JUST CHOOSE THE RIGHT CHOICE Since it is known that the magnetic permeability of the steel rod and the core of the electromagnet and the cross-sections are the same, the steps to be passed in order to find the current value that will lift the steel rod with a weight of G = 1 ton are correctly expressed in which option. μrpol=1;μrnuve=200;μr steel=200 1-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=10-3/(200μoS); 2-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=2*10-3/μoS); 3-) S=75.10-4; m2 Rnuve+Rsteel=(2*500+2*100)10-3/(200μoS) ;Rpoly=2*10-3/(200μoS); 4-)Rtotal=Rnuve+Rcelik+Rpoly; 5-) Φ is found from G=Φ2/(μoS). 6-) Φ is found from G=0.5o2/(μoS). 7-) There is I current from Φ=NI/Rtotal. 8-) Φ=NIRtotal I current is found.   A. 2,4,5,7 B. 2,4,6,8 C. 3,4,6,8 D. 1,4,5,7

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JUST CHOOSE THE RIGHT CHOICE

Since it is known that the magnetic permeability of the steel rod and the core of the electromagnet and the cross-sections are the same, the steps to be passed in order to find the current value that will lift the steel rod with a weight of G = 1 ton are correctly expressed in which option. μrpol=1;μrnuve=200;μr steel=200

1-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=10-3/(200μoS);

2-) S=75.10-4; m2 Rnuve+Rsteel=(450+2*75+500)10-3/(200μoS) ;Rpoly=2*10-3/μoS);

3-) S=75.10-4; m2 Rnuve+Rsteel=(2*500+2*100)10-3/(200μoS) ;Rpoly=2*10-3/(200μoS);

4-)Rtotal=Rnuve+Rcelik+Rpoly;

5-) Φ is found from G=Φ2/(μoS).

6-) Φ is found from G=0.5o2/(μoS).

7-) There is I current from Φ=NI/Rtotal.

8-) Φ=NIRtotal I current is found.

 

A. 2,4,5,7
B. 2,4,6,8
C. 3,4,6,8
D. 1,4,5,7
a=500 mm
b=50 mm
c=150 mm
d=100 mm
1=1 mm
N=1200 turns
μ-200μο
b
polymer
polymer
steel
bar
u=200µ0
b
G
Orio07=r
Transcribed Image Text:a=500 mm b=50 mm c=150 mm d=100 mm 1=1 mm N=1200 turns μ-200μο b polymer polymer steel bar u=200µ0 b G Orio07=r
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