{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 30

# Chapter 30 - 30.1 a 2 M di 1 dt 3 25 10 4 H(830 A/s 0.270 V...

This preview shows pages 1–4. Sign up to view the full content.

30.1: a) V, 0.270 /s) A 830 ( H) 10 25 . 3 ( ) / ( 4 1 2 = × = = - dt di M ε and is constant. b) If the second coil has the same changing current, then the induced voltage is the same and V. 270 . 0 1 = ε 30.2: For a toroidal solenoid, . 2 / and , / 1 1 0 B 1 2 2 2 r A i N i N M B π μ = Φ Φ = So, . 2 / 2 1 0 r N AN M π μ = 30.3: a) H. 1.96 A) (6.52 / Wb) 0320 . 0 ( ) 400 ( / 1 2 2 = = Φ = i N M B b) When Wb. 10 7.11 (700) / H) (1.96 A) 54 . 2 ( / A, 54 . 2 3 1 2 B 2 1 - × = = = Φ = N M i i 30.4: a) H. 10 6.82 s) / A 0.242 ( / V 10 65 . 1 ) / ( / 3 3 2 - - × = - × = = dt di M ε b) A, 20 . 1 , 25 1 2 = = i N Wb. 10 3.27 25 / H) 10 (6.82 A) 20 . 1 ( / 4 3 2 1 2 - - × = × = = Φ N M i B c) mV. 2.45 s) / A (0.360 H) 10 82 . 6 ( / and s / A 360 . 0 / 3 2 1 2 = × = = = - dt Mdi dt di ε 30.5: Ωs. 1 A)s / (V 1 AC)s / (J 1 A / J 1 A / Nm 1 A / Tm 1 A / Wb 1 H 1 2 2 2 = = = = = = = 30.6: For a toroidal solenoid, ). / ( / / dt di i N L B ε = Φ = So solving for N we have: turns. 238 s) / A (0.0260 Wb) (0.00285 A) (1.40 V) 10 6 . 12 ( ) / ( / 3 = × = Φ = - dt di i N B ε 30.7: a) V. 10 4.68 s) / A (0.0180 H) 260 . 0 ( ) / ( 3 1 - × = = = dt di L ε 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 30.8: a) H. 130 . 0 m) 120 . 0 ( 2 ) m 10 80 . 4 ( ) 1800 ( ) 500 ( 2 / 2 5 2 0 2 0 m m = × = = - π μ πr A N μ K K L b) Without the material, H. 10 2.60 H) 130 . 0 ( 500 1 1 4 m m - × = = = K L K L

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
30.9: For a long, straight solenoid: . / / and / 2 0 0 l A N μ L l NiA μ i N L B B = = Φ Φ = 30.10: a) Note that points a and b are reversed from that of figure 30.6. Thus, according to Equation 30.8, s. / A 00 . 4 H 0.260 V 04 . 1 - = = = - - L V V dt di a b Thus, the current is decreasing. b) From above we have that . ) s / A 00 . 4 ( dt di - = After integrating both sides of this expression with respect to , t we obtain A. 4.00 s) (2.00 A/s) (4.00 A) 0 . 12 ( s) / A 00 . 4 ( = - = - = i t i 30.11: a) H. 0.250 A/s) (0.0640 / V) 0160 . 0 ( ) / ( / = = = dt di L ε b) Wb. 10 4.50 (400) / H) (0.250 A) 720 . 0 ( / 4 - × = = = Φ N iL B 30.12: a) J. 540 . 0 2 / A) (0.300 H) 0 . 12 ( 2 1 2 2 = = = LI U b) W. 2 . 16 ) 180 ( A) 300 . 0 ( 2 2 = = = R I P c) No. Magnetic energy and thermal energy are independent. As long as the current is constant, constant. = U 30.13: πr Al N μ LI U 4 2 1 2 2 0 2 = = turns. 2850 A) 0 . 12 ( ) m 10 00 . 5 ( J) (0.390 m) 150 . 0 ( 4 4 2 2 4 0 2 0 = × = = - μ π AI μ πrU N 30.14: a) J. 10 1.73 h) / s 3600 h/day (24 W) 200 ( 7 × = × = = Pt U b) H. 5406 A) (80.0 J) 10 73 . 1 ( 2 2 2 1 2 7 2 2 = × = = = I U L LI U 30.15: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability μ is used in place of . 0 μ
30.16: a) free space: J. 3619 ) m 0290 . 0 ( 2 T) (0.560 V 2 V 3 0 2 0 2 = = = = μ μ B u U b) material with J. 04 . 8 ) m 0290 . 0 ( ) 450 ( 2 T) (0.560 V 2 V 450 3 0 2 0 m 2 m = = = = = μ μ K B u U K 30.17: a) 3 2 6 0 2 0 0 2 m 1 . 25 T) (0.600 J) 10 60 . 3 ( 2 2 Volume 2 = × = = = = μ μ μ B U B Vol U u . b) T. 9 . 11 T 4 . 141 m) (0.400 J) 10 60 . 3 ( 2 2 2 3 6 0 0 2 = = × = = B Vol U B μ μ 30.18: a) . mT 35 . 4 ) m 0690 . 0 ( 2 ) A 50 . 2 ( ) 600 ( 2 0 0 = = = π μ π μ r NI B b) From Eq. (30.10), . m / J 53 . 7 2 ) T 10 35 . 4 ( 2 3 0 2 3 0 2 = × = = - μ μ B u c) Volume . m 10 52 . 1 ) m 10 50 . 3 ( ) m 0690 . 0 ( 2 2 V 3 6 2 6 - - × = × = = π π rA d) . J 10 14 . 1 ) m 10 52 . 1 ( ) m / J 53 . 7 ( 5 3 6 3 - - × = × = = uV U e) . H 10 65 . 3 ) m 0690 . 0 ( 2 ) m 10 50 . 3 ( ) 600 ( 2 6 2 6 2 0 2 0 - - × = × = = π μ π μ r A N L J 10 14 . 1 ) A 50 . 2 ( ) H 10 65 . 3 ( 2 1 2 1 5 2 6 2 - - × = × = = LI U same as (d). 30.19: a) . s / A 40 . 2 H 50 . 2 V 00 . 6 0 When . = = = - = dt di i L iR dt di ε b) When . A/s 800 . 0 H 50 . 2 ) 00 . 8 ( ) A 500 . 0 ( V 00 . 6 A 00 . 1 = - = = dt di i c) At A. 413 . 0 ) 1 ( 8.00 V 00 . 6 ) 1 ( s 200 . 0 s) (0.250 H) 50 . 2 / 00 . 8 ( ) / ( = - = - = = - - e e R ε i t t L R d) As A. 750 . 0

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 33

Chapter 30 - 30.1 a 2 M di 1 dt 3 25 10 4 H(830 A/s 0.270 V...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online