Register to reply 
Magnetic field of a dipole in co ordinate free form 
Share this thread: 
#1
Jan2913, 10:20 AM

P: 23

Hello everybody!
I am writing this topi because i got stuck in this!I have a cylindrical magnet with 1,5mm Radius,2mm thickness and Br 1,38 Tesla! I want to calculate the magnetic field in a distance s = [0 0 0.01](in meters) ,that means in 1cm distance while my magnet's position is α = [0 0 0]. the vector form is [x y z] . using the equation found at http://en.wikipedia.org/wiki/Dipole#Magnitudefor Vector Form i got a result of [0 0 46.5498] Tesla which is impossible! for the magnetic moment calculation i used the type m =π*Br*d^2*l/(4 μ0) where d is the diameter of my magnet and l the thickness Can not spot my mistake since i expect to have uT as a result!Any opinion aprreciated!Thanx in advance! 


#2
Jan2913, 12:42 PM

Mentor
P: 11,589

I get m=0.01553 Am^2 (WolframAlpha) and 3.11mT (WolframAlpha), using your formulas and values and assuming the magnet is aligned with the zaxis.



#3
Jan2913, 12:52 PM

Mentor
P: 11,619




#4
Jan2913, 02:08 PM

P: 23

Magnetic field of a dipole in co ordinate free form
i used matlab to do my calculations!!!more specific!!!!
MagnetLoc = [0 0 0]; Sensors = [0 0 0.01] R = magnetLoc  Sensors'; rH = R./norm(R); theta = acos(R(3)/norm(R)); gamma = atan(R(2)/norm(R)); m1 = 0.0155171294871; % magnitude of magnetic moment m m = [m1*sin(theta)*cos(gamma) m1*sin(theta)*sin(gamma) m1*cos(theta)]; M = 1.2566370614 * (10^6); %vacuum perneability (μ0) A = M/(4*pi*(norm(R)^5)); C = (3 * dot( rH , m) * rH)'  ((norm(R)^2)*m); Field = A*C 


#5
Jan2913, 04:18 PM

Mentor
P: 11,619

OK, I'll move this over to the Matlab forum and maybe someone there can check whether you've implemented the equation properly.



#6
Jan2913, 04:55 PM

Mentor
P: 11,589

The first part (3 * dot( rH , m) * rH) should use R I think. Otherwise, you have an expression which grows (with R>0) with 1/R^5.



#7
Jan2913, 05:01 PM

P: 23

xmm!the equation says to use unit vector of R!if i use R i get even bigger magnitude!
What do you mean with your second recommendation? 


#8
Jan2913, 05:49 PM

Mentor
P: 11,589

(3 * dot( rH , m) * rH) does not depend on the magnitude of R. For a constant direction, your expression can be simplified to c/R^5 (neglecting the second term here). That is wrong, a dipole field is proportional to 1/R^3. 


#9
Jan2913, 06:17 PM

P: 23

ok you are right!when i use R instead of rH i get 0.0031 Tesla!
What i want to do is find the position of a magnet in an area 1cm to 3cm(see it as a cube) away from my sensor! In order to do that i calculate the theoritical value of the magnetic field at a specific position (here at 1cm) and then i use least square algorithm for the relationship : (BtheoriticalBexperiment) in order to minimize this and find the best solution!the problem is that even 3100uT is not even close to the value which my sensor gives to me at 1cm distance which is approximately 1000uT according to my sensor! 


#10
Jan3013, 09:02 AM

Mentor
P: 11,589

 1.38 Tesla could be the magnetic field at some specific point, not everywhere in the magnet
 higher moments (quadruple, ...) might influence the value a bit Do you get the same ratio measured/calculated for other distances? 


#11
Jan3013, 09:17 AM

P: 23

no is completely differenT!!!it drives me crazy!!!!can not find where i make the mistake..



#12
Jan3013, 09:34 AM

Mentor
P: 11,619

(By the way, we have a rule against using textmessage abbreviations here. Now you know.) 


#13
Jan3013, 09:38 AM

Mentor
P: 11,589

Can you give some examples for "different"? It might be possible to use a different formula to fit the data.



#14
Jan3013, 09:40 AM

P: 23

ok let me get some measurements again and i will post them as soon as i get them !!!



#15
Jan3113, 09:14 AM

P: 23

ok got the measurements!after calibration and removing the earth magnetic field i got :
for 1cm : x = 173uT y = 74.4733 z = 2048,1 uT for 2cm : x = 19.7439 y = 53,2893 z = 402,8459 uT for 3cm : x = 3,4647 y = 4,2611 z = 141,9412 uT 


#16
Feb113, 11:06 AM

Mentor
P: 11,589

2048/402.8=5.08 < 2^{3}, your field drops slower than a dipole field.
402.8/141.9=2.84 < 1.5^{3}=3.375, same here. The last value fits to the theoretic dipole prediction (~115μT), which is a hint that higher orders of the field could be relevant for 1cm and 2cm. I neglected the small x and ycomponents. 


#17
Feb313, 11:06 AM

P: 23




#18
Feb313, 12:12 PM

Mentor
P: 11,589

Your magnet is not a perfect, pointlike dipole. Those deviations can be expressed as quadrupole moment, sextupole moment, ...
Alternatively, measure more points, and find some effective formula as calibration. 


Register to reply 
Related Discussions  
Can a Magnetic Dipole Form a Circle?  General Physics  9  
Magnetic field at a dipole (how many atoms surround the dipole?)  Introductory Physics Homework  0  
Griffiths E&M 3.33 write efield of dipole moment in coordinate free form  Advanced Physics Homework  4  
Write the Magnetic Field of a dipole in coordinatefree form?  Advanced Physics Homework  2  
Question about a magnetic dipole in an inhomogeneous magnetic field.Please help ASAP  Advanced Physics Homework  3 