How Does the Biot-Savart Law Apply to a Wire Shaped in Two Half Circles?

[Kyousik]

Homework Statement

"A wire is formed into the shape of two half circles connected by equal-length straight sections. A current I flows in the circuit clockwise as shown. Determine (a) The magnitude and the direction of the magnetic field at the center, C, and (b) the magnetic dipole moment of the circuit"

http://img225.imageshack.us/img225/4218/untitled1mm6.jpg

Homework Equations

Biot-Savart Law -> dB = (Mu * I * dL) / (4 * Pi * R^2)

The Attempt at a Solution

The answer is

( Mu * I * [R1 + R2] )/ (4 * R1 * R2)

Not sure how they get it, I'm seeing it as two half circles and you minus them from each other.

B = ( Mu * I * [2 * Pi * R1 / 2] ) / (4 * Pi * R1^2)
B1 = (Mu * I) / (2 * R1)

B2 = (Mu * I) / (2 * R2)

Bt = B1 + B2 ?

 

[variation]

Hello,

Do you know the magnetic field in the center of a circle loop with radius a and current i ?


Regards

 

[Kyousik]

B = (Mu * I) / (2 * Pi * a); where R = a; I = Ienclosed.

 

[variation]

What i know is
$$B=\frac{\mu_0I}{2a}$$

The magnetic field in the question is the superposition of contributions of two half-circle loops.


Best regards

 

[WilcoRogers]

This question uses the Biot-Savart Law:

$$B = \frac{\mu_0 I}{4\pi}\int \frac{d\vec{l} \times \hat{r}}{r^2}$$

In this case, use cylindrical co-ordinates to find your field for the two radii. For the straight parts, the current is parallel to the r-hat vector, and as such the B-field is zero on those sections. For the two radii:

$$dl = rd\phi \hat{\phi}; \hat{r} = \hat{s}$$

Use the cross product and integrate over the angle to get your expression. Do this for the two different radii and add together to give your final magnetic field.