Question

QUESTION 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? Id) what amount of current would the induced voltage push thru the solenoid wires ?] 0.3343 V 0 1.275V 0816V O 102 V O 0.000597 V

Answer 1

Number of turn = Joo turn 2.) = 0.16om length of solenoid = 160mm Number of turn per unit length= = 3500 turnim 560 0.16 - Valtage applied 10.2 Volt current par parsing = GA Resistance of solenoid = 10-2 6 1.7 Ameer a.) Magnetic field = MONI = 417 x 10-7 x 3500 x 6 B = 2-637 x 10 - 2 Tesla B = 0.026 37 Tesla As clockwise current in Right end, means south end Direction of magnetic field south to North inside the solenoid. direction is - x durection to left) Hence Right to

is - b.) As current steady means induced emf is zero before and during turn off. after turn off emf a induced and it It try to stop decrease in current. Hence Electric field will in the same direction of Currento C.) AS dt = 125 m sec 0.125 sec As turn off the switch Zero in 0.125 sec. / magnetic field 0.02637 Tesla becomes radius of coil solenoid = 30mm > = during this process 0.03m induced emf dd dt 0.02637 X U (0.0312 0.125 e= 0.000597 volt d.) induced current = induced Resistance emf •597 x 10-3 - 1.7 = 0.351 x 10-3 Ampere induced current = 0.351 mA