Let's solve the problem step by step as requested. --- ## Step 1: Find the magnetic field strength at point P ### **Given:** - Currents: \( I_1 = 5\,A \) (out of the page), \( I_2 = 8\,A \) (into the page) - Distances: \( a = 7\,cm = .07\,m \), \( b = 2\,cm = .02\,m \) - Point P is at (a, b) - Use right-hand rule for current directions. ### **Magnetic Field due to a Long Straight Wire:** \[ B = \frac{\mu_ I}{2\pi r} \] Where: - \( \mu_ = 4\pi \times 10^{-7} \,\text{T}\cdot\text{m/A} \) - \( I \) = current - \( r \) = perpendicular distance from wire #### **Find \( B_1 \) (due to wire 1 at P):** - Distance from wire 1 to P: \( b = .02\,m \) - Direction: Out of the page (right-hand rule: circular field, at P it points left, i.e., negative x-direction) \[ B_1 = \frac{\mu_ I_1}{2\pi b} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times .02} = \frac{2 \times 10^{-6}}{.02} = 1 \times 10^{-4} \,\text{T} \] Direction: Negative x-axis (\(-\hat{i}\)) #### **Find \( B_2 \) (due to wire 2 at P):** - Distance from wire 2 to P: \( a = .07\,m \) - Direction: Into the page (at P, field points downward, negative y-direction) \[ B_2 = \frac{\mu_ I_2}{2\pi a} = \frac{4\pi \times 10^{-7} \times 8}{2\pi \times .07} = \frac{3.2 \times 10^{-6}}{.07} \approx 4.57 \times 10^{-5} \,\text{T} \] Direction: Negative y-axis (\(-\hat{j}\)) **Magnetic Field Vector at P:** \[ \vec{B}_P = -1 \times 10^{-4} \hat{i} - 4.57 \times 10^{-5} \hat{j} \, \text{T} \] #### **Explanation (End of Step 1):** We found the magnetic field at point P due to each wire using the Biot-Savart law for straight wires, taking care of direction (using the right-hand rule). --- ## Step 2: Find the force on a 1 m wire carrying 5 A (out of the page) at P ### **Formula for Force on a Current-Carrying Wire:** \[ \vec{F} = I \, \vec{L} \times \vec{B} \] Where: - \( I = 5\,A \) - \( \vec{L} = 1\,m \) (out of the page, along \(+\hat{k}\)) - \( \vec{B}_P = -1 \times 10^{-4} \hat{i} - 4.57 \times 10^{-5} \hat{j} \) \[ \vec{F} = 5 \left[ \hat{k} \times (-1 \times 10^{-4} \hat{i} - 4.57 \times 10^{-5} \hat{j}) \right] \] Recall: - \( \hat{k} \times \hat{i} = \hat{j} \) - \( \hat{k} \times \hat{j} = -\hat{i} \) So, \[ \vec{F} = 5 \left[ (-1 \times 10^{-4}) \hat{j} + (4.57 \times 10^{-5}) (-\hat{i}) \right] \] \[ = 5 \left[ -1 \times 10^{-4} \hat{j} - 4.57 \times 10^{-5} \hat{i} \right] \] \[ = -2.285 \times 10^{-4} \hat{i} - 5 \times 10^{-4} \hat{j} \, \text{N} \] **Magnitude:** \[ |\vec{F}| = \sqrt{(2.285 \times 10^{-4})^2 + (5 \times 10^{-4})^2} \approx \sqrt{5.22 \times 10^{-8} + 2.5 \times 10^{-7}} \approx \sqrt{3.02 \times 10^{-7}} \approx 5.5 \times 10^{-4} \,\text{N} \] #### **Explanation (End of Step 2):** We calculated the force using the vector cross product, considering the current direction and the field at P. The force is in the negative x and y directions. --- ## **Final Answers:** ### **a) Magnetic field at P:** \[ \boxed{ \vec{B}_P = -1 \times 10^{-4} \hat{i} - 4.57 \times 10^{-5} \hat{j} \, \text{T} } \] ### **b) Force on a 1 m, 5 A wire at P (out of the page):** \[ \boxed{ \vec{F} = -2.285 \times 10^{-4} \hat{i} - 5 \times 10^{-4} \hat{j} \,\text{N} } \] or magnitude: \[ \boxed{ 5.5 \times 10^{-4} \,\text{N} } \]