Let's answer part (a) step-by-step: --- ## (a) Step 1: **Relate emf to the nonconservative electric field in each resistor** Given: - The total emf around the loop: \(\mathcal{E}_{\text{loop}} = \oint \vec{E}' \cdot d\vec{l} = 48\,\text{V}\), counterclockwise (CCW). - The loop has two resistors in series: \(4\,\Omega\) and \(20\,\Omega\). - The steady current through the loop: \(I = 2\,\text{A}\). **Explanation:** The nonconservative electric field \(\vec{E}'\) is responsible for driving the current due to the changing magnetic flux (Faraday's Law). Since the resistors are in series, the same current flows through both. --- ### Step 2: **Find the magnitude and direction of \(\vec{E}'\) in each resistor** **Total resistance:** \(R_{\text{tot}} = 4\,\Omega + 20\,\Omega = 24\,\Omega\) **Voltage (emf) drop across each resistor (Ohm's Law):** - For \(4\,\Omega\) resistor: \(V_4 = IR = 2\,\text{A} \times 4\,\Omega = 8\,\text{V}\) - For \(20\,\Omega\) resistor: \(V_{20} = 2\,\text{A} \times 20\,\Omega = 40\,\text{V}\) **The emf (nonconservative work per unit charge) across each resistor is equal to the voltage drop:** - Across \(4\,\Omega\): \(8\,\text{V}\) - Across \(20\,\Omega\): \(40\,\text{V}\) **Magnitude of \(\vec{E}'\):** \(\vec{E}' = \frac{\text{emf across resistor}}{\text{length of resistor}}\) But without lengths, just state the emf values. **Direction:** Both fields point in the direction of current (CCW). **Explanation:** The nonconservative electric field \(\vec{E}'\) in each resistor is responsible for the voltage (emf) drop across it, and points in the direction of the current, which is CCW. --- ## **Final Answer (a):** - **In the \(4\,\Omega\) resistor:** \(\vec{E}'\) produces an emf of \(8\,\text{V}\), directed CCW. - **In the \(20\,\Omega\) resistor:** \(\vec{E}'\) produces an emf of \(40\,\text{V}\), directed CCW. **The direction of \(\vec{E}'\) in both resistors is counterclockwise (CCW), matching the direction of the induced emf and current.**

## Experiment 2 Solutions ### (a) Step 1: **Relate emf to the nonconservative electric field in each resistor** Given: - Total emf around the loop: \(\mathcal{E}_{\text{loop}} = 48\,\text{V}\) (CCW) - Resistors in series: \(R_1 = 4\,\Omega\) and \(R_2 = 20\,\Omega\) - Steady current through the loop: \(I = 2\,\text{A}\) **Explanation:** The nonconservative electric field \(\vec{E}'\) drives the current due to the changing magnetic flux (according to Faraday's Law). Since the resistors are in series, the same current flows through both. --- ### Step 2: **Find the magnitude and direction of \(\vec{E}'\) in each resistor** **Total resistance:** \[ R_{\text{tot}} = R_1 + R_2 = 4\,\Omega + 20\,\Omega = 24\,\Omega \] **Voltage (emf) drop across each resistor (using Ohm's Law):** - For \(4\,\Omega\) resistor: \[ V_1 = I \cdot R_1 = 2\,\text{A} \times 4\,\Omega = 8\,\text{V} \] - For \(20\,\Omega\) resistor: \[ V_2 = I \cdot R_2 = 2\,\text{A} \times 20\,\Omega = 40\,\text{V} \] **Magnitude of \(\vec{E}'\):** - Across \(4\,\Omega\): \(E'_1 = 8\,\text{V}\) - Across \(20\,\Omega\): \(E'_2 = 40\,\text{V}\) **Direction:** Both fields point in the direction of current (CCW). **Explanation:** The nonconservative electric field \(\vec{E}'\) in each resistor is responsible for the voltage (emf) drop across it, directed CCW to match the current flow. --- ## **Final Answer (a):** - **In the \(4\,\Omega\) resistor:** \(\vec{E}'\) produces an emf of \(8\,\text{V}\), directed CCW. - **In the \(20\,\Omega\) resistor:** \(\vec{E}'\) produces an emf of \(40\,\text{V}\), directed CCW. ### (b) Potential across each resistor **Using the principle of potential difference:** \[ V_{\text{in}} - V_{\text{out}} = IR \] 1. **For \(4\,\Omega\) resistor:** - Let \(V_a = 0\) (point where charge enters). - Then, \(V_b = V_a - 8\,\text{V} = -8\,\text{V}\). 2. **For \(20\,\Omega\) resistor:** - Entering at point \(b\) where \(V_b = -8\,\text{V}\). - Exiting at point \(c\): \[ V_c = V_b - 40\,\text{V} = -8\,\text{V} - 40\,\text{V} = -48\,\text{V} \] **Explanation:** The potential difference across each resistor corresponds to the voltage drops calculated previously, indicating how the electric potential changes as charge moves through the circuit. --- ## **Final Answer (b):** - **Potential across \(4\,\Omega\) resistor:** \(V_a = 0\,\text{V}\) to \(V_b = -8\,\text{V}\) - **Potential across \(20\,\Omega\) resistor:** \(V_b = -8\,\text{V}\) to \(V_c = -48\,\text{V}\) ### (c) Charge accumulation - **At point \(a\):** Highest potential, marked with \(++\) - **At point \(b\):** Lower potential, marked with \(--\) **Explanation:** Charge accumulation at points \(a\) and \(b\) indicates the potential difference where positive and negative charges build up. --- ## **Final Answer (c):** - **Point \(a\)**: \(++\), potential \(0\,\text{V}\) - **Point \(b\)**: \(--\), potential \(-8\,\text{V}\) ### (d) Conservative electric fields **Within each resistor, the conservative electric field \(\vec{E}_c\) is zero.** - **Direction:** Not applicable, as \(\vec{E}_c = 0\). --- ## **Final Answer (d):** - **Conservative electric fields:** \(\vec{E}_c = 0\) in both resistors. ### (e) Net electric fields **Net electric fields within each resistor:** - Both resistors have a net electric field caused only by the nonconservative field \(\vec{E}'\), directed CCW. **Comparison to Experiment 1:** In Experiment 1, the electric field was conservative from the battery, whereas here it is nonconservative due to the solenoid. --- ## **Final Answer (e):** - **Net electric fields:** Directed CCW in both resistors. ### Experiment 3 If we replace the two-piece wire with a uniform loop, the potential would vary continuously around the loop. The nonconservative electric field would cause a continuous drop in potential that corresponds to the induced current. **Explanation:** In a uniform material loop, the induced emf will still create a changing electric field leading to a consistent potential drop around the entire loop. --- ## **Final Answer (Experiment 3):** - **Potential varies continuously around the loop due to the nonconservative electric field from the induced current.**