Let's solve the problem step by step. ### **Step 1: Express the average induced emf (\( \epsilon_{avg} \))** **Formula:** The average induced emf in an inductor is given by Faraday's Law: \[ \epsilon_{avg} = -L \frac{\Delta I}{\Delta t} \] - \( L \) = inductance - \( \Delta I \) = change in current - \( \Delta t \) = time interval **Explanation:** When the switch is opened, the current drops from \( I \) to in time \( \Delta t \), causing a change in magnetic flux that induces an emf in the inductor. --- ### **Step 2: Calculate the numerical value** **Given:** - \( L = 4.5\, \text{H} \) - \( I = .65\, \text{A} \) - \( \Delta t = .85\, \text{s} \) - \( \Delta I = - .65 = -.65\, \text{A} \) **Plug into the formula:** \[ \epsilon_{avg} = -L \frac{\Delta I}{\Delta t} \] \[ \epsilon_{avg} = -4.5 \times \frac{-.65}{.85} \] \[ \epsilon_{avg} = -4.5 \times (-.7647) \] \[ \epsilon_{avg} = 3.44\, \text{V} \] **Explanation:** The negative sign indicates the direction of emf opposes the change in current (Lenz's Law). The magnitude is \( 3.44\, \text{V} \). --- ### **Final Answer:** **a.** The magnitude of the average induced emf is: \[ |\epsilon_{avg}| = L \frac{|\Delta I|}{\Delta t} \] **b.** The numerical value is: \[ \boxed{3.44\ \text{V}} \] **c.** **Direction:** The induced emf produces a current that opposes the decrease in current (Lenz's Law). Thus, it tries to maintain the original current direction. If the original current was clockwise, the induced current will also be clockwise.