Need help on sources of magnetic fields

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Homework Help Overview

The discussion revolves around calculating the magnetic field at a specific point due to a current loop consisting of three wires, focusing on the contributions from vertical and horizontal segments. The subject area is magnetism, particularly the application of the Biot-Savart Law and Ampere's Law in determining magnetic fields from current-carrying conductors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the Biot-Savart Law to find the magnetic field at the origin but questions their application and the correctness of their answer. Some participants clarify the contributions of the vertical wires and suggest a different approach to simplify the calculation. Others introduce a related scenario involving a straight wire and derive expressions for magnetic field contributions using integration.

Discussion Status

Participants are actively exploring different methods to calculate the magnetic field, with some providing alternative approaches and clarifications. There is no explicit consensus yet, but several productive lines of reasoning are being discussed.

Contextual Notes

Participants are working under the constraints of applying specific laws of electromagnetism and are considering the geometry of the setup, including distances and angles relevant to the magnetic field calculations.

andrew410
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Consider the current loop in the figure below.
Figure: http://east.ilrn.com/graphing/bca/user/appletImage?dbid=1450953670

Determine the magnetic field (in terms of a, I and d) at the origin.

So I will use mu as the permeability of free space. Since the two vertical wires have opposite directions in current they cancel out, right? If so, then all I have to calculate is the horizontal wire. Well...I use the biot-savart law, but can't seem to get the correct answer. My answer is ((mu*I)/(4*pi*d))*(2*cos(pi/4)). Did I apply the law wrong or am I just totally wrong?

Any help would be great, Thx!
 
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Since the two vertical wires have opposite directions in current they cancel out, right?

Actually, no. The wire with the current going up contributes the same magnetic field at point O as the wire going down (use the right hand grip rule). So determine the magnetic field due to the wire with the current going up (or down), double it, and add it to the magnetic field due to the horizontal wire.

In fact you can simplify the calculation even more with some clever tricks. The magnitude of the magnetic field at O due to the three wires in the diagram is the same as:

- The magnitude of the magnetic field due to one infinitely long wire perpendicular distance a from the point O (use Ampere's Law - easy), MINUS;
- The magnitude of the magnetic field due to a wire of length 2d & perpendicular distance a from the point O (use Biot Savart Law), MINUS;
- The magnitude of the magnetic field due to a wire of length 2a & perpendicular distance d from the point O (use Biot Savart Law again. In fact you can use the result of the previous step but swap a & d).

(think about why this is true).
 
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s_a said:
- The magnitude of the magnetic field due to one infinitely long wire perpendicular distance a from the point O (use Ampere's Law - easy), MINUS;

The magnitude is (mu*I)/(2*pi*a).

How do you get the other two magnitudes?
 
Suppose you have a straight wire carrying current I, of length 2L, and the midpoint M of the wire is a perpendicular distance D from a point O where you wish to find the magnetic field.

Suppose OM is the line going from point O to point M (the line OM meets the wire at 90 degrees). Suppose you have another line OT, going from the point O to the wire and meeting it at point T (T is anywhere on the wire). Let @ be the angle TOM (the angle between OT and OM). NB: |@| < arctan(L/D)

Using the Biot Savart Law, the magnetic field contribution dB, due to the current element at point T is:

dB = μ/(4*pi) * I (dl x ř)/r^2 (NB: ř = r/|r|)
= (μ/(4*pi) * I / r^2) * sin(pi/2 - @) dl
= (μ/(4*pi) * I / r^2) cos@ dl

now dl = D*tan(d@) = D*sec^2(@)*d@, and r = D/cos@, so:
dB = (μ/(4*pi) * I / D^2) cos^3(@) * D * sec^2(@) d@
= μ/(4*pi) * I/D * cos@ d@

To find B, integrate dB from @=0 to @=arctan(L/D), and then double the result to get the answer (by symmetry).

B = 2 * μ/(4*pi) * I/D * Integral{0 -> arctan(L/D)} cos@ d@
= μ/(2*pi) * I/D * sin(arctan(L/D))
= μ/(2*pi) * IL /(D * sqrt(D^2 + L^2) )

Use this result. Side note: What do you notice when L is very large?
 
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