# Magnetic field in the open loop

#### tomfrank

1. Homework Statement
I am trying to find the magnetic field at the point in the middle of the picture can someone give a hint on how to approach the problem?

2. Homework Equations

3. The Attempt at a Solution

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#### ApexOfDE

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any help?

#### berkeman

Mentor
1. Homework Statement
I am trying to find the magnetic field at the point in the middle of the picture can someone give a hint on how to approach the problem?

2. Homework Equations

3. The Attempt at a Solution
The hint is to use the Biot-Savart law... (Hint: it often simplifies the integration to make your variable of integration theta...)

#### ApexOfDE

you should divide the wire into some parts and find m.f created by these parts.

#### tomfrank

would the left and the right side cancel out....

So I have to do it just for the bottom side???

#### tomfrank

so thus this integral work :

B= ($$\mu$$*I)/(4*pi*x)$$\int$$ between(-pi/4) and (pi/4) of cos ($$\theta) d\theta$$

#### fluidistic

Gold Member
would the left and the right side cancel out....

So I have to do it just for the bottom side???
I believe so (though I might be wrong). In all cases you should prove it via arithmetic so don't use this shortcut if you are not sure.

#### ApexOfDE

so thus this integral work :

B= ($$\mu$$*I)/(4*pi*x)$$\int$$ between(-pi/4) and (pi/4) of cos ($$\theta) d\theta$$
No. Biot-Savart:

$$dB = \frac{\mu_0 I}{4 \pi} \frac{dl sin\theta}{r^2}$$

With $$\theta$$ is angle between dl and r. imo, you can use symmetry property to make life easier.

#### tomfrank

so i stil do not quite get it....does the two side cancel out?

$$dB = \frac{\mu_0 I}{4\pi r^2} {dl sin\theta}$$

will the angle be (-pi/4) to (pi/4)

#### ApexOfDE

No. r is changing with theta. Can you find relationship between sin(theta) and r? and between r and x?

About two sides, you can say that they cancel out.

#### tomfrank

i see how you are looking at the picture.

basically the

sin (theta) dl = cos (theta') dl

and

1/r^2 = cos^2(theta')/ x^2

when i put all of them in the integral reduce down to

$$dB = \frac{\mu_0 I}{4\pi x} {dl cos\theta}$$

i do not know yet what boundary to use for the integral

#### ApexOfDE

1/r^2 = cos^2(theta')/ x^2
from this, i guess theta and theta' as shown in fig below:

http://img149.imageshack.us/img149/6568/88285645.jpg [Broken]

is that right?

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#### rl.bhat

Homework Helper
dB = μo/4π*I*di*sinθ/r^2
where sinθ = x/r = x/sqrt( x^2 + l^2)
Take the limits of integration from -x/2 to +x/2.

#### tomfrank

yes the picture with the angle is exactly what I was thinking....what are the limits for the integral angles???

#### berkeman

Mentor
yes the picture with the angle is exactly what I was thinking....what are the limits for the integral angles???
The limits come from the geometry of the diagram. You really should be able to figure that part out.

And on your PM question to me, know, the B-field contributions from the two sides on the left and right do not cancel out.

#### tomfrank

i did to -45 to 45....will that work?

#### berkeman

Mentor
i did to -45 to 45....will that work?
Based on your initial drawing, yes.

thank you