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851_PartUniversity Physics Solution

# 851_PartUniversity Physics Solution - ELECTROMAGNETIC...

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29-1 E LECTROMAGNETIC I NDUCTION 29.1. IDENTIFY: 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. SET UP: 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 . EXECUTE: (a) ΔΦ = 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 EVALUATE: This induced potential is certainly large enough to be easily detectable. 29.2. IDENTIFY: B t ΔΦ = Δ E . cos B BA Φ= . B Φ is the flux through each turn of the coil. SET UP: i 0 = ° . f 90 = ° . EXECUTE: (a) 54 2 8 ,i cos0 (6.0 10 T)(12 10 m )(1) 7.2 10 Wb. B BA −− = × × ° The total flux through the coil is 85 ,i (200)(7.2 10 Wb) 1.44 10 Wb B N × . ,f cos90 0 B BA Φ == ° . (b) 5 4 if 3.6 10 V 0.36 mV 0.040 s NN t Φ− Φ × × = Δ = E . EVALUATE: The average induced emf depends on how rapidly the flux changes. 29.3. IDENTIFY and SET UP: 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) EXECUTE: The magnitude of the average emf induced in the coil is av . B I N t ΔΦ = Δ E Initially, i cos . B B AB A = The final flux is zero, so fi av . BB NBA N tt Φ− Φ Δ Δ E The average induced current is av . NBA I R Rt Δ E The total charge that flows through the coil is . NBA NBA QI t t R tR ⎛⎞ =Δ= Δ= ⎜⎟ Δ ⎝⎠ EVALUATE: The charge that flows is proportional to the magnetic field but does not depend on the time . t Δ (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. 29.4. IDENTIFY and SET UP: Apply the result derived in Exercise 29.3: /. QN B AR = In the present exercise the flux changes from its maximum value of B BA to zero, so this equation applies. R is the total resistance so here 60.0 45.0 105.0 . R = Ω EXECUTE: 5 42 (3.56 10 C)(105.0 ) says 0.0973 T. 120(3.20 10 m ) NBA QR QB RN A ×Ω = = × EVALUATE: A field of this magnitude is easily produced. 29.5. IDENTIFY: Apply Faraday&s law. SET UP: Let + z be the positive direction for A ! . Therefore, the initial flux is positive and the final flux is zero. EXECUTE: (a) and (b) 2 3 0 (1.5 T) (0.120 m) 34 V. 2.0 10 s B t π ΔΦ =− =+ Δ× E Since E is positive and A ! is toward us, the induced current is counterclockwise. EVALUATE: The shorter the removal time, the larger the average induced emf. 29

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29-2 Chapter 29 29.6. IDENTIFY: Apply Eq.(29.4). / . IR = E SET UP: // . B d dt AdB dt Φ= EXECUTE: (a) () 54 4 ( ) (0.012 T/s) (3.00 10 T/s ) B Nd d d N ABN A t t dt dt dt Φ == = + × E ( ) 44 3 43 3 (0.012 T/s) (1.2 10 T/s ) 0.0302 V (3.02 10 V/s ) .
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851_PartUniversity Physics Solution - ELECTROMAGNETIC...

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