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# A very small circular loop of radius r and carrying a current i1 is placed in the x-y plane with its centre on x-axis at the point C(a,0). A square loop of side length 2l carrying a current i2 is fixed in the y-z plane with the centre of the loop at the origin . Calculate the torque exerted by the square loop on the circular loop?

12 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