Physics-
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Easy

Question

A rectangular loop with a sliding connector of length l = 1.0 m is situated in a uniform magnetic field B = 2T perpendicular to the plane of loop. Resistance of connector is r = 2capital omega. Two resistance of 6capital omega and 3capital omega are connected as shown in figure. The external force required to keep the connector moving with a constant velocity v = 2m/s is

  1. 6 N    
  2. 4 N    
  3. 2 N    
  4. 1 N    

The correct answer is: 2 N


    Motional emf e equals B v l rightwards double arrow e equals 2 cross times 2 cross times 1 equals 4   V
    This acts as a cell of emf E equals 4   Vand internal resistance r equals 2 capital omega.
    This simple circuit can be drawn as follows

    Current through the connector i equals fraction numerator 4 over denominator 2 plus 2 end fraction equals 1   A
    \magnetic force on connectorF subscript m end subscript equals B i l equals 2 cross times 1 cross times 1 equals 2   N
    (Towards left)

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