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A little solenoid electromagnet is 110mm long and 30mm diameter ; it has 30? res

ID: 1262595 • Letter: A

Question

A little solenoid electromagnet is 110mm long and 30mm diameter ; it has 30? resistance and inductance 90mH (with an iron rod in it).
is powered it with 4 dry cells (6V) with a switch.
. . . a) Calculate the current that should have been flowing in the solenoid long after the switch was closed.
. . . b) Calculate the solenoid Area, and determine the magnetic field that it must have been producing.
. . . surprised? that is with the iron rod filling its core ... its inductance is 1/9 as much without the iron rod.
. . . c) using its resistance and its inductance, how long should it have taken for the current to rise to 125mA?
. . . d) if its B

A little solenoid electromagnet is 110mm long and 30mm diameter ; it has 30? resistance and inductance 90mH (with an iron rod in it). is powered it with 4 dry cells (6V) with a switch. . . . a) Calculate the current that should have been flowing in the solenoid long after the switch was closed. . . . b) Calculate the solenoid Area, and determine the magnetic field that it must have been producing. . . . surprised? that is with the iron rod filling its core ... its inductance is 1/9 as much without the iron rod. . . . c) using its resistance and its inductance, how long should it have taken for the current to rise to 125mA? . . . d) if its B AA changed by (5/8) A?.018 TmA^2 in that time (c), what average epsilon was induced?

Explanation / Answer

A)

supply voltage=6v

inductance XL=2*3.14*50*90*10-3

=28.26OHMS

R=30OHMS

Z=30+j28.26

=41.21ohms

I=V/Z

=6/41.21

=0.1455A

B)

L=mu*N2*A/L

A=3.14*900*10-6/4

=706.5*10-6m2

90*10-3=80.66*10-16*N2

N=3.34*106TURNS

N*PI=LI

=90*10-3*0.1455

=13.095/3.34*106

=3.92*10-6

flux=BA

B=3.92*10-6/706.5*10-6

=5.54mT

C)

time constant of circuit=L/R

=90/30

=3milliseconds

flux=BA

=0.1125wb

induced emf=0.1125/3

=0.0375*103VOLTS

=37.5volts

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