# Homework Help: Magnetic field at a point above a metal strip.

1. Apr 4, 2014

### Ruitker

1. The problem statement, all variables and given/known data
An infinitely long thin metal strip of width w=12cm carries a current of I=10A that is uniformly distributed across its cross section. What is the magnetic field at point P a distance a=3cm above the center of the strip?

2. Relevant equations

dB = (μ0*dI)/(2*∏*r)

3. The attempt at a solution

I evaluate the integral: ∫(μ0*I)/(2*∏*w)*(dr/r)

I am unsure of the limits and cannot get the correct answer.

2. Apr 4, 2014

### rude man

Your implied approach may be the best way. Place the strip in the x-z plane with its mid-height at y = 0 ( the height is arbitrarily small). The plane runs from - infinity < x < +infinity. The width is in the z direction (into & out of plane of page).

Consider a thin substrip of the strip of width dz and current dI = 10A*dz/0.12m and use Ampere's law for that substrip. The observation point is at P(0, 0.03m, 0).

Then add all the vector contributions to the field at P.

P.S. take advantage of symmetry about the z axis of the current distribution.

Last edited: Apr 4, 2014
3. Apr 5, 2014

### Ruitker

Thanks for your help. I am still confused how to form the integral.

4. Apr 5, 2014

### majormaaz

You're trying to integrate dB, right? It forms the total magnetic field at Point P, yeah? You have to take into account the magnetic contributions of the metal strip to get that B value.

To form the integral, you have to figure out which 'side' of the magnetic strip your magnetic contributions (your dB's) start from and which 'side' they end at. Would you start at the -∞ of the strip and integrate to the +∞ side of the strip?

Just things to think about.

5. Apr 5, 2014

### Ruitker

I tried this:
S we have B times the closed integral of ds (the B is taken out of the integral because the magnetic field is uniform at the circumference of our circle we drew as that is the amperian loop) which is equal to nu times the enclosed current. The closed integral of ds would just be s, so we have B x s=u(nu) x I (The "x's" representing multiplication). The s is just the circumference of the circle, which is 2 x pi x r (which in this case is "a"). So now we have B x (2 x pi x a)=u x I. If we solve for B, we get B=(u x I)/(2 x pi x a).

I plugged in my values but the answer is wrong.

6. Apr 5, 2014

### rude man

It's more complex than that.

Each thin strip generates its own dB field at P. dB is a vector. You have to vectorially sum all the dB's across the 12 cm width.

Pair the strips up: for each dB at one side of the middle of the sheet combine that dB with the dB from the opposite side. For example, the strip at +3 cm from the middle is combined with the dB at -3 cm. Integrate from -6 cm to + 6 cm across the width of the sheet.

7. Apr 5, 2014

### Ruitker

Would these be correct then?

Limits of +0.06 and -0.06 => ∫(μI)/(2∏w)*(dr/r) == (μI)/(2∏a)ln(w/2) ? It can't be because can't evaluate ln(-w/2)

8. Apr 5, 2014

### rude man

You're not treating dB as a vector. I see no vector algebra in your posts.

9. Apr 5, 2014

### Ruitker

I'm sorry, i have no idea what im doing. I entered an answer to the online homework grader but it was wrong. It doesn't give me a solution either :/. What equation should i have used?

10. Apr 5, 2014

### rude man

First, draw a picture so we can talk constructively:
x axis out of page, y axis up, z axis to the right. x ranges from -infinity to + infinity, z ranges from =w/2 to + w/2, and your point P is at (0,y0,0)
w = 12 cm
y0 = 3 cm
Assume I points in the +x direction.

Now take a slice dz at +z. What is the current in this slice?
Draw the B field at P due to this slice with the correct magnitude and direction.

Then take the slice dz opposite to the first slice, i.e. at -z. Draw the B vector for it at P also.
Then form the resultant. Where does it point? What's its magnitude dB?

Then integrate all the dB's with the appropriate limits of integration.
Check your answer: if w << y0, what is the B field at P? That's just a simple application of Ampere's law for a wire.

11. Apr 5, 2014

### Ruitker

This was the image we were given: http://imgur.com/QJ0rEmH

12. Apr 5, 2014

### Ruitker

I've come up with this:

B = -(μ*I)/(∏*w) * (arctan(w/2a))

Is this correct?

13. Apr 5, 2014

### rude man

Yes, but since B is a vector you should also indicate the direction. The magnitude B is correct.

it is.

14. Apr 5, 2014

### Ruitker

So, it would be j^.

I would include the negative sign in the answer.

15. Apr 5, 2014

### Ruitker

I have the correct answer, my calculator was in degrees instead of radians :( i can't believe i didn't notice for a while. Thanks for all of your help.

16. Apr 5, 2014

### rude man

You didn't define where j is pointing. Neither did the illustration you provided.

So describe the direction of B with respect to the illustration, or define your coordinate system?

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