University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

402

Pages:

5
Author:

Jabin F.

#Induced Current: ε = -μ₀a * ln(s₁/s₂) * (dI/dt) I_induced = ε / R = -(μ₀a / 2πR) * ln(s₁/s₂) * (dI/dt) #Induced Current always opposes the change of flux. Here we have a fitting wire: → I↑, Φ(t)↓ ∴ B_y↓ Φ_loop ∝ B_wire B_applied = 0 B_wire ↓ Φ_loop ↓ Induced tries to increase Φ_loop Induced tries to increase magnetic field B_induced is generated by I_induced B_applied = 0, so B_induced = 0, direction of current is anticlockwise. Q.4 Bi = 10 tesla Bf = 9 tesla R = 1m, r = 1cm Φ = ΔΦm / R = (Φi - Φf) / R = (Biπr² - Bfπr²) / R Φ = πr²(Bi - Bf) / R = 3.14 × (1×10⁻²)² × (10⁻⁹) / 1 = 3.14 × 10⁻¹⁴ Coulomb. Due: 12/5 Page-62 Q = (Φi - Φf) / R = (Bπr² - 0) / R = πr²B / R Q = NA / B = (2 × 10⁵ × 20) / (40 × 0.01) = 10⁶ T Q.4 Φ = (Φi - Φf) / R Φi = Bπr² |_(t=0, to=1 sec) = B₀e^(-t/to)πr² |_(t=0, to=1 sec) = B₀e^(0)πr² = B₀πr² = 0.01 × 3.14 × (2 × 10⁻²)² = 3.14 × 4 × 10⁻⁶ Φf = Bπr² |_(t=∞, to=1) = B₀e^(-t/to)πr² |_(t=∞, to=1) = 0 Φ = (3.14 × 4 × 10⁻⁶ - 0) / (1) = 3.14 × 4 × 10⁻⁶ = 6.28 × 10⁻⁶ C = 6.28 µC Ques.3 Page 65 B = μ₀I / 2πx If x is increasing, magnetic field will decrease. Hence flux will also decrease. Time-dependent: B = μ₀I / 2πx (a) Φm = μ₀IA / 2π * ln(s + a) / s (b) ε = -dΦm/dt = -μ₀IA / 2π * (1/(s + a) - 1/s) * ds/dt = μ₀IA²v / 2πs(s + a) Induced tries to increase B or Φm B_induced = counter clockwise c) Wire = μ₀I / 2πx Φm = constant Here, there is no variable. Φm = μ₀IA / 2π * ln(s + a) ε = dΦm/dt = 0 Option(b) for direction of current flow: Two methods: V × B → direction of current flow (right-hand rule) ∮ E · dl = -dΦ/dt (Faraday's law) Diagram: Rectangular loop with current I flowing clockwise Magnetic field B pointing out of the page V × B → direction of induced current (anticlockwise) ∮ E · dl = -dΦ/dt (Faraday's law) Induced current tries to increase Φ B_induced is parallel to B_applied B_induced = 0, direction of current = anticlockwise