Ampere's Law: Find |H| at (6cm, 9cm, 0) in Toroid

[Natalie89]

1. Homework Statement

A toroid of a circular cross section whose center is at the origin and the axis the same as the z-axis has 1000 turns with po=10cm, a=1cm. If the toroid carries a 100mA current, find |H| at (6cm, 9cm, 0).



2. Homework Equations

How do I calculate |H| at certain points?



3. The Attempt at a Solution

Knowing that H = NI / 2*pi*po, and po - a < p < po + a, I have found p to be between 9cm and 11 cm.

I assumed H = H1 + H2.

If H1 = ((1000)(0.1) / 2 * pi * 0.9) * (0.06x + 0.09y) = 10.61x + 15.92y
Similarly, H2 is the same except p = 0.11 m. Therefore H2=8.68x + 13.02y.

After adding the two, and then finding the magnitude I get 34.8 A/m, but the answer is 147.1 A/m.

 

[kuruman]

I don't know what H1 and H2 are and you don't explain. Look, you have a formula that gives H as a function of the radius. Everywhere at that radius the field is the same. So, at what radius is point (6 cm, 9 cm, 0)?

 

[Natalie89]

H1 and H2 are H, except one is calculated at 9cm and one is calculated at 11cm. H, I'm assuming, is the sum.

 

[kuruman]

No it is not the sum. At a given radius r

$$H=\mu_{0}\frac{N I}{2\pi r}$$

The above expression says "You give me the radius, I will give you H."

So, at what radius is point (6 cm, 9 cm, 0)?

 

[Natalie89]

Ok thank you! That helps a lot. Even if I do find 'r', what do I do with the a and p values?

 

[kuruman]

What do the a and p values represent?