Let's solve the problem step by step: --- ### **Given:** - **Cross-sectional area,** \( A = 2.03 \text{ cm}^2 = 2.03 \times 10^{-4} \text{ m}^2 \) - **Turns per cm,** \( n = 87.8 \text{ turns/cm} = 8780 \text{ turns/m} \) - **Current in solenoid,** \( i(t) = (0.173 \text{ A/s}^2)t^2 \) - **Secondary winding turns,** \( N_2 = 5.0 \) - **Current at instant,** \( i = 3.2 \text{ A} \) --- ### **Step 1: Find the time \( t \) when the current is 3.2 A** Given \( i(t) = (0.173)t^2 \), \[ 3.2 = 0.173 t^2 \implies t^2 = \frac{3.2}{0.173} \implies t = \sqrt{\frac{3.2}{0.173}} \] \[ t = \sqrt{18.5} \approx 4.30 \text{ s} \] --- ### **Step 2: Find the rate of change of current at this instant** \[ \frac{di}{dt} = 2 \times 0.173 \times t = 0.346 \times t \] At \( t \approx 4.30 \) s: \[ \frac{di}{dt} = 0.346 \times 4.30 \approx 1.49 \text{ A/s} \] --- ### **Step 3: Find the magnetic flux through the secondary winding** The **magnetic field inside the solenoid**: \[ B = \mu_0 n i \] where \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A. **Magnetic flux through one turn of the secondary winding:** \[ \Phi_B = B \cdot A = (\mu_0 n i) \cdot A \] --- ### **Step 4: Find the induced emf in the secondary winding** The emf induced in the secondary winding is: \[ |\mathcal{E}| = N_2 \left| \frac{d\Phi_B}{dt} \right| \] Since \( A, n, \mu_0 \) are constants: \[ \frac{d\Phi_B}{dt} = \mu_0 n A \frac{di}{dt} \] \[ |\mathcal{E}| = N_2 \mu_0 n A \frac{di}{dt} \] --- ### **Step 5: Substitute values** - \( N_2 = 5.0 \) - \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A - \( n = 8780 \) turns/m - \( A = 2.03 \times 10^{-4} \) m² - \( \frac{di}{dt} = 1.49 \) A/s \[ |\mathcal{E}| = 5.0 \times (4\pi \times 10^{-7}) \times 8780 \times 2.03 \times 10^{-4} \times 1.49 \] Calculate step by step: 1. \( 4\pi \times 10^{-7} \approx 1.257 \times 10^{-6} \) 2. \( 5.0 \times 1.257 \times 10^{-6} \approx 6.285 \times 10^{-6} \) 3. \( 6.285 \times 10^{-6} \times 8780 \approx 0.05515 \) 4. \( 0.05515 \times 2.03 \times 10^{-4} \approx 1.1206 \times 10^{-5} \) 5. \( 1.1206 \times 10^{-5} \times 1.49 \approx 1.67 \times 10^{-5} \) --- ### **Final Answer** \[ \boxed{|\mathcal{E}| = 1.67 \times 10^{-5} \ \mathrm{V}} \] Or, in microvolts: \[ \boxed{|\mathcal{E}| = 16.7\ \mu\text{V}} \] --- **Summary:** The magnitude of the induced emf in the secondary winding at the instant the current in the solenoid is 3.2 A is \[ \boxed{1.67 \times 10^{-5} \ \mathrm{V}} \] or \[ \boxed{16.7\,\mu\text{V}} \]

Let's carefully work through this problem step-by-step, including all detailed calculations and explanations. --- ### **Given Data:** - Cross-sectional area of the solenoid: \[ A = 2.03\, \text{cm}^2 = 2.03 \times 10^{-4}\, \text{m}^2 \] - Turns per centimeter: \[ n_{cm} = 87.8\, \text{turns/cm} \] - Convert turns per cm to turns per meter: \[ n = 87.8\, \text{turns/cm} \times 100\, \text{cm/m} = 8780\, \text{turns/m} \] - Current in the solenoid as a function of time: \[ i(t) = 0.173\, \text{A/s}^2 \times t^2 \] - The secondary winding has: \[ N_2 = 5\, \text{turns} \] - Find the emf when current \( i = 3.2\, \text{A} \). --- ### **Step 1: Find the time \( t \) when \( i(t) = 3.2\, \text{A} \).