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PHY_solution-chapter30

PHY_solution-chapter30 - 30.1 a V 0.270/s A 830 H 10 25 3 4...

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Unformatted text preview: 30.1: a) V, 0.270 /s) A 830 ( H) 10 25 . 3 ( ) / ( 4 1 2 = × = =- dt di M ε and is constant. If the second coil has the same changing current, then b) the induced voltage is the same and V. 270 . 1 = ε 30.2: For a toroidal solenoid, . 2 / and , / 1 1 B 1 2 2 2 r A i N i N M B π μ = Φ Φ = So, . 2 / 2 1 r N AN M π μ = 30.3: a) H. 1.96 A) (6.52 / Wb) 0320 . ( ) 400 ( / 1 2 2 = = Φ = i N M B When b) 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 . / 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: 30.7: a) V. 10 4.68 s) / A (0.0180 H) 260 . ( ) / ( 3 1- × = = = dt di L ε Terminal b) 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 . m) 120 . ( 2 ) m 10 80 . 4 ( ) 1800 ( ) 500 ( 2 / 2 5 2 2 m m = × = =- π μ πr A N μ K K L Without the material, b) H. 10 2.60 H) 130 . ( 500 1 1 4 m m- × = = = K L K L 30.9: For a long, straight solenoid: . / / and / 2 l A N μ L l NiA μ i N L B B = ⇒ = Φ Φ = 65f00646f7964c93e91cb481e6d4e6c52774adf4.doc Page 1 of 28 turns. 238 s) / A (0.0260 Wb) (0.00285 A) (1.40 V) 10 6 . 12 ( ) / ( / 3 = × = Φ =- dt di i N B ε 30.10: Note that points a) 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. From above we b) 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) . 12 ( s) / A 00 . 4 ( =- = ⇒ ∆- = ∆ i t i 30.11: a) H. 0.250 A/s) (0.0640 / V) 0160 . ( ) / ( / = = = dt di L ε b) Wb. 10 4.50 (400) / H) (0.250 A) 720 . ( / 4- × = = = Φ N iL B 30.12: a) J. 540 . 2 / A) (0.300 H) . 12 ( 2 1 2 2 = = = LI U b) W. 2 . 16 ) 180 ( A) 300 . ( 2 2 = Ω = = R I P No. Magnetic energy and thermal energy are independent. c) As long as the current is constant, constant. = U 30.13: πr Al N μ LI U 4 2 1 2 2 2 = = turns. 2850 A) . 12 ( ) m 10 00 . 5 ( J) (0.390 m) 150 . ( 4 4 2 2 4 2 = × = = ⇒- 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 ....
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PHY_solution-chapter30 - 30.1 a V 0.270/s A 830 H 10 25 3 4...

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