Figure shown a metallic square frame of edge an in a vertical plane. A uniform magnetic field B exists in the space in a direction 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 displace the corners at a uniform speed u.
(a) Find the induced emf in the frame at the instant when the angles at these corners reduced to 60°.
(b) Find the induced current in the frame at this instant if the total resistance of the frame is R.
(c) Find the total charge which flows through a side of the frame by the time the square is deformed into a straight line.
Given:
Edge length of square frame =a
Magnetic field intensity B
Speed of corners of rhombus =u
(a) when the angles at the corner reduce to 60°
We know that,
motional emf produced due to a conductor of length l moving with velocity v in a magnetic field B is given by
Motional emf is produced in each side of rhombus and the effective length of each side is the length perpendicular to velocity of corners.
From the fig. the effective length of each side is
Since velocity is perpendicular to magnetic field the equation of emf induced in each side given by
Total emf induced in all four side is
Therefore, induced emf in the frame when the angles at the corner reduces to 60° is 2uBa
(b) total resistance of the frame =R
hence current flowing in the frame is
Therefore, current flowing in the frame at this instant is 2uBa/R
(c) initially the frame is in form of square of side a and area 
at this time flux through this frame is given by
Finally, when the frame reduces to straight line flux passing through the frame reduces to zero
Average emf induced in the frame in time t is given by
The current flowing through the frame is then given by
Where R is the resistance of frame
Hence the charge (Q) flowing through the side of frame in time t is
Therefore total charge flowing through side of frame by the time the frame reduces to straight line is 