Calculating a large toroid's magnetic field

[annamal]

Homework Statement

A current of 1.5 A flows through the windings of a large, thin toroid with 200 turns per meter and a radius of 1 meter. If the toroid is filled with iron for which $X = 3\times 10^3$, what is the magnetic field within it?

Relevant Equations

$B_0 = \frac{\mu_0(N)(I)}{2\pi r}$
$B_{net} = B_0 + B_m = (1+X)\times B_0$

$B_0 = \frac{\mu_0(N)(I)}{2\pi r}$
$\frac{N}{2\pi r} = 200$
$B_{net} = B_0 + B_m = (1+X)\times B_0$
Plugging in the numbers:
$B_0 = 4\pi\times 10^{-7}(200)(1.5) = 3.8\times 10^{-4}$

$B_{net} = (1+X)\times B_0 = (1+ 3\times 10^3)\times 3.8\times 10^{-4}$ = 1.13 T

But the answers says it is 0.18 T.

Where did I go wrong?

 

[TSny]

Relevant Equations:: $B_0 = \frac{\mu_0(N)(I)}{2\pi r}$

$B_0 = 4\pi\times 10^{-7}(200)(1.5) = 3.8\times 10^{-4}$

Looks like you forgot to divide by $2 \pi r$

 

[annamal]

Looks like you forgot to divide by $2 \pi r$

I am given N is 200 turns per meters which should be $\frac{N}{2\pi r}$, so that should be taken into account with the substitution of 200 in the equation

 

[kuruman]

I am given N is 200 turns per meters which should be $\frac{N}{2\pi r}$, so that should be taken into account with the substitution of 200 in the equation

You are given 200 which is $N$ not $\frac{N}{2\pi r}$. Take a good look at the derivation for the field inside a toroid and at what the symbols stand for.

 

[TSny]

I am given N is 200 turns per meters which should be $\frac{N}{2\pi r}$, so that should be taken into account with the substitution of 200 in the equation

Yes, you are correct! I mistakenly took the 200 to be the total number of turns even though it clearly states "200 turns per meter". Looks like whoever wrote the answers made the same mistake as I did. Sorry about that. I agree with your answer.

 

[annamal]

You are given 200 which is $N$ not $\frac{N}{2\pi r}$. Take a good look at the derivation for the field inside a toroid and at what the symbols stand for.

200 is the turns per meter so it should equal n = $\frac{N}{2\pi r}$

$B_0 = \frac{\mu_0(N)(I)}{2\pi r}$ where N is the number of times the toroid is rotated. which means N/Length = 200 where length = $2\pi r$

 

[kuruman]

Yes, you are correct! I mistakenly took the 200 to be the total number of turns even though it clearly states "200 turns per meter". Looks like whoever wrote the answers made the same mistake as I did. Sorry about that. I agree with your answer.

Me too. Sorry about the confusion.