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## 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=[tex]\mu[/tex]

_{0}I([tex]\pi[/tex]r)/4[tex]\pi[/tex]r

^{2}(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 (B

_{1}=(4[tex]\pi[/tex]x10

^{-7})(3.8)/(4[tex]\pi[/tex])(.1070

^{2}([tex]\pi[/tex].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 ([tex]\sqrt{(1.114e-5)^2+(2.935e-5)^2}[/tex]) 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.

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