Let's solve the problem step by step as requested: --- ## Step 1: **Find the Magnetic Field from Each Wire at Point P** **Given Data:** - Both wires have current \( I = 3.00\, \text{A} \) - Distance between wires: \( d_1 = 8.00\, \text{m} \) - Distance from wires to point \( P \): \( d_2 = 3.00\, \text{m} \) - Point \( P \) is on the perpendicular bisector of the wires ### Magnetic Field Due to a Long Straight Wire The magnetic field at a distance \( r \) from a long straight wire is: \[ B = \frac{\mu_ I}{2\pi r} \] where \(\mu_ = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m}/\text{A}\). Since both wires are the same distance from \( P \), both contribute equally in magnitude. - \( r = d_2 = 3.00\, \text{m} \) - \( I = 3.00\, \text{A} \) \[ B = \frac{4\pi \times 10^{-7} \times 3.00}{2\pi \times 3.00} = \frac{12\pi \times 10^{-7}}{6\pi} = 2 \times 10^{-7}\, \text{T} \] **Explanation:** Both wires produce a magnetic field of \( 2 \times 10^{-7} \, \text{T} \) at \( P \), but the direction of each field must be determined next. --- ## Step 2: **Find the Net Magnetic Field (Magnitude and Direction) at P** ### (a) **Direction Using Right-Hand Rule** - Assume **one wire has current into the page** and **one out of the page** (if both are same, the direction analysis still applies, but they point in opposite sense). - By the right-hand rule, at point \( P \) (to the right of the wires), the magnetic field due to the upper wire will be **downward** and due to the lower wire will be **upward**. Since P is equidistant from both wires and on the perpendicular bisector: - The horizontal components (x-direction) of the magnetic fields **cancel**. - The vertical components (y-direction) **add**. Let’s calculate the **vertical component**: - The angle between each wire-P line and the y-axis: \(\tan\theta = \frac{d_2}{d_1/2} = \frac{3.00}{4.00} = .75\) \(\theta = \tan^{-1}(.75) \approx 36.87^\circ\) The vertical (y) component is \( B_y = B \sin\theta \), and both add: \[ B_{\text{net}} = 2 B \sin\theta \] \[ B_{\text{net}} = 2 \times 2 \times 10^{-7} \times \sin(36.87^\circ) \] \[ \sin(36.87^\circ) \approx .6 \] \[ B_{\text{net}} = 4 \times 10^{-7} \times .6 = 2.4 \times 10^{-7}\, \text{T} \] **Explanation:** The net magnetic field is the sum of the vertical components from both wires. The horizontal components cancel due to symmetry. --- ## **Final Answer** **(a) Magnitude:** \[ \boxed{2.4 \times 10^{-7}\,\text{T}} \] **(b) Direction:** - **Along the positive y-axis** - **Relative to the positive x-axis: \(+90^\circ\)** --- ### **Summary Table** | Part | Value | Explanation | |-----------|----------------------------|--------------------------------------------------| | (a) Magnitude | \(2.4 \times 10^{-7}\, \text{T}\) | Vertical components add, horizontal cancel. | | (b) Direction | \(+90^\circ\) (positive y-axis) | Net field points upward along y-axis. |