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?

Question

AskiitianExpert Shine · Answer

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

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