91Ó°ÊÓ

The magnetic field inside a long solenoid can be calculated using the formula \( B = \mu_0 n i(t) \), where \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \) is the permeability of free space, \( n \) is the number of turns per meter, and \( i(t) \) is the current at time \( t \). First, we convert the number of turns per centimeter to turns per meter: \( n = 90.0 \times 100 = 9000 \text{ turns/m} \). Substituting these into the formula, the magnetic field \( B(t) = \mu_0 \cdot 9000 \cdot (0.160t^2) \).

Next, solve for \( t \) when \( i(t) = 3.20 \text{ A} \). The current as a function of time is given as \( i(t) = 0.160t^2 \), which can be rearranged to solve for \( t \):\[ 3.20 = 0.160t^2 \]Solve for \( t \):\[ t^2 = \frac{3.20}{0.160} = 20 \]\[ t = \sqrt{20} = 4.47 \text{ s} \]

The emf \( \epsilon \) induced in a loop is given by Faraday's law of electromagnetic induction: \( \epsilon = -N \frac{d\Phi}{dt} \), where \( N = 5 \) is the number of turns in the secondary winding, and \( \Phi = BA \) is the magnetic flux through one loop. The rate of change of magnetic flux is necessary:\[ \frac{d\Phi}{dt} = A \frac{dB}{dt} \]Plug in the area in meters squared \( A = 2.00 \times 10^{-4} \, \text{m}^2 \) and the rate of change of \( B \). Since \( B = \mu_0 n i(t) \), we find \( \frac{dB}{dt} = \mu_0 n \frac{di}{dt} \).The derivative \( \frac{di}{dt} = 0.160 (2t) \). At \( t = 4.47s \), \( \frac{di}{dt} = 0.160 \times 2 \times 4.47 \).\[ \frac{di}{dt} = 1.43 \text{ A/s} \]Calculate \( \frac{dB}{dt} = \mu_0 \cdot 9000 \cdot 1.43 \).Finally, the emf is:\[ \epsilon = -5 \cdot (2.00 \times 10^{-4}) \cdot \mu_0 \cdot 9000 \cdot 1.43 \]

Substitute in known values:\[ \epsilon = -5 \cdot (2.00 \times 10^{-4}) \cdot (4\pi \times 10^{-7}) \cdot 9000 \cdot 1.43 \]Calculate,\[ \epsilon = -5 \cdot 2.85 \times 10^{-6} \]\[ \epsilon = -1.425 \times 10^{-5} \text{ V} \]The magnitude of \( \epsilon \) is \( 1.425 \times 10^{-5} \text{ V} \) or 14.25 \( \mu\text{V} \).