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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 mutually 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. 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?

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The power from external sources is required to overcome the resisting force of 75\times 10^{-6} N

The velocity at which the rod moves is given as v=12 cm/s = 0.12 m/s

The power is calculated as,

P=Fv =75\times10^{-3} \times 0.12 =9 \times 10 ^{-3} W =9 mW

So, When key K is open, there is no current flow in the rod, thus no power is required.

Posted by

Pankaj Sanodiya

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