Magnetic Field of Tortoidal Coil | Lecture Note - Edubirdie

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

2024
Views:

260

Pages:

5
Author:

Jabin F.

Magnetic field of toroidal Coil:- (endless solenoid) into the Page (-2) Amperian loop Inner radius = a outer radius = b current = I Total no. of turns = N for r < a (inside) Ienc. = 0 B = 0 for a < x < b Ienc. = NI ∮B⋅dl = μ₀NI B(2πr) = μ₀NI B = μ₀NI / 2πr for x > b (outside) Ienc. = 0 B = 0 Toroidal Coil: B = μ₀NI / 2πr for a < x < b B = 0 other wise (rb) Ques. 43 (Due): 58 Steady current electrically neutral (i) If there is no charge E = 0 B = μ₀I/2πr (ii) If charge is present: E ≠ 0 E = μ₀I/4πr Here E ⊥ B ∇⋅(E × B) = 0 (option-d) check for option-c (d = z) vec ∇⋅(E × B) = (∂/∂x)(EyBz - EzBy) + (∂/∂y)(EzBx - ExBz) + (∂/∂z)(ExBy - EyBx) Ques. 43 Bz>0 = μ₀k/2 (ŷ × ẑ) = μ₀k/2 (x̂) if z < 0 Bz>0 = μ₀k/2 (ŷ × (-ẑ)) = μ₀k/2 (-x̂) Ques. 45 same as above Ques. (43) Ques. 45 B = μ₀k/2 or Wb/m² Tesla

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