## Step 1: Find the Magnetic Field from Each Wire at Point P We use the formula for the magnetic field from a long straight wire at distance \( r \): \B = \frac{\mu_ I}{2\pi r} \] Where: - \( \mu_ = 4\pi \times 10^{-7} \, \mathrm{T \cdot m/A} \) - \( I = 40.\,\mathrm{A} \) - Distance from each wire to point P is \( .60\,\mathrm{m} \) **Calculation for one wire:** \[ B = \frac{4\pi \times 10^{-7} \times 40.}{2\pi \times .60} = \frac{160\pi \times 10^{-7}}{2\pi \times .60} = \frac{160 \times 10^{-7}}{1.2} = \frac{160}{1.2} \times 10^{-7} = 133.33 \times 10^{-7} = 1.33 \times 10^{-5} \, \mathrm{T} \] **Explanation:** Both wires contribute equally in magnitude since they are the same distance away from P and carry equal current. --- ## Step 2: Find the Total Magnetic Field Vector The wires are located symmetrically about the y-axis, and point P is directly above the midpoint between them. By the right-hand rule, the fields due to both wires at P will have equal magnitudes, and their horizontal components will cancel, leaving only the upward (y-direction) components. **The vertical (y) component of each:** \[ B_y = B \cos() = B \] Since both are pointing in the same direction (out of the page or into the page depending on current direction, but for symmetry, let's consider both add in the y-direction): \[ B_{\text{total}} = 2B \cos(\theta) \] But here, since P is directly above the midpoint, \(\theta = \) and \(\cos() = 1\), so: \[ B_{\text{total}} = 2 \times 1.33 \times 10^{-5} = 2.66 \times 10^{-5}\, \mathrm{T} \] **Explanation:** The symmetry means the horizontal (x) components cancel, and the vertical (y) components add. --- ## **Final Answer** - **Magnitude:** \( 2.66 \times 10^{-5}\, \mathrm{T} \) - **Direction:** Upward along the y-axis (or \(90.^\circ\) counterclockwise from the positive x-axis) --- **Summary Table:** | Step | Result | Explanation | |------|--------|-------------| | 1 | \( B_\text{each} = 1.33 \times 10^{-5}\, \mathrm{T} \) | Field from one wire at P | | 2 | \( B_\text{total} = 2.66 \times 10^{-5}\, \mathrm{T} \), \(90.^\circ\) | Sum of both fields, direction upward |

# Calculation of Magnetic Field at Point P ## Step 1: Find the Magnetic Field from Each Wire at Point P We use the formula for the magnetic field from a long straight wire at a distance \( r \): \[ B = \frac{\mu_0 I}{2\pi r} \] Where: - \( \mu_0 = 4\pi \times 10^{-7} \, \mathrm{T \cdot m/A} \) - \( I = 40 \, \mathrm{A} \) - Distance from each wire to point P is \( r = 0.60 \, \mathrm{m} \) **Calculation for one wire:** \[ B = \frac{4\pi \times 10^{-7} \times 40}{2\pi \times 0.60} \] Simplifying: \[ B = \frac{160\pi \times 10^{-7}}{1.2\pi} = \frac{160 \times 10^{-7}}{1.2} = \frac{160}{1.2} \times 10^{-7} = 133.33 \times 10^{-7} = 1.33 \times 10^{-5} \, \mathrm{T} \] **Explanation:** Both wires contribute equally in magnitude since they are the same distance away from point P and carry equal current. --- ## Step 2: Find the Total Magnetic Field Vector The wires are symmetrically located about the y-axis, and point P is directly above the midpoint between them. By the right-hand rule, the fields from both wires at P will have equal magnitudes, and their horizontal components will cancel, leaving only the vertical (y-direction) components. **The vertical (y) component of each:** Since both contribute in the same direction, we can simply add their contributions: \[ B_{\text{total}} = 2B \] Thus, \[ B_{\text{total}} = 2 \times 1.33 \times 10^{-5} = 2.66 \times 10^{-5} \, \mathrm{T} \] **Explanation:** The symmetry of the setup means the horizontal (x) components cancel out, and the vertical (y) components sum directly. --- ## **Final Answer** - **Magnitude:** \( 2.66 \times 10^{-5} \, \mathrm{T} \) - **Direction:** Upward along the y-axis (or \( 90^\circ \) counterclockwise from the positive x-axis) --- **Summary Table:** | Step | Result | Explanation | |------|--------|-------------| | 1 | \( B_{\text{each}} = 1.33 \times 10^{-5} \, \mathrm{T} \) | Field from one wire at P | | 2 | \( B_{\text{total}} = 2.66 \times 10^{-5} \, \mathrm{T} \), \(90^\circ\) | Sum of both fields, direction upward |