Biot-Savare Law- split loop

[GDGirl]

Homework Statement

A circular conducting ring or radius R = 10.7 cm is connected to two exterior straight wires ending at two ends of a diameter (see Figure). The current splits into uneven portions, with I1 = 3.8 A passing through the top semicircle, and I2 = 10 A passing through the lower semicircle. What is B at the center of the ring?
HELP: Apply the Biot-Savart Law to each semicircle. Adding the two resulting B fields, being careful to keep track of their signs.
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/Knox/phys130a/spring/homework/14/02/P28_31.jpg [Broken]

Homework Equations

B=$$\mu$$₀I($$\pi$$r)/4$$\pi$$r² (for each half circle)

The Attempt at a Solution

I used the equation I put above, and found that the B field for the first current was 1.114e-5 (B₁=(4$$\pi$$x10^-7)(3.8)/(4$$\pi$$)(.1070² ($$\pi$$.107)) and the second current gave me 2.935e-5. From there I tried simply adding them, to give me a total of 4.049e-5. This is incorrect. So I figured they might be vectors, and tried using pythagorean on them ($$\sqrt{(1.114e-5)^2+(2.935e-5)^2}$$) and got 3.139e-5. this is also incorrect.
I figure I'm doing something wrong, but I'm not sure what. The hint says to be sure to pay attention to the signs of the B fields, but I don't see where I would get something other than a positive sign.

 

[rl.bhat]

What is the method of finding the direction of the magnetic field due to current carrying conductor?
Since the directions of the current in two semicircles is not the same, the signs of the fields must be different.

 

[GDGirl]

What is the method of finding the direction of the magnetic field due to current carrying conductor?
Since the directions of the current in two semicircles is not the same, the signs of the fields must be different.

But aren't the directions the same? Both currents are going to the right.

 

[rl.bhat]

In semicircle direction should be either clockwise or counterclockwise.

 

[GDGirl]

Oh. Okay, I got it now, thanks. :)