tailieunhanh - Physics exercises_solution: Chapter 30

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 30 | a 2 M di1 dt x 10 4 H 830 A s V and is constant. b If the second coil has the same changing current then the induced voltage is the same and 1 V. For a toroidal solenoid M N2OB2 i1 and OB2 u0N1i1 A 2xr. So M y0 AN1 N2 2xr. a M N2OB2 i1 400 Wb A H. b When i2 A OB1 i2M N1 A H 700 x 10 3Wb. a M s2 di dt x 10 3 V A s x 10-3H. b N2 25 i1 A OB2 i1M N2 A x 10-3H 25 x 10-4 Wb. c di2 dt A s and 1 Mdi2 dt x 10 3 H A s mV. 1H 1 Wb A 1Tm2 A 1 Nm A2 1 J A2 1 J AC s 1 V A s 1 Qs. For a toroidal solenoid L NOB i s di dt . So solving for N we have x 10-3 V A N i OB di dt --------------------------- 238 turns. Wb A s a L di1 dt H A s x 10-3V. 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. _ . r 2 500 1800 2 x 10-5 m2 n a L K unN A 2nr ------------0------------------------- H. K m 0 2n m b Without the material L LK H x 10-4 H. Km Km 500 For a long straight solenoid L NOB i and OB p0NiA l L p0N2 A l. a Note that points a and b are reversed from that of figure . Thus according to Em iation 30 8 b a 4 00 A s Thiiq ftip ci irrpnt 1 q dpctpaQmu to t quaiivti dt l 0 260 H . - a s. a itus ute cutten l i s viecteasng. b From above we have that di - A s dt. After integrating both sides of this expression with respect to t we obtain Ai - A s At i A - A s s A. a L s di dt V A s . b OB iL N A H 400 x 10-4 Wb. a U 1 LI2 H A 2 2 J. b P 12R A 2 180 Q W. c No. Magnetic energy and thermal energy are independent. As long as the current is constant U constant. U 1 LI2 N2Al2 2 4nr N . 4nrU 1 4n m J 7 - 2850 turns. AI2 x 10 Air .