The aluminum foil in a package containing 75 square feet of foil weighs 12 ounces. aluminum has a density of 2.70 g/ cm3. what is the approximate thickness of the foil in mm?

Solution:

Since we know that 1 ounce=28.4g/oz

So for 12 oz  X  28.4 g/oz  =  340.8 g

And,

 340.8 g  X  1cm3/2.7g  =  126.22 cm3

Again,

75 ft2  X  12 in/ft  X  12 in/ft  X  2.54 cm/in  X  2.54 cm/in  =  69,677.28 cm2  

(Here ft^2=ft*ft)

Now,

69,677.28 cm2  X Thickness  =  126.22 cm3

Finally Solve for Thickness: 126.22 cm3 / 69,677.28 cm2  =  0.00181 cm

Thus for mm we have to:

0.00181 cm  X  10 mm/1 cm  =  0.0181 mm.

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it