Magnetic field at the centre of current carrying spiral

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Homework Help Overview

The problem involves calculating the magnetic field at the center of a spiral formed by a thin insulated wire carrying a current. The spiral has a specified number of turns, with defined inner and outer radii.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the magnetic field using different interpretations of the number of turns per unit area versus per unit width. Questions arise regarding the integration limits and the correctness of the derived expressions.

Discussion Status

There is ongoing exploration of the problem, with participants attempting to clarify their reasoning and check assumptions about the number of turns in relation to the spiral's geometry. Some participants express uncertainty about the final answer and seek confirmation or correction from others.

Contextual Notes

Participants note a potential typo regarding the designation of inner and outer radii, which may affect the interpretation of the problem. The discussion also highlights the importance of understanding how the number of turns is defined in relation to the spiral's dimensions.

Titan97
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Homework Statement


A thin insulated wire forms a plane spiral of N turns carrying a current ##i##. The inner radius is ##b## an outer radius is ##a##. Find magnetic field at centre of spiral

Homework Equations


$$B=\frac{n\mu_0i}{2R}$$

The Attempt at a Solution


For an area of ##\pi(b^2-a^2)##, number of turns is N.
So for an elemental area of ##2\pi r dr##,number of turns ##n## is ##\frac{2rNdr}{(b^2-a^2)}##

After substing ##n##, $$dB=\frac{\mu_0 I 2Ndr}{2(b^2-a^2)}$$
But on integrating, I am getting wrong answer

Instead of taking number of turns per unit area, if I took number turns per unit width, I got the correct answer. That is if $$n=\frac{N}{b-a}$$.

Why does the answer change?
 
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Titan97 said:
The inner radius is ##b## an outer radius is ##a##.
Did you mean to have ##a## as the inner radius and ##b## as the outer radius?
$$dB=\frac{\mu_0 I 2Ndr}{2(b^2-a^2)}$$
OK. What did you get for your final answer for ##B##?
 
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typo. Inner radius is ##a## and outer radius is ##b##.

Integrating from a to b, $$B=\frac{\mu_0 i N}{b+a}$$ This is not the answer given.
 
Titan97 said:
Integrating from a to b, $$B=\frac{\mu_0 i N}{b+a}$$ This is not the answer given.

That looks right to me. It has the correct limit for ##a## →##b##. Hopefully someone else will confirm this or else point out our mistake.

What answer was given?
 
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Ah, good. Thank you. Now I see the mistake. There are, in fact, a constant number of turns per unit radial distance. Just picture the individual wraps of wire. The wire has a fixed thickness. So, there will be a fixed number of turns per unit radial distance. So, it is not the area ##2 \pi r dr## that determines the number of wraps of wire in a distance ##dr##, but rather its just the length ##dr## itself that determines the number of wraps. Sorry for not catching that.
 
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