PHY_solution-chapter30

PHY_solution-chapter30 - 30.1: a) V, 0.270 /s) A 830 ( H)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 30.1: a) V, 0.270 /s) A 830 ( H) 10 25 . 3 ( ) / ( 4 1 2 = = =- dt di M and is constant. If the second coil has the same changing current, then b) the induced voltage is the same and V. 270 . 1 = 30.2: For a toroidal solenoid, . 2 / and , / 1 1 B 1 2 2 2 r A i N i N M B = = So, . 2 / 2 1 r N AN M = 30.3: a) H. 1.96 A) (6.52 / Wb) 0320 . ( ) 400 ( / 1 2 2 = = = 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- = = = = 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 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-- = = = 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 30.5: s. 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 = = So solving for N we have: 30.7: a) V. 10 4.68 s) / A (0.0180 H) 260 . ( ) / ( 3 1- = = = dt di L 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 = = =- r A N 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 L l NiA i N L B B = = = 65f00646f7964c93e91cb481e6d4e6c52774adf4.doc Page 1 of 28 turns. 238 s) / A (0.0260 Wb) (0.00285 A) (1.40 V) 10 6 . 12 ( ) / ( / 3 = = =- dt di i N B 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 b) Wb. 10 4.50 (400) / H) (0.250 A) 720 . ( / 4- = = = 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 = = = R I P No. Magnetic energy and thermal energy are independent. c) As long as the current is constant, constant. = U 30.13: r Al N 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 = = = - AI rU 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 is used in place of ....
View Full Document

This note was uploaded on 02/25/2009 for the course PHY 127 taught by Professor Mr.yang during the Spring '08 term at SUNY Stony Brook.

Page1 / 28

PHY_solution-chapter30 - 30.1: a) V, 0.270 /s) A 830 ( H)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online