What current would it need to carry in order to experience a very modest 1.0×10−3n⋅m torque?

By using the formula:

Tau=(I)(A)(B)Sinθ 

Get the area first of the circle:

A= πr^2 

A= π(0.08m)^2 

A= 0.02011 m^2 

Then solve, 

B(earth)= 5x10^-5 T 

θ = 0, since the angle of the wire is oriented for maximum torque in the earth's field. Which means the angle is 0. 

Reorder the formula to solve for the charge (I): 

I = (tau)/(A)(B)Sinθ 

I = (tau)/(A)(B) 

I = (1.0x10^-3) / (0.02011)(5x10^-5) 

I = 994 A

I'm not sure what your asking buddy.. but i know that you can weld two peices of steel by pure force and pressure.