Question

QUESTION 4 Our big lecture-table solenoid has 560 turs, each with 60 mm diameter, spaced tightly to only be 160 mm Long (left to right) The standard power supply only needs 10.2 Volt to push 60 A current thru the solenoid wire, clockwise as seen from the right We can fit a small flat coil, 200 turns with 25 mm diameter inside that solenoid cavity (so their axes align along x) When turning off the power supply, we adjust the current to zero in 125 millisec (a) show that the magnetic field in the solenoid space used to be 0.02639 T leftward (-x).] b) What magnetic flux (B-A) pierces the small coll?] [c) what direction is the induced electric field, inside the solenoid cylinder : 1) before turnoff ; 2) during turnoff: 3) after turnoff complete d) what voltage magnitude was induced in the small coil? 0.3343 V O 00829 V 0.02073 V O 002073 V O 0.1194 V 2.073 V

Answer 1

Magnetic field along the axis of Solenoid Ba MONI L 4TX10tx 560% 6 160 x103 Or = 0.026387 flux Hirough small coil x area small loop 0.02638 x T 25 X10 1.3x105 The direction of 3 in along tâj. Now, Gx 25 at 1912 = -8x دیا of 11 fon oley aloz En Ez 2 at B (-) mm ( 253 - ohne ] So, the electric field before doing turnoff. Induced &mf in along Z- direction. Small loop w dt do dt AIN- (-0-02638) 2 XT 125x103 (15 pin) Kov { 0.0207 V.