### Magnetic Field Between Parallel Wires**Figure Description:** The diagram illustrates two long, straight, and parallel wires that are 12.0 cm apart. - Wire \( I_1 \) carries a current of 9.0 A directed into the page, represented by an "X".- Wire \( I_2 \) carries a current of 12.0 A directed out of the page, represented by a dot.**Problem Statement:** Find the magnitude and direction of the magnetic field at a point midway between the wires.**Choices:**1. \( 0.5 \times 10^{-5} \, \text{T, upward} \)2. \( 0.5 \times 10^{-5} \, \text{T, downward} \)3. \( 3.5 \times 10^{-5} \, \text{T, upward} \)4. \( 3.5 \times 10^{-5} \, \text{T, downward} \)5. \( 7.0 \times 10^{-5} \, \text{T, upward} \)6. \( 7.0 \times 10^{-5} \, \text{T, downward} \)7. \( 1.0 \times 10^{-5} \, \text{T, upward} \)8. \( 1.0 \times 10^{-5} \, \text{T, downward} \)9. \( 1.2 \times 10^{-5} \, \text{T, upward} \)10. \( 1.2 \times 10^{-5} \, \text{T, downward} \)11. \( 8.4 \times 10^{-5} \, \text{T, upward} \)12. \( 8.4 \times 10^{-5} \, \text{T, downward} \)13. \( 3.0 \times 10^{-5} \, \text{T, upward} \)14. \( 3.0 \times 10^{-5} \, \text{T, downward} \)15. \( 0 \)**Note:** The direction is determined using the right-hand rule for magnetic fields around current-carrying wires.

Answered: Image /qna-images/answer/f13f618f-190a-43ce-9ab0-ff22ad1d0b2e.jpg

Answered: 12.0 cm X 1₁ = 9.0A I₂ = 12.0A Two…

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