Chapter 30 HW Solutions

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

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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 0 . 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 0 . 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 : 0 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 0 di L dt - = E 6.00 V 2.40 A/s 2.50 H di dt L = = = E (b) 0 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 - - = = = 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 - - - = - = - = - = E (d) Final steady state means and 0, so 0. di t iR dt → ∞ - = E 1

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6.00 V 0.750 A 8.00 i R = = = 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 . 30.21. I DENTIFY : / / (1 ), t i R e τ - = - E with / . L R τ = The energy stored in the inductor is 2 1 2 . U Li = S ET U P : The maximum current occurs after a long time and is equal to / . R E E XECUTE : (a) max / i R = E so max /2 i i = when 1 2 (1 ) t/τ e - - = and 1 2 . t/τ e - = ( 29 1 2 / ln . - = 3 ln2 (ln2)(1.25 10 H) 17.3 s 50.0 L R - × = = = (b) 1 max max 2 when 2. U U i i = = / 1 1 2, e - - = so / e 1 1 2 0.2929. - = - = ln(0.2929)/ 30.7 s. t L = - = E VALUATE : 5 / 2.50 10 s 25.0 s. L R
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