Magnetic moment and magnetic field and dot product

In summary: 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?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?
  • #1
thee qs
34
0

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) ,
 
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  • #2
What are the units of the numbers associated with the magnetic moment and the magnetic field? Without units any numerical answer is meaningless.
 
  • #3
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 , referring 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 I am 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
thee qs said:

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
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 can't be right

3606 mJ?
 
Last edited:
  • #7
ohhh its a programming language , i recently bought a mathlab book, and installed the full program. I've been using mathematica alpha, and kinda like it . anyway i dled Texlive and going to check it out after the exam electricity and magnetism next week .
 
  • #8
thee qs said:
ohhh its a programming language
It is not. It is a text typesetting language.
 
  • #9
thee qs said:
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
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
thee qs said:
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
ty!
 
  • #13
nope , correct answer is -1 apparently ,


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

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

Related to Magnetic moment and magnetic field and dot product

1. What is magnetic moment?

Magnetic moment is a measure of the strength and direction of the magnetic field created by a spinning or orbiting electric charge. It is a vector quantity, meaning it has both magnitude and direction.

2. How is magnetic moment different from magnetic field?

Magnetic moment refers to the strength and direction of a magnetic field created by a charged particle, while magnetic field refers to the region around a magnet or electric current in which magnetic forces can be detected.

3. What is the dot product in relation to magnetic moment and magnetic field?

The dot product is a mathematical operation used to calculate the angle between two vectors. In the context of magnetic moment and magnetic field, the dot product can be used to determine the strength of the magnetic field at a specific point in space.

4. How is the dot product used to calculate the magnetic moment?

The dot product between the magnetic moment vector and the magnetic field vector is equal to the product of their magnitudes and the cosine of the angle between them. This can be used to calculate the magnetic moment when the magnitude and direction of the magnetic field are known.

5. Can the dot product be negative in the context of magnetic moment and magnetic field?

Yes, the dot product can be negative if the angle between the magnetic moment vector and the magnetic field vector is greater than 90 degrees. This indicates that the two vectors are pointing in opposite directions and the resulting magnetic moment is in the opposite direction of the magnetic field.

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