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
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 currentcarrying 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 ....
View
Full
Document
This note was uploaded on 03/18/2010 for the course PHYS 241 taught by Professor Milsom during the Fall '08 term at Arizona.
 Fall '08
 Milsom
 Magnetism, Work

Click to edit the document details