A long solenoid (𝑛=800 turns/m) has a cross-sectional area of 0.51m2 and a current given by 𝐼=5.6+6.3𝑡2A, where 𝑡 is in seconds. A circular coil (𝑁=900 turns) with a cross-sectional area of 0.36m2 is inside and coaxial with the solenoid. What is the magnitude of the emf induced in the coil at 𝑡=1.4s?

To solve this problem, we need to use Faraday's law of electromagnetic induction, which states that the induced emf in a circuit is equal to the negative rate of change of magnetic flux through the circuit.

The magnetic field inside a solenoid is given by B = μ0 * n * I, where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.

First, we need to find the magnetic field inside the solenoid at time t = 1.4s. The current at this time is I = 5.6 + 6.3 * (1.4)^2 = 14.68 A.

So, the magnetic field B = μ0 * n * I = 4π * 10^-7 T*m/A * 800 turns/m * 14.68 A = 0.0147 T.

The magnetic flux through the coil is given by Φ = B * A, where A is the area of the coil. So, Φ = 0.0147 T * 0.36 m^2 = 0.0053 Wb.

The rate of change of the magnetic flux is given by dΦ/dt. Since the current and therefore the magnetic field is changing with time, we need to differentiate the magnetic flux with respect to time.

The current as a function of time is given by I = 5.6 + 6.3t^2. So, the magnetic field as a function of time is B = μ0 * n * I = 4π * 10^-7 T*m/A * 800 turns/m * (5.6 + 6.3t^2).

Differentiating this with respect to time gives dB/dt = 4π * 10^-7 T*m/A * 800 turns/m * 2 * 6.3t = 0.0063t T/s.

The rate of change of the magnetic flux is then dΦ/dt = dB/dt * A = 0.0063t T/s * 0.36 m^2 = 0.0023t Wb/s.

At t = 1.4s, this is dΦ/dt = 0.0023 * 1.4 = 0.0032 Wb/s.

Finally, the magnitude of the induced emf in the coil is given by Faraday's law as |ε| = N * |dΦ/dt|, where N is the number of turns in the coil.

So, |ε| = 900 turns * 0.0032 Wb/s = 2.88 V.