What is the B-field at the center of a semicircle using the Biot-Savart Law?

[pinkfishegg]

Homework Statement

Use the Biot-Savart Law to find the magnetic field strength at the center of the semicircle in fig 35.53

Homework Equations

B_current=(μ/4π)*(IΔsXr^{^})/r²
B_wire=μI/2πd

The Attempt at a Solution

The solution from the back of the book is
B=μI/4πd

It looks like they just added the two wires together and the the B-field of the loop in the middle is zero. I don't understand what the cross product would give you zero though. The example that derived the B-field of a wire put a point on the y-axis and used that to find R and determine the cross-product. Why would that lead to a cross product of zero for the arc?

 

[andrewkirk]

You need to integrate over all the infinitesimal wire segments. Where is your integral attempt?

The arc does not produce a zero magnetic field.

 

[pinkfishegg]

I'm confused on how to set up the integral because I don't know where to set P, and therefore how to know what R should be. In the example where they derived the B-Field of the wire, the Set P on the Y axis. I'm not sure how to figure that out.

 

[andrewkirk]

The Biot Savart law says the mag field at a point P is

$$ \frac{\mu_0}{4\pi} \int_C \frac{I d\mathbf l\times\mathbf{\hat{r'}}}{|\mathbf{r'}|^2}$$

where $d\mathbf{l}$ is a vector pointing in the direction of conventional current, along the incremental bit of wire, $\mathbf{r'}$ is the vector from that bit of wire to P, and $C$ is the wire, which should be a circuit but once you start the integration you'll see that it doesn't matter that what they've drawn isn't a circuit.

Do the integrations separately for the straight and the curved bits of wire.

It doesn't matter where in the number plane you put the wire and the point. What matters is their position relative to one another. All vectors in the calculation are relative.