Magnetic field due to current loop

1. Oct 28, 2009

exitwound

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

2. Relevant equations

$$\vec B = \frac{\mu i \phi}{4 R}$$

3. The attempt at a solution

The field due to the curve R_1 is:

$$\vec B = \frac{\mu i \phi}{4 R}$$

$$\vec B = \frac{(1.26x10^-6)(.281)(\pi)}{4 (.023)} = -1.21x10^-5 \hat k$$

The field due to the curve R_2 is:

$$\vec B = \frac{(1.26x10^-6)(.281)(\pi)}{4 (.066)} = 4.21x10^-6 \hat k$$

Add 'em up and you get $-7.88x10^-6 k$ but apparently it's wrong.

What the heck am I doing wrong?? I even derived the formula separate from the book to ensure I knew what was going on.

2. Oct 28, 2009

Troels

First of all, what's the $$\phi$$ in the formula? The fraction of the circle? Note that the formula for a full loop is

$$\vec B = \frac{\mu_0 i}{2 R} \hat z$$

so you have already taken into account that you are dealing with only half a loop by multiplying with 1/2, getting a 4 in the denominator.

Now, the field due to the outer loop is:

$$\vec B_2 = \frac{\mu_0 i}{4 R_2} \hat z$$

and the inner loop:

$$\vec B_1 = -\frac{\mu_0 i}{4 R_1} \hat z$$

$$\vec B=\frac{\mu_0 i}{4 R_2} \hat z +\left( -\frac{\mu_0 i}{4 R_1} \hat z\right)=-\frac{\mu i}{4}\left(R_1^{-1}-R_2^{-1} \right) \hat z$$

Inserting the numericals I get:
(EDIT: The radii should of course be in meters!)
$$\vec B=-\frac{4\pi\times10^{-7}\cdot 0.281}{4}\left(0.023^{-1}-0.066^{-1}\right)=-2.5\times 10^{-6}\hat z\,\mathrm{T}$$
(with a little help from Wolfram Alpha)

So, you must have punched wrong on the calculatur somewhere :) Try to construct a complete expression before inserting numericals, that greatly reduces that risk

Last edited: Oct 28, 2009
3. Oct 28, 2009

exitwound

Phi is the partial angle of the circle. In this case, it's pi.

In your calculations, there's no variable for how far around the circle the loop goes. You need to have a phi to calculate whether or not you're going 360 degrees around the point, or 180, or 90 etc.

4. Oct 28, 2009

Troels

Oh but there is - it's the very first paragraph of my reply :)

No.

You need to multiply with the fraction of the circle you are dealing with. A full circle gives one value, a half circle gives half that value. Straight forward

5. Oct 28, 2009

exitwound

The equation should be:

$$\vec B = \frac{\mu i \phi}{4\pi R}$$

So when $\pi$ is put into Phi, the equation comes out as

$$\vec B = \frac{\mu_0 i}{4 R} \hat z$$

which is one half the full circle equation that you posted.

6. Oct 28, 2009

Troels

Oh yes, and I have to edit it again I'm affraid - seems I messed up my units on the radii - sorry :) :)

7. Oct 28, 2009

exitwound

Guess what, I submitted -2.5e-6 and it's wrong too.

8. Oct 28, 2009

Troels

Come to think of it, it says "magnitude" so a personal guess of mine is you should prob. omit the minus...

9. Oct 28, 2009

exitwound

Thats the correct answer. No negative. This is why I HATE HATE HATE this crappy online submission system. If that was handed in to a homework grader, I might have gotten 9/10 on it. But online would give me 0/10.

I complained to the dean about it, but he said they don't have enough money to pay staff to check 900 students' homeworks. They have enough money to constantly upgrade the football stadium though (Penn state). Did I Mention i HATE this format of homework?