A circuit in the form of a circular turn of wire of radius b is placed at the ce

Question

A circuit in the form of a circular turn of wire of radius b is placed at the center of a larger turn of radius a, b << a. The small circiut is fixed so that it is free to rotate about one of its diameters, this diameter being located in the plane of the larger circuit. The circuits carry the steady currents Ib and Ia respectively.

(b) In what direction is the torque when Ib and Ia circulate in the same sense?

Please show all work and thanks in advance.

Explanation / Answer

a)

here , at the centre of the large circle ,

magnetic field is

B = u0*Ia/(2a)

Now , for angle theta between the planes ,

Torque = (u) X B

Torque = Ib*pi*b^2 * (u0*Ia/(2a)) * sin(theta)

Torque = u0*pi*Ib*Ia*b^2*sin(theta)/(2a)

b) for current in same sense in both sense ,

the torque is downwards

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