Q6.14 (e) Figure 6.20 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=15cmB=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.

(e) How much power is required (by an external agent) to keep the rod moving at the same speed   (=12cm\: s^{-1})   when Kis closed? How much power is required when K is open?

Answers (1)

Force on the rod 

  F=7.5*10^{-2}

Speed of the rod 

  v=12cm/s=0.12m/s

Power required to keep moving the rod at the same speed 

P=Fv=7.5*10^{-2}*0.12=9*10^{-3}=9mW

Hence required power is 9mW.

When the key is open, no power is required to keep moving rod at the same speed.

Exams
Articles
Questions