesla and answer has to be in milliJoules ,

thus i just need to find the convert ratio /constant /formula to convert from tesla to joules?


the answer from the dot product , which would be 3-2= 1 amp per meter square millitesla??
1 ampere per ((square meter) millitesla) =
1000 m^-2 kg^-1 s^2 A^2 , since
 
  • #11
kuruman
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the answer from the dot product , which would be 3-2= 1 amp per meter square millitesla??
And 1 amp per meter square millitesla is 1 milli-Joule :smile:
 
  • #12
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ty!!!!!
 
  • #13
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nope , correct answer is -1 apparently ,


and that question counted for 1% of my final grade....

so why -1 ?
 
  • #14
Orodruin
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Did you look up the expression for the potential energy? It is ##-\vec \mu \cdot \vec B##, not ##\vec \mu \cdot \vec B## ...
 

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