{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw 20 - yang(ey942 – ohw20 – turner –(56705 1 This...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: yang (ey942) – ohw20 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A toroid having a rectangular cross section ( a = 1 . 85 cm by b = 4 . 11 cm) and inner radius 1 . 69 cm consists of N = 450 turns of wire that carries a current I = I sin ω t , with I = 73 . 3 A and a frequency f = 67 . 1 Hz. A loop that consists of N ℓ = 12 turns of wire links the toroid, as in the figure. b a N R N l Determine the maximum E induced in the loop by the changing current I . Correct answer: 0 . 761396 V. Explanation: Basic Concept: Faraday’s Law E =- d Φ B dt . Magnetic field in a toroid B = μ N I 2 π r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = μ N I 2 π r . So, the flux through the loop of wire is Φ B 1 = integraldisplay B dA = μ N I 2 π sin( ω t ) integraldisplay b + R R a dr r = μ N I 2 π a sin( ω t ) ln parenleftbigg b + R R parenrightbigg . Applying Faraday’s law, the induced emf can be calculated as follows E =- N ℓ d Φ B 1 dt =- N ℓ μ N I 2 π ω a ln parenleftbigg b + R R parenrightbigg cos( ω t ) =-E cos( ω t ) where ω = 2 πf was used. The maximum magnitude of the induced emf , E , is the coefficient in front of cos( ω t ). E =- N ℓ d Φ B 1 dt =- N ℓ μ N I ω 2 π a ln bracketleftbigg b + R R bracketrightbigg =- (12 turns) μ (450 turns) × (73 . 3 A) (67 . 1 Hz) (1 . 85 cm) × ln bracketleftbigg (4 . 11 cm) + (1 . 69 cm) (1 . 69 cm) bracketrightbigg =- . 761396 V |E| = 0 . 761396 V . 002 10.0 points The two-loop wire circuit is 71 . 619 cm wide and 47 . 746 cm high. The wire circuit in the figure is located in a magnetic field whose magnitude varies with time according to the expression B = (0 . 001 T / s) t and its direction is into the page. Assume The resistance per length of the wire is 0 . 145 Ω / m. B B P Q 23 . 873 cm 47 . 746 cm 47 . 746 cm yang (ey942) – ohw20 – turner – (56705) 2 When the magnetic field is 0 . 8 T, find the magnitude of the current through middle leg PQ of the circuit. Correct answer: 274 . 402 μ A. Explanation: Let : ℓ = 0 . 47746 m , A l = 1 2 ℓ 2 = 0 . 227968 m 2 / s , A r = ℓ 2 = 0 . 455936 m 2 / s , δ = 0 . 145 Ω / m , and d B dt = d dt α t = α = 0 . 001 T / s . Basic Concept: Faraday’s Law is E =- d Φ B dt , where Φ B = A B . Ohm’s Law is V = I R . Solution: The instantaneous value of the magnetic field ( B = 0 . 8 T) is not germane to this problem. B B P Q ℓ 2 ℓ ℓ I r I l I PQ The resistance for the wire is proportional to the length of the wire. For a length of 47 . 746 cm, the resistance is R = δ ℓ = (0 . 145 Ω / m) (0 . 47746 m) = 0 . 0692317 Ω ....
View Full Document

{[ snackBarMessage ]}

Page1 / 12

hw 20 - yang(ey942 – ohw20 – turner –(56705 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online