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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?`
8 years ago

10 Points
```										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

```
8 years ago
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