AskiitianExpert Shine
Last Activity: 15 Years ago
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