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

2024
Views:

114

Pages:

5
Author:

Jabin F.

# EMF (Electromotive Force): Work done per unit charge (i) Loop is moving, but magnetic field is constant. (ii) Loop is stationary, but magnetic field is time-varying. (Page-210) #(i) Loop is moving but magnetic field is constant: Area vector will take direction of motion. On pulling out, area is changing, but B is constant. Find EMF in the loop & current in the loop with d/dt? 1. Ɛ = ∮(v × B)⋅dl (motional emf) 2. Ɛ = -dΦm/dt where Φm = magnetic flux = ∫ₛ B⋅dS = ∫ₛ BdScosθ Φm could change by change of (i) B (ii) area, (iii) angle between B and area vector. From (1): Ɛ = ∮(v × B)⋅dl = vB ∫₀^d dl = vBd #(ii) If loop is pushed into the magnetic field: A (+ve), D (-ve) Ɛ = VBl I = VBl/RL (from B to c) Graph of time and current: A square-wave graph showing the current as a function of time. Both push-pull motion of the loop. I = ±VBl/RL #(iii) Loop is rotating with angular speed ω: B₀ is constant, normal to the loop. ωt = θ Area under the magnetic field: A = l²cos(ωt) Φm = BA = Bl²cos(ωt) Ɛ = -dΦm/dt = Bl²ωsin(ωt) Graph of time and current: A sinusoidal graph showing the current as a function of time. Ɛ = Ɛ₀sin(ωt), where Ɛ₀ = Bl²ω I = Ɛ₀sin(ωt)/RL Power dissipated due to radiation: P = I²RL = ε²/RL = B₀²a²l²ω²sin²(ωt) / RL (4) Average power:

= B₀²a²l²ω²

= B₀²a²l²ω² / 2RL (5) Area Under the Magnetic Field: a²l² = πr² At t = 0, loop is in the y-z plane, Φ(t=0) = B₀πr² Φ(t) = B₀πr²cos(ωt) Similarly, by above approaches:

= (B₀²π²r⁴ω²) / (2RL) = (9x10⁻⁵)²(π²)(10⁻⁴)²(10²) / (2x10) = 180 watts

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