I am using 1.8x1.8 m magnetic coil 30 A/m with 220 V and 60 Hz
How to calculate current I

 

Measure the resistance, because that is a small coil without much inductance, unless the description just neglected to mention the details, which we have none. I=E/R
Not a word about the construction, core, number of turns, or wire size.

 

I am using 1.8x1.8 m magnetic coil 30 A/m with 220 V and 60 Hz
How to calculate current I

This page lists all the calculators available on Encyclopedia Magnetica
For example: formula and calculator for loop

 

I am using 1.8x1.8 m magnetic coil 30 A/m with 220 V and 60 Hz
How to calculate current I

What is the intended purpose/application?

 

Measure the resistance, because that is a small coil without much inductance, unless the description just neglected to mention the details, which we have none. I=E/R
Not a word about the construction, core, number of turns, or wire size.

Number of turns is 12

 

OK, so the length of the wire is 12 x 4 x 1.8 meters. And now the assumption is that there is no magnetic material in the core. So while there are formulas for predicting the inductance, I do not have them handy, other than to predict that the inductive reactance at 60H will be quite low, and not affect the current much. The wire resistance per length is published in tables for many wire sizes, and as we do not know the wire size the TS will need to look that up.
The calculations for the flux density are posted, so that can be calculated. The current will be close to what is predicted by the supply voltage and the 60 Hz frequency for the calculated resistance, which depends on the wire size. At 60 Hz the inductive reactance will not be significant. The current will be limited by the resistance.
I am not sure what the "30A/ M tells me, unless that is the intended current density.
That whole process should be obvious to a design engineer.
And this looks a whole lot like homework for one who was not paying attention in class.

 

Resistance is nothing to do when in AC. In DC yes, then i=V/R. But in AC the X(L)>>R thus the i=V/X(L) where x(L)=2pi()fL. If want very exact, may add the active component by Pythagorus: Z=sqrt(X(L)^2+R^2). But R and X ratio rather sharp differs by moving the core in and out, thus the most exact method is to measure. LATR+Ammeter+Vmeter+magnet. What is permissible current may dedicate seeing the wire diameter. A=pi()r^2/4 and permissibl;e is 2 A/mm2 when bibbin is perfctly cool, and 3...3.5 A/mm2 when it is hot on the edge, and up to 5 A/mm2 for rare short applications like lock unlocking. One minute work and one hour to cool.

 

Resistance is nothing to do when in AC. In DC yes, then i=V/R. But in AC the X(L)>>R thus the i=V/X(L) where x(L)=2pi()fL. If want very exact, may add the active component by Pythagorus: Z=sqrt(X(L)^2+R^2). But R and X ratio rather sharp differs by moving the core in and out, thus the most exact method is to measure. LATR+Ammeter+Vmeter+magnet. What is permissible current may dedicate seeing the wire diameter. A=pi()r^2/4 and permissibl;e is 2 A/mm2 when bibbin is perfctly cool, and 3...3.5 A/mm2 when it is hot on the edge, and up to 5 A/mm2 for rare short applications like lock unlocking. One minute work and one hour to cool.

Did you calculate the inductive reactance of that coil at 60 Hz??? The TS was not asking for an exact answer, based on the lack of information provided.

 

I am using 1.8x1.8 m magnetic coil 30 A/m with 220 V and 60 Hz
How to calculate current I

Number of turns is 12

I hope, TS already calculated, that 12-turn 1.8 x 1.8 m rectangular coil
has magnetic field strength H = 30 A/m in its center
at current I = 5A.

 

I hope, TS already calculated, that 12-turn 1.8 x 1.8 m rectangular coil
has magnetic field strength H = 30 A/m in its center
at current I = 5A.

May I know how did u get I = 5A

 

May I know how did u get I = 5A

Using this formula

then divide result 60 A by 12 (number of turns).

 

OK, my approximation says that is 288 feet of wire, 12 turns of 24 ft each and if the wire is #16, reasonable for 5 amps, and about 4 ohms per thousand feet, then the resistance will be a bit over ONE ohm. So the inductance of that coil must be a fair amount.

 

Using this formula
View attachment 285828
then divide result 60 A by 12 (number of turns).

Using this formula
View attachment 285828
then divide result 60 A by 12 (number of turns).

by using the above formula and inserting x=12 a=1.8 and b=1.8 but still i was not able to get the answer 60

 

Using this formula
View attachment 285828
then divide result 60 A by 12 (number of turns).

i also inserted by using x=0

 

by using the above formula and inserting x=12 a=1.8 and b=1.8 but still i was not able to get the answer 60

x is location from the center of the loop (the centre is located at point x = 0),
so we use x=0.

i also inserted by using x=0

Double check your calculation. With x=0 answer should be 60.
This answer is for one turn (loop).
For 12 turns we should divide answer by 12.
So, I=60/12=5A
ADDED:
Formula for Excel:
=30/((2/PI())*(1.8*1.8/SQRT(1.8*1.8+1.8*1.8))*(1/(1.8*1.8)+1/(1.8*1.8)))
Answer:
59.9789196651...