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

2024
Views:

474

Pages:

6
Author:

Jabin F.

Faraday's Law E.F produced by changing mag field Here, Ē is non-electrostatic field because ∇×Ē ≠ 0 is non-electrostatic ∇·Ē = 0 If Ē is electrostatic field ∇×Ē = 0 ∇·Ē ≠ 0 Electric field like these Be cause ∇×Ē ≠ 0 ∇·Ē = 0 ques.28 page-62 ques.27 page-62 B(r,t) = B₀e^(-αt)êz α (> 0) uniform and non-constant B(t) E = -dΦm/dt = -∫s (dB/dt) · d⃗s = ∫∫₀^(2π) B₀(-α)e^(-αt)êz · r dr dφ êz ε = ... B₀απr²e^(-αt) (option-a) Ques. A thin Conducting square loop of Side L is placed in the first quadrant of the xy-plane with one of the Vertices at the origin. If a changing magnetic field B(t) = B₀(5zyt² + 2zty + 3y²t²) is applied where B₀ is constant, then the magnitude of the induced Electromotive force in the loop is- (a) 4B₀L (b) 3B₀L (c) 2B₀L (d) B₀L B = B₀(5zyt² + 2zty + 3y²t²) y z=0 plane x (option-d) Ques. 32 Similar Ques.19 Page-61 I(t) = I₀e^(-st) (non-steady current) Page-63 I(t) = I₀e^(-st) ∫∫ B₀(3y²) dy dz 00 -SB₀ ∫∫ (y²) dy dz = -B₀L B₀L ε = ε = -B₀L (option-d) ∮c ⃗e · d⃗τ = -∫ (∂⃗B/∂t) · d⃗x #Inside: Ein = 3H₀N (∂I/∂t) πr² ⇒ Fin = -μ₀N²γ (∂I/∂t) πr² ϕ̂ (i) Note Applied Current is along + direction then induced current would be along - direction Direction of induced Electric field same as direction of induced current I = I₀cosωt Ein = 3μ₀N²I₀ωsinωt ϕ̂ (ii) #induced Electric field E ∝ -(∂I/∂t) #Outside: ∮⃗E · d⃗l = -∫ (∂⃗B/∂t) · d⃗z = -μ₀N (∂I/∂t) πR² Eout = -μ₀NR² (∂I/∂t) ϕ̂ (iii) Eout = μ₀NR² ωsinωt ϕ̂ (iv) at surface R