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Unformatted text preview: PHYS 241 Ch. 29 Homework Solutions 29.1. I DENTIFY : Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. S ET U P : The flux through a coil of N turns is Φ = NBA cos φ , and by Faraday’s law the magnitude of the induced emf is E = d Φ /dt . E XECUTE : (a) ∆Φ = NBA = (50)(1.20 T)(0.250 m)(0.300 m) = 4.50 Wb (b) E = d Φ /dt = (4.50 Wb)/(0.222 s) = 20.3 V E VALUATE : This induced potential is certainly large enough to be easily detectable. 29.3.I DENTIFY and S ET U P : Use Faraday’s law to calculate the average induced emf and apply Ohm’s law to the coil to calculate the average induced current and charge that flows. (a) E XECUTE : The magnitude of the average emf induced in the coil is av . B I N t ∆Φ = ∆ E Initially, i cos . B BA BA φ Φ = = The final flux is zero, so f i av . B B NBA N t t Φ- Φ = = ∆ ∆ E The average induced current is av . NBA I R R t = = ∆ E The total charge that flows through the coil is . NBA NBA Q I t t R t R = ∆ = ∆ = ∆ E VALUATE : The charge that flows is proportional to the magnetic field but does not depend on the time . t ∆ (b) The magnetic stripe consists of a pattern of magnetic fields. The pattern of charges that flow in the reader coil tell the card reader the magnetic field pattern and hence the digital information coded onto the card. (c) According to the result in part (a) the charge that flows depends only on the change in the magnetic flux and it does not depend on the rate at which this flux changes. 29.7.I DENTIFY : Calculate the flux through the loop and apply Faraday’s law. S ET U P : To find the total flux integrate B d Φ over the width of the loop. The magnetic field of a long straight wire, at distance r from the wire, is 2 I B r μ π = . The direction of B r is given by the right-hand rule. E XECUTE : (a) When 2 i B r μ π = , into the page. (b) . 2 B i d BdA Ldr r μ π Φ = = (c) ln( / ). 2 2 b b B B a a iL dr iL d b a r μ μ π π Φ = Φ = = ∫ ∫ (d) ln( ) . 2 B d L di b a dt dt μ π Φ = = E (e) 7 (0.240 m) ln(0.360/0.120)(9.60 A/s) 5.06 10 V. 2 μ π- = = × E E VALUATE : The induced emf is proportional to the rate at which the current in the long straight wire is changing 29.15. I DENTIFY and S ET U P : The field of the induced current is directed to oppose the change in flux. E XECUTE : (a) The field is into the page and is increasing so the flux is increasing. The field of the induced current is out of the page. To produce field out of the page the induced current is counterclockwise. (b) The field is into the page and is decreasing so the flux is decreasing. The field of the induced current is into the page. To produce field into the page the induced current is clockwise....
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