Question

A conducting rod PQ of length 20 cm and resistance 0.1 ohm rests on two smooth parallel rails of negligible resistance AA' and CC'. It can slide on the rails and the arrangement is positioned between the poles of a permanent magnet producing uniform magnetic field B = 2.4 T. The