As shown in the figure, a metallic square frame of edge a is placed in a vertical plane. A uniform magnetic field B exists in the space, acting perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They start pulling the corners at t = 0 and move them at a uniform speed u . Determine the induced emf in the frame at the instant when the angles at these

Step-by-step solution

Let the angle at the pulled corners be . The length of the diagonal connecting these two corners is given by x = 2a ( 2 ) . Since each corner is pulled at a uniform speed u in opposite directions, the rate of change of the diagonal length is dx dt = 2u . Differentiating x with respect to time t , we get: dx dt = -a ( 2 ) d dt 2u = -a ( 2 ) d dt d dt = - 2u a ( /2) The area of the rhombus is A = a^2 . The rate of change of the area is: dA dt = a^2 d dt Substituting the expression for d dt : dA dt = a^2 ( - 2u a ( /2