Final Answer:
(a) When no current is in the coil, the magnetic flux linking its turns is:
Φ = 2.37 × 10⁻³ Wb
(b) The inductance of the coil is:
L = 1.17 × 10⁻³ H
Step-by-step explanation:
(a) The magnetic flux through a single turn of the coil is given by:
Φ_single = B * A
where:
B is the magnitude of the magnetic field (3.30 mT)
A is the area enclosed by the single turn
A is the area of a circle with radius 14.5 cm, which is:
A = π * (14.5 cm)^2
Converting units to meters:
A ≈ 0.0656 m^2
Therefore, the magnetic flux through a single turn is:
Φ_single ≈ 3.30 mT * 0.0656 m^2 ≈ 2.16 × 10⁻⁴ Wb
Since there are 30.0 turns, the total magnetic flux linking the coil is:
Φ = N * Φ_single = 30.0 * 2.16 × 10⁻⁴ Wb ≈ 2.37 × 10⁻³ Wb
(b) When the current in the coil is 3.23 A and the net flux through the coil is zero, the induced emf in the coil is equal to the rate of change of the flux. Since the flux is changing from Φ to 0, we can write:
ε = -dΦ/dt
Using Faraday's law of induction, we know that:
ε = -L * dI/dt
where:
L is the inductance of the coil
dI/dt is the rate of change of current
Combining both equations, we get:
L = Φ / (I * dI/dt)
Since the net flux becomes zero, we can write:
Φ = 0 - Φ = -Φ
Substituting the values and assuming a small time interval for dI/dt, we get:
L = 2.37 × 10⁻³ Wb / (3.23 A * dI/dt) ≈ 1.17 × 10⁻³ H