Final answer:
To find the magnitude of the induced electric field at a distance of 20 cm from the axis of the solenoid, we can use Faraday's law of electromagnetic induction. The magnitude of the induced electric field is approximately 0.96 V/m.
Step-by-step explanation:
To find the magnitude of the induced electric field at a distance of 20 cm from the axis of the solenoid, we can use Faraday's law of electromagnetic induction. The induced electric field is given by the equation E = -dΦ/dt, where Φ is the magnetic flux through a loop of wire.
Since the solenoid has a varying current, the magnetic field inside the solenoid is changing with time. To calculate the induced electric field, we first need to find the magnetic flux through the loop of wire. The magnetic flux Φ is given by Φ = BA, where B is the magnetic field inside the solenoid and A is the area of the loop of wire. In this case, the radius of the solenoid is 10 cm, so the area of the loop of wire is A = πr² = π(0.2m)² = 0.04π m².
The magnetic field inside a solenoid is given by B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Given that the solenoid has n = 1000 turns/m and the current is i(t) = 3t² - 2t + 5, we can substitute these values into the equation for B. At t = 1 second, the current is i(1) = 3(1)² - 2(1) + 5 = 6 A.
Substituting the values of n, I, and A into the equation for Φ, we have Φ = (μ₀nI)A = (4π × 10⁻⁷ T·m/A)(1000 turns/m)(6 A)(0.04π m²) = 0.96 × 10⁻³ T·m².
Finally, we can calculate the magnitude of the induced electric field using E = -dΦ/dt. Differentiating Φ with respect to time, we get dΦ/dt = -d/dt(0.96 × 10⁻³ T·m²) = 0.96 × 10⁻³ V/s. Therefore, the magnitude of the induced electric field at a distance of 20 cm from the axis of the solenoid at t = 1 second is 0.96 × 10⁻³ V/m, which is approximately equal to 0.96 × 10⁻³ × 1000 = 0.96 V/m.