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Q6.14 (d)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.

 (d) What is the retarding force on the rod when K is closed?

Answers (1)

best_answer

Induced emf = 9mV (calculated in a part of this question)

The resistance of loop with rod = 9m\Omega

Induced Current

 i = \frac{e}{R}=\frac{9mV}{9m\Omega }=1A

Now,'

Force on the rod 

F= Bil= 0.5*1*0.15=7.5*10^{-2}

Hence retarding force when k is closed is 7.5*10^{-2}.

Posted by

Pankaj Sanodiya

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