Let's solve each part step by step. --- ### 1. **What is the magnitude of the electric field at location D?** **Given:** - EMF of each battery, \( \mathcal{E} = 1.5 \) V - Length of each wire, \( L = 22 \) cm \( = 0.22 \) m - Diameter of wire, \( d = 0.0007 \) m → radius \( r = 0.00035 \) m - Electron density, \( n = 7 \times 10^{28} \) electrons/m³ - Electron mobility, \( \mu_e = 6 \times 10^{-5} \) (m/s)/(V/m) - Two batteries in series, so total voltage \( V = 1.5 + 1.5 = 3.0 \) V - The circuit has two wires (top and bottom, each 22 cm). **Assumption:** The total resistance is dominated by the wires, not the battery internal resistance. #### **Step 1: Calculate the total length of the wire in the loop** \[ L_\text{total} = 2 \times 0.22\,\text{m} = 0.44\,\text{m} \] #### **Step 2: Find the electric field in the wire** The electric field in the wire (assuming uniform field): \[ E = \frac{V}{L_\text{total}} \] \[ E_D = \frac{3.0\,\text{V}}{0.44\,\text{m}} = 6.82\,\text{V/m} \] **Final Answer:** \[ \boxed{E_D = 6.8\ \text{V/m}} \] --- ### 2. **How many electrons per second enter the positive end of battery #2?** #### **Step 1: Calculate the cross-sectional area of the wire** \[ A = \pi r^2 = \pi (0.00035)^2 = 3.85 \times 10^{-7}\ \text{m}^2 \] #### **Step 2: Find the drift velocity** \[ v_d = \mu_e E = (6 \times 10^{-5}) \times 6.82 = 4.09 \times 10^{-4}\ \text{m/s} \] #### **Step 3: Find the current** \[ I = n e A v_d \] where \( e = 1.6 \times 10^{-19} \) C \[ I = (7 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (3.85 \times 10^{-7}) \times (4.09 \times 10^{-4}) \] Calculate step by step: 1. \( 7 \times 10^{28} \times 1.6 \times 10^{-19} = 1.12 \times 10^{10} \) 2. \( 1.12 \times 10^{10} \times 3.85 \times 10^{-7} = 4.312 \times 10^{3} \) 3. \( 4.312 \times 10^{3} \times 4.09 \times 10^{-4} = 1.76\ \text{A} \) #### **Step 4: Number of electrons per second** \[ \text{Electrons/sec} = \frac{I}{e} = \frac{1.76}{1.6 \times 10^{-19}} = 1.1 \times 10^{19}\ \text{electrons/s} \] **Final Answer:** \[ \boxed{1.1 \times 10^{19}\ \text{electrons/s}} \] --- ### 3. **If the cross-sectional area of both wires were increased by a factor of 4, what would be the magnitude of the electric field at location D?** - Current must stay the same for the same voltage. - Resistance: \( R = \frac{\rho L}{A} \) - For a given voltage, \( I = \frac{V}{R} \) - If \( A \rightarrow 4A \), then \( R \rightarrow \frac{R}{4} \), so for the same current, \( E \) (which is \( V/L \)) must decrease by 4. \[ E_\text{new} = \frac{E_D}{4} = \frac{6.8}{4} = 1.7\ \text{V/m} \] **Final Answer:** \[ \boxed{E_D = 1.7\ \text{V/m}} \] --- ## **Summary of Answers** 1. **Magnitude of electric field at D:** \(\boxed{6.8\ \text{V/m}}\) 2. **Electrons per second entering positive end of battery #2:** \(\boxed{1.1 \times 10^{19}\ \text{electrons/s}}\) 3. **New electric field at D (with area ×4):** \(\boxed{1.7\ \text{V/m}}\)