Question

A horizontal metal frame, ABCD, moves with a uniform (horizontal) velocity of 0.2 m/s, into a uniform field of 10^-2T, acting vertically downwards. AB = 0.1m and AD = 0.2m and the resistance R of the frame is 5Ω . As shown, the sides AB and CD enter the field in a direction normal to the field boundary. Find the magnitude of current induced in the metal frame when 

(i) CD just enters the field 

(ii) the whole frame is moving through the field 

(iii) CD just moves out of the field on the other side? 

Draw a sketch graph showing the variation of the induced current, the force required to move the frame with a constant velocity, and the power spend by the external agency

Answer 1

(i) The sides AD and BC are parallel to V. Hence no emf in induced in them. For the side CD, moving in the field,

(ii) As the whole frame in moving through the field from time t₁ to t₂ say the flux through the frame is constant. Hence there is no induced current during this interval. [Alternatively the induced emf in the arms AB and CD are in opposite directions, the resultant emf in the frame is zero.] 

(iii) When CD just moves out of the field region, on the other side, the magnitude of the induced emf, in AB, moving through the field, is again equal to 2 x 10^-4 V. Hence magnitude of induced current, in the frame, is same, but its direction is now opposite to that in the first case (The flux is now decreasing with time). The required sketches, therefore, have the forms shown.