Question

A long solenoid (n = 1500 turns/m) has a cross-sectional area of ​​0.40 m2 and a current given by I = (4.0 + 3.0t2) A, where t is in seconds. A flat circular coil (N = 300 turns) with a cross sectional area of ​​0.15 m2 is inside and coaxial with the solenoid. What is the magnitude of the induced emf in the coil at t = 2.0 s?

Answer 1

The magnitude of the emf induced in the coil at t = 2.0 s is 0.254 V.

Identify the key concepts:

Faraday's law of electromagnetic induction: A changing magnetic flux through a loop of wire induces an electromotive force (emf) in the loop.

Magnetic flux: The product of the magnetic field strength and the area perpendicular to the field through which it passes.

Solenoid: A coil of wire that produces a magnetic field when a current flows through it.

Gather the given information:

Number of turns per meter of solenoid (n): 1500 turns/m

Cross-sectional area of solenoid (A_solenoid): 0.40 m²

Current in solenoid (I): 3.0t A

Time (t): 2.0 s

Number of turns in the coil (N): 300 turns

Cross-sectional area of coil (A_coil): 0.15 m²

Calculate the magnetic field strength inside the solenoid:

B = μ₀ * n * I

B = (4π × 10⁻⁷ T·m/A) * (1500 turns/m) * (3.0t A)

B = 5.654 × 10⁻³t T

Calculate the magnetic flux through the coil:

Φ = B * A_coil

Φ = (5.654 × 10⁻³t T) * (0.15 m²)

Φ = 8.481 × 10⁻⁴t T·m²

Calculate the rate of change of magnetic flux:

dΦ/dt = 8.481 × 10⁻⁴ T·m²/s

Apply Faraday's law to find the induced emf:

emf = -N * (dΦ/dt)

emf = -(300 turns) * (8.481 × 10⁻⁴ T·m²/s)

emf = -0.254 V

Take the magnitude of the emf:

|emf| = 0.254 V