Chapter 30 HW Solutions

Chapter 30 HW Solutions - PHYS 241 Ch. 30 Homework...

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Unformatted text preview: PHYS 241 Ch. 30 Homework Solutions 30.9. I DENTIFY and S ET U P : Apply / . L di dt = E Apply Lenz’s law to determine the direction of the induced emf in the coil. E XECUTE : (a) 3 ( / ) (0.260 H)(0.0180 A/s) 4.68 10 V L di dt- = = = × E (b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the + terminal at . a E VALUATE : The induced emf is directed so as to oppose the decrease in the current. 30.15. I DENTIFY : A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy. (a) S ET U P : The magnetic field inside a solenoid is . B nI μ = E XECUTE : 7 (4 10 T m/A)(400)(80.0 A) = 0.161 T 0.250 m B π- × ⋅ = (b) S ET U P : The energy density in a magnetic field is 2 . 2 B u μ = E XECUTE : 2 4 3 7 (0.161 T) 1.03 10 J/m 2(4 10 T m/A) u π- = = × × ⋅ (c) S ET U P : The total stored energy is U = uV . E XECUTE : 4 3 4 2 ( ) (1.03 10 J/m )(0.250 m)(0.500 10 m ) 0.129 J U uV u lA- = = = × × = (d) S ET U P : The energy stored in an inductor is 2 1 2 . U LI = E XECUTE : Solving for L and putting in the numbers gives 5 2 2 2 2(0.129 J) 4.02 10 H (80.0 A) U L I- = = = × E VALUATE : An inductor stores its energy in the magnetic field inside of it. 30.19. I DENTIFY : Apply Kirchhoff’s loop rule to the circuit. i ( t ) is given by Eq.(30.14). S ET U P : The circuit is sketched in Figure 30.19. di dt is positive as the current increases from its initial value of zero. Figure 30.19 E XECUTE : R L v v-- = E ( 29 ( / ) 0 so 1 R L t di iR L i e dt R--- = =- E E (a) Initially ( t = 0), i = 0 so di L dt- = E 6.00 V 2.40 A/s 2.50 H di dt L = = = E (b) di iR L dt-- = E (Use this equation rather than Eq.(30.15) since i rather than t is given.) Thus 6.00 V (0.500 A)(8.00 ) 0.800 A/s 2.50 H di iR dt L-- Ω = = = E (c) ( 29 ( 29 ( / ) (8.00 /2.50 H)(0.250 s) 0.800 6.00 V 1 1 0.750 A(1 ) 0.413 A 8.00 R L t i e e e R-- Ω- =- =- =- = Ω E (d) Final steady state means and 0, so 0. di t iR dt → ∞ →- = E 1 6.00 V 0.750 A 8.00 i R = = = Ω E E VALUATE : Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final value of / . R E The slope of the current in the figure, which is di/dt , decreases with t ....
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This note was uploaded on 03/18/2010 for the course PHYS 241 taught by Professor Milsom during the Fall '08 term at Arizona.

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Chapter 30 HW Solutions - PHYS 241 Ch. 30 Homework...

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