(1)The magnetic field due to a straight current carrying wire at a point on its perpendicular bisector is given by- (please see attachment)

B ={UoI/(4IId)}*(

cos@1 -

Cos@2), where Uo stands for mu-not and II stands for pi)

Put Io/4II = 10^-7, d = 10cm = 0.1 m, @1=45 and @2 = 135 degree.

The field due to four sides will be four times of the above value and will be directed inward.