## Problem Restatement **Given:** - Two wires cross at a 30° angle, but do not touch. - Each wire carries a 5. A current (directions as shown in the figure). - Points 1 and 2 are each 4. cm from the intersection, equally distant from both wires. **Asked:** What are the **magnitude and direction** of the magnetic fields at points 1 and 2? --- ## Step 1: Magnetic Field from a Long Straight Wire The **magnetic field** at distance \( r \) from a long, straight wire carrying current \( I \) is: \[ B = \frac{\mu_ I}{2\pi r} \] where \( \mu_ = 4\pi \times 10^{-7}~\text{T}\cdot\text{m/A} \), \( I = 5.~\text{A} \), \( r = 4.~\text{cm} = .040~\text{m} \). \[ B_{\text{each wire}} = \frac{4\pi \times 10^{-7} \times 5.}{2\pi \times .040} = \frac{2 \times 10^{-6}}{.040} = 5. \times 10^{-5}~\text{T} \] **Explanation:** We calculate the magnetic field produced at a perpendicular distance of 4. cm from each wire, using the Biot–Savart law for an infinite straight wire. --- ## Step 2: Vector Addition of Fields and Direction **Since the two wires cross at a 30° angle:** - At each point, the magnetic field due to each wire is perpendicular to the wire and follows the right-hand rule. - At points 1 and 2, the distance to both wires is the same, so the magnetic field magnitude from each wire is the same: \( 5. \times 10^{-5}~\text{T} \). - The angle between the two wires (and thus between the two magnetic field vectors at the points) is 30°. The **resultant magnetic field** magnitude is: \[ B_{\text{total}} = \sqrt{B_1^2 + B_2^2 + 2B_1B_2\cos\theta} \] where \( \theta = 30^\circ \): \[ B_{\text{total}} = \sqrt{(5. \times 10^{-5})^2 + (5. \times 10^{-5})^2 + 2 (5. \times 10^{-5}) (5. \times 10^{-5}) \cos 30^\circ} \] \[ = 5. \times 10^{-5} \sqrt{2 + 2\cos 30^\circ} \] \[ \cos 30^\circ = \frac{\sqrt{3}}{2} \] \[ 2 + 2\cos 30^\circ = 2 + 2 \times \frac{\sqrt{3}}{2} = 2 + \sqrt{3} \] \[ B_{\text{total}} = 5. \times 10^{-5} \sqrt{2 + \sqrt{3}} \] \[ 2 + \sqrt{3} \approx 2 + 1.732 = 3.732 \] \[ \sqrt{3.732} \approx 1.931 \] \[ B_{\text{total}} \approx 5. \times 10^{-5} \times 1.931 = 9.66 \times 10^{-5}~\text{T} \] **Explanation:** We added the two magnetic field vectors, separated by 30°, using vector addition. The direction can be determined using the right-hand rule and will be at an angle bisecting the 30° between the two field vectors. --- ## **Final Answer** **Magnitude:** \[ \boxed{9.7 \times 10^{-5}~\text{T}} \] **Direction:** Perpendicular to the plane formed by the two wires, at an angle bisecting the 30° between the two magnetic field vectors at points 1 and 2. The precise direction (into or out of the page) depends on the application of the right-hand rule for each current direction as shown in the figure.