Question

A square frame carrying a current I = 0.90 A is located in the same plane as a long straight wire carrying a current I₀ = 5.0 A. The frame side has a length a = 8.0 cm. The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is η = 1.5 times greater than the side of the frame. Find: 

(a) Ampere force acting on the frame; 

(b) the mechanical work to be performed in order to turn the frame through 180° about its axis, with the currents maintained constant.

Answer 1

(a) As is clear from the condition, Ampere's forces on the sides (2) and (4) are equal in magnitude but opposite in direction. Hence the net effective force on the frame is the resultant of the forces, experienced by the sides (1) and (3).

Now, the Ampere force on (1),

So, the resultant force on the frame = F₁ - F₃ , (as they are opposite in nature.)

(b) Work done in turning the frame through some angle,  where Φ_f is the flux through the frame in final position, and Φ_i that in the the initial position.