Final answer:
The magnitude of the induced emf in the coil at t=1s is 4.43 mV.
Step-by-step explanation:
The magnitude of the induced emf can be found using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of the magnetic flux through the coil. The magnetic flux can be calculated by multiplying the magnetic field inside the solenoid by the area of the coil.
Given that the solenoid has 400 turns per meter and carries a current given by I=0.30sin(120πt)A, we can calculate the magnetic field inside the solenoid at t=1s using the expression B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns per meter, and I is the current.
Substituting the values into the equation, we get B = (4π×10⁻⁷ T·m/A)(400/m)(0.3sin(120π(1))) ≈ 60.377 μT.
The area of the coil can be calculated using the formula A = πr², where r is the radius of the coil.
Substituting the values into the equation, we get A = π(0.06m)² = 0.113 m².
Finally, we can calculate the magnitude of the induced emf using the formula emf = -dΦ/dt, where dΦ/dt is the rate of change of the magnetic flux.
Substituting the values into the equation, we get emf = -(d/dt)(BA) = -(d/dt)(60.377μT)(0.113m²) = 4.43 mV.