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Hi
by symmetry we can conclude that the net magnetic feild due to the square loop at (a,o) is in the x direction So, B(net)=4*[(u0*i2/4*pi*x)*(2cosy)]*cos(90-y) where x=(a2+l2)1/2 y=tan-1(l/a) cos(90-y) is done as it gives x-component) SO B=u0*i2*a*l/(pi*(x 3/2)) now assuming B remains uniform throughout circular loop, torque=mxB where, m=magnetic moment of circular loop
SO B=u0*i2*a*l/(pi*(x 3/2))
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