Question

Question 2 --/2 Our big lecture-table solenoid has 560 turns, 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 6.0 A current thru the solenoid wire, clockwise as seen from the right When turning off the power supply, we adjust the current to zero in 125 millisec. * read this * → [a) show that the magnetic field in the solenoid space used to be 0.02639 T , leftward (-x).] [b) what direction is the induced electric field, inside the solenoid cylinder : 1) before turnoff; 2) during turnoff; 3) after turnoff complete c) what voltage magnitude was induced in the solenoid's wire coil? [d) what amount of current would the induced voltage push thru the solenoid wires?) 0.000597 v 2 1.275 V 3 0.3343 V 4 10.2 V 81.6 V

Answer 1

N- d gas 560 turns 60mm 30 mm 160mm 2 N e o L6 =560 =) 3 500 turns m -7 x3600 x 6 twen off (a) B = NonI = 4ITXU 0.02639T b) Induced electric field :- before turn off > No direction during - Rightwards - after torn able - No direction C) induced --NA /B₂-Bi 32 - 3) - - 560 XT1 (0.03)*x60 -0.02639) 0.125 0.04178 0. 3343/v 0.125 Voltage (d) Induced Current = Vinduced = 0.3343 (10.216) 10. 1966 A