Physics-
General
Easy

Question

A conductor A B O C D moves along its bisector with a velocity of 1 blank m divided by s through a perpendicular magnetic field of 1 blank w b divided by m to the power of 2 end exponent, as shown in fig. If all the four sides are of 1 m length each, then the induced emf between points A and D is

  1. 0  
  2. 1.41 blank v o l t  
  3. 0.71 blank v o l t  
  4. None of the above  

The correct answer is: 1.41 blank v o l t


    There is no induced emf in the part A B and C D because they are moving along their length while emf induced between B and C blank i. e. comma between A and D can be calculated as follows

    Induced emf between B and C equals Induced emf between A and B equals B v open parentheses square root of 2 l close parentheses equals 1 cross times 1 cross times 1 cross times square root of 2 equals 1.41 blank v o l t

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