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Q6.14 (a) Figure shows a metal rod PQ  resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod=15cm ,  B=0.50T, resistance of the closed loop containing the rod=9.0m\Omega . Assume the field to be uniform. Suppose K is open and the rod is moved with a speed of 12cm\: s^{-1} in the direction shown. Give the polarity and magnitude of the induced emf.

                

            

Answers (1)

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Given

Length of the rod

 l=15cm=0.15m

Speed of the rod

  v=12cm/s=0.12m/s

Strength of the magnetic field

B=0.5T

induced emf in the rod

e=Bvl=0.5*0.12*0.15=9*10^{-3}V

Hence 9mV emf is induced and it is induced in a way such that P is positive and Q is negative.

Posted by

Pankaj Sanodiya

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