** \[ i(t) = 0.173\, t^2 \] Set \( i(t) = 3.2 \): \[ 3.2 = 0.173\, t^2 \] Solve for \( t^2 \): \[ t^2 = \frac{3.2}{0.173} \approx 18.50 \] Calculate \( t \): \[ t = \sqrt{18.50} \approx 4.30\, \text{s} \] --- ### **Step 2: Find the rate of change of current \( \frac{di}{dt} \) at \( t = 4.30\, \text{s} \).** Differentiating \( i(t) \): \[ \frac{di}{dt} = 2 \times 0.173 \times t = 0.346 \times t \] At \( t = 4.30\, \text{s} \): \[ \frac{di}{dt} = 0.346 \times 4.30 \approx 1.49\, \text{A/s} \] --- ### **Step 3: Find the magnetic flux \( \Phi_B \) through the secondary winding.** The magnetic flux per turn is given by: \[ \Phi_B = B \times A \] where the magnetic field \( B \) inside the solenoid: \[ B = \mu_0 n i \] with: - \( \mu_0 = 4\pi \times 10^{-7}\, \text{T}\cdot \text{m/A} \) Thus: \[ \Phi_B = \mu_0 n i \times A \] --- ### **Step 4: Find the induced emf \( \mathcal{E} \) in the secondary winding.** The emf induced in the secondary coil is: \[ \mathcal{E} = N_2 \times \left| \frac{d\Phi_B}{dt} \right| \] Since \( \Phi_B = \mu_0 n i A \), and \( A \), \( n \), \( \mu_0 \) are constants, their derivatives are zero. The time variation comes from \( i(t) \): \[ \frac{d\Phi_B}{dt} = \mu_0 n A \frac{di}{dt} \] Therefore: \[ \boxed{ \mathcal{E} = N_2 \times \mu_0 n A \times \frac{di}{dt} } \] --- ### **Step 5: Plug in the known values** - \( N_2 = 5 \) - \( \mu_0 = 4\pi \times 10^{-7} \approx 1.257 \times 10^{-6}\, \text{T}\cdot \text{m/A} \) - \( n = 8780\, \text{turns/m} \) - \( A = 2.03 \times 10^{-4}\, \text{m}^2 \) - \( \frac{di}{dt} \approx 1.49\, \text{A/s} \) Calculate step-by-step: \[ \mathcal{E} = 5 \times (1.257 \times 10^{-6}) \times 8780 \times 2.03 \times 10^{-4} \times 1.49 \] --- ### **Step 6: Compute each part** - \( 5 \times 1.257 \times 10^{-6} = 6.285 \times 10^{-6} \) - \( 6.285 \times 10^{-6} \times 8780 \approx 6.285 \times 8780 \times 10^{-6} \) Calculate \( 6.285 \times 8780 \): \[ 6.285 \times 8780 \approx 55,179 \] So, \[ 6.285 \times 10^{-6} \times 8780 \approx 55,179 \times 10^{-6} = 0.055179 \] - Multiply this by \( A = 2.03 \times 10^{-4} \): \[ 0.055179 \times 2.03 \times 10^{-4} \approx (0.055179 \times 2.03) \times 10^{-4} \] Calculate \( 0.055179 \times 2.03 \): \[ 0.055179 \times 2.03 \approx 0.1118 \] Therefore: \[ 0.1118 \times 10^{-4} = 1.118 \times 10^{-5} \] - Now multiply by \( \frac{di}{dt} \approx 1.49 \): \[ 1.118 \times 10^{-5} \times 1.49 \approx 1.666 \times 10^{-5} \] --- ### **Final result:** \[ \boxed{ \mathcal{E} \approx 1.67 \times 10^{-5}\, \text{V} } \] or approximately: \[ \boxed{ \mathcal{E} \approx 16.7\, \mu\text{V} } \] --- ### **Summary:** **The magnitude of the emf induced in the secondary winding when the current in the solenoid is 3.2 A is approximately:** \[ \boxed{ \boxed{1.67 \times 10^{-5}\ \text{V}} } \] **or about 16.7 microvolts.**

The induced emf in the secondary coil arises from the time-varying magnetic flux linked with it, which depends on the solenoid's magnetic field \( B = \mu_0 n i \). As current \( i(t) \) changes over time, the flux \(\Phi_B = B A\) varies, producing an emf given by Faraday’s law: \(\mathcal{E} = N_2 \frac{d\Phi_B}{dt}\). Since \(\Phi_B = \mu_0 n i A\), its rate of change involves \(\frac{di}{dt}\). Therefore, the emf is proportional to the number of turns, the rate of change of current, and the magnetic flux per turn.