Determining the B-field in center of current carrying loop

In summary: The actual B-field at the center is very different because the infinitesimal dl's closer to the center have a larger contribution to the B-field as they are closer to the center.In summary, it is incorrect to apply Ampere's law in this situation because it does not meet the necessary conditions. Instead, Biot-Savart's law should be used to calculate the B-field in the center of the circular loop of current.
  • #1
kaashmonee
7
1

Homework Statement


Determine the B-field inside the middle of a circular loop of current.

Homework Equations


Attempt at using Ampere's law: ##\oint \vec{B} \cdot d \vec{l} = \mu_0 i##

The Attempt at a Solution


##\oint B \cdot R d \theta = \mu_0 i \Rightarrow BR(2 \pi) = \mu_0 i \Rightarrow B = \frac{\mu_0 i}{2 \pi R}## How do I get the ##2 \pi## term to cancel out when doing this with Ampere's law?
 
Physics news on Phys.org
  • #2
kaashmonee said:
How do I get the ##2 \pi## term to cancel out when doing this with Ampere's law?
You don't. Although Ampere's law is always valid, the symmetry of the problem is not high enough to allow calculation of the B-field using it. You have to use the Law of Biot-Savart.
 
  • #3
To be more specific, you have gone about using Ampere's law in a way different from what it is saying. Instead of looking at the B-field in a point and summing up the contributions from all parts of the loop, what you have done is to integrate the B-field along the loop and considered the current as a current enclosed by that loop. This procedure will give you the B-field at a distance ##R## from a straight long infinite wire carrying a current ##i##, not the B-field in the center of a loop carrying the current ##i##. I believe this should serve as a good warning that you cannot apply formulae blindly without understanding what they are saying.

As @kuruman said, try applying Biot-Savart's law instead and be careful in what it says to interpret the result correctly.
 
  • #4
Awesome, thank you for your replies! But I'm still wondering why it would be incorrect to take a tiny dl of the current carrying loop, draw an amperian loop around it, and add up the upward facing components of the B field?
 
  • #5
kaashmonee said:
Awesome, thank you for your replies! But I'm still wondering why it would be incorrect to take a tiny dl of the current carrying loop, draw an amperian loop around it, and add up the upward facing components of the B field?
Again, this is not what Ampere's law is telling you. Ampere's law is about the magnetic field at the Amperian loop and the current that passes through it. It is not about the current at the loop and the magnetic field inside it. What you need is Biot-Savart's law that tells you what the contribution to the magnetic field at a given point is from a small current carrying segment somewhere else. You indeed get the full magnetic field by summing those contributions, i.e., integrating.
 
  • #6
kaashmonee said:
But I'm still wondering why it would be incorrect to take a tiny dl of the current carrying loop, draw an amperian loop around it, and add up the upward facing components of the B field?
Whatever Amperian loop you use, it has to enclose the current and it has to pass through the point where you want to find the B-field (center of the loop) and add the tangential component of the B-field around the complete loop, not just a "tiny" element dl. That's what ##\oint \vec B \cdot d\vec l## is saying you should do. Ampere's law is useful for finding the B-field when two conditions are met, (a) ## \vec B## is either parallel or perpendicular to ##d \vec l## in which case ##\vec B \cdot d\vec l= B~dl## or ##\vec B \cdot d\vec l= 0## and (b) ##B## is constant on the loop so that it can be taken out of the integral, in which case ##\oint dl=C## where ##C## is the perimeter of the loop. In this problem, if you consider an Amperian loop that encloses the current, neither of these conditions is met and that's why you cannot use Ampere's law. I repeat that the fact that you cannot use Ampere's law in this case does not invalidate Ampere's law. If you could do the line integral, say numerically, you would indeed get a number that is equal to ##\mu_0i##.

Now to complete the picture for you, if you make the radius of the Amperian loop very very tiny so that it is much less than the radius of the wire loop, the circle locally will look like a long straight wire much like a straight line drawn parallel to the surface of the Earth looks "straight" at small distances above it. In this approximation, your calculation will give what you got in your original post. However, that is the field very near the circumference of the loop and very far from the center of the wire loop, which is the point where you want to find the B-field.
 

Related to Determining the B-field in center of current carrying loop

1. How is the B-field calculated in the center of a current carrying loop?

The B-field in the center of a current carrying loop can be calculated using the formula B = (μ0*I)/(2*r), where μ0 is the permeability of free space, I is the current in the loop, and r is the radius of the loop.

2. What is the direction of the B-field in the center of a current carrying loop?

The B-field in the center of a current carrying loop is perpendicular to the plane of the loop and follows the right-hand rule. This means that if you curl your fingers in the direction of the current, your thumb will point in the direction of the B-field.

3. Is the B-field stronger inside or outside of the current carrying loop?

The B-field is stronger inside the current carrying loop. As you move further away from the center, the B-field decreases in strength.

4. Can the B-field be affected by changing the current or the radius of the loop?

Yes, the B-field in the center of a current carrying loop is directly proportional to the current and inversely proportional to the radius of the loop. This means that increasing the current or decreasing the radius will result in a stronger B-field.

5. How is the B-field affected if the current is reversed in direction?

If the direction of the current is reversed, the direction of the B-field will also be reversed. This means that the B-field will now point in the opposite direction, following the right-hand rule.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
363
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
143
  • Introductory Physics Homework Help
Replies
3
Views
182
  • Introductory Physics Homework Help
Replies
4
Views
914
Replies
2
Views
921
  • Introductory Physics Homework Help
Replies
4
Views
83
Replies
4
Views
969
  • Introductory Physics Homework Help
Replies
12
Views
609
  • Introductory Physics Homework Help
Replies
1
Views
728
Back
Top