Wrong B-Field Result: What Should I Have Done?

In summary: I'm not sure how that would change the answer though.Your guess would be that the current is supposed to be uniformly distributed across the disc, so each ring element width dr carries a current ##\frac {Idr}R##. I'm not sure how that would change the answer though.
  • #1
rmrribeiro
5
2
Homework Statement
I need to calculate de B (in a position along the axis) field created by a disk with current I.
The parametera were:
Radius of the disk: a=0,007 m
Current: I=1,07 A
Miu0= 1,2566E-6 NA-2
The position: Z=0,0593m
Relevant Equations
THe b filed along the a spire (or loop)
B= (Miu0 . I) (a^2/(a^2+z^2)^(3/2)
I integrated B within the limits of a (from 0 to 0.007)
teh result was 3.64E-10 T and it was wrong. the correcto one would be 5.8 E-4 T and it is a major diference (aprox 1 million times )
Waht shoud I have done?

Regards
 
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  • #2
Is it possible that you put in the wrong form of ##dA## (differential area element) when integrating over the disk?

##dA = r dr d \theta## for your double integral.
 
  • #3
rmrribeiro said:
Homework Statement:: I need to calculate de B (in a position along the axis) field created by a disk with current I.
The parametera were:
Radius of the disk: a=0,007 m
Current: I=1,07 A
Miu0= 1,2566E-6 NA-2
The position: Z=0,0593m
Relevant Equations:: THe b filed along the a spire (or loop)
B= (Miu0 . I) (a^2/(a^2+z^2)^(3/2)

I integrated B within the limits of a (from 0 to 0.007)
teh result was 3.64E-10 T and it was wrong. the correcto one would be 5.8 E-4 T and it is a major diference (aprox 1 million times )
Waht shoud I have done?
Please show your work so we can see what you actually did.
 
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  • #4
Imagem WhatsApp 2023-03-12 às 17.47.04.jpg
 
  • #5
My approach was that a disk is a “infinite” number of circular loops with radius between the values of ‘a’
 
  • #6
Your approach is clearly wrong because it produces a result with the wrong units. You're multiplying an expression which has units of T with da which has units of m, so your answer is going to be T•m, not just T as you want.

What does it mean for a disk to carry a current ##I##? It makes sense to talk about a current going around a ring. There's only one path the charges can follow, but that's not the case for a disk. Can you post the actual problem statement?
 
  • #7
The B field in a loop with current I is guiven by (at a guiven distance along the axis)
B=(μ0 . I) (a^2/(a^2+z^2)^(3/2)

If we have a continuos set of rings from 0 to 0.007 m, each with a current I= 1.07 A, what would be the B field in a point at 0.0593 m along the axis?
 
  • #8
If each infinitesimal ring carried a finite current of 1.07 A, there would be an infinite current flowing around the center of the disk because you would need an infinite number of rings. The field at z would be infinite.
 
  • #9
I realised that, but believed it to be some sort of missusing of words. ANd the suposed correct solution is B(0,0593)=5.8E-4 T
 
  • #10
Well, without a well-posed problem, you're just guessing.

In any case, the supposed answer seems to be unrealistically large given the numbers in the problem. If you calculate the field at ##z## for a ring of radius ##a## carrying current ##I##, the field is many orders of magnitude smaller than the supposed answer to the problem. I don't see how spreading the current around on a disk is going to change that.
 
Last edited:
  • #11
rmrribeiro said:
I realised that, but believed it to be some sort of missusing of words. ANd the suposed correct solution is B(0,0593)=5.8E-4 T
My guess would be that the current is supposed to be uniformly distributed across the disc, so each ring element width dr carries a current ##\frac {Idr}R##.
 

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