# Magnetic field at the centre of current carrying spiral

1. Dec 8, 2015

### Titan97

1. The problem statement, all variables and given/known data
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

2. Relevant equations
$$B=\frac{n\mu_0i}{2R}$$

3. 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 $$n=\frac{N}{b-a}$$.

2. Dec 8, 2015

### TSny

Did you mean to have $a$ as the inner radius and $b$ as the outer radius?
OK. What did you get for your final answer for $B$?

3. Dec 9, 2015

### Titan97

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.

4. Dec 9, 2015

### TSny

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.

5. Dec 9, 2015

6. Dec 9, 2015

### TSny

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.