Let's break down the problem: ### **Given Data** - **Coil:** - Number of turns, \( N = 160 \) - Radius, \( r = 2.5\,\text{cm} = .025\,\text{m} \) - Resistance, \( R = 6.8\,\Omega \) - **Solenoid:** - Turns per cm, \( n = 180\,\text{turns/cm} = 18,000\,\text{turns/m} \) - Diameter, \( d = 3.8\,\text{cm} \) (not needed, coil fits inside) - **Current change in solenoid:** - Initial current, \( I_ = 1.9\,\text{A} \) - Final current, \( I = \) - Time interval, \( \Delta t = 33\,\text{ms} = .033\,\text{s} \) --- ## **Step 1: Magnetic Field Inside Solenoid** \[ B = \mu_ n I \] where \( \mu_ = 4\pi \times 10^{-7}\,\text{T}\cdot\text{m/A} \). --- ## **Step 2: Change in Magnetic Flux Through Coil** \[ \Phi_B = N \cdot (B \cdot A) \] where \( A = \pi r^2 \) is the area of the coil. --- ## **Step 3: Induced EMF (Faraday's Law)** \[ \mathcal{E} = -N \frac{d\Phi_B}{dt} \] \[ \Phi_B = B \cdot A = (\mu_ n I) (\pi r^2) \] So, \[ \frac{d\Phi_B}{dt} = \mu_ n \pi r^2 \frac{dI}{dt} \] --- ## **Step 4: Induced Current** \[ I_\text{induced} = \frac{\mathcal{E}}{R} \] --- ## **Step 5: Plug in Values** ### **Find \( \frac{dI}{dt} \):** \[ \frac{dI}{dt} = \frac{ - 1.9}{.033} = -57.58\,\text{A/s} \] ### **Calculate Induced EMF:** \[ \mathcal{E} = N \cdot \mu_ n \pi r^2 \left| \frac{dI}{dt} \right| \] \[ N = 160 \] \[ \mu_ = 4\pi \times 10^{-7} \] \[ n = 18,000 \] \[ r = .025 \] \[ \pi r^2 = \pi \times (.025)^2 = .0019635 \] Now plug these in: \[ \mathcal{E} = 160 \times (4\pi \times 10^{-7}) \times 18,000 \times .0019635 \times 57.58 \] Calculate step by step: - \( 4\pi \times 10^{-7} = 1.2566 \times 10^{-6} \) - \( 1.2566 \times 10^{-6} \times 18,000 = .022619 \) - \( .022619 \times .0019635 = 4.44 \times 10^{-5} \) - \( 4.44 \times 10^{-5} \times 57.58 = .002557 \) - \( .002557 \times 160 = .4091 \) V --- ## **Induced Current** \[ I_\text{induced} = \frac{\mathcal{E}}{R} = \frac{.4091}{6.8} = .0602\,\text{A} \] --- ### **Final Answer** \[ \boxed{.0602\,\text{A}} \] --- **The correct induced current is approximately \( .060\,\text{A} \) (not 23.63 A). Please double-check your answer.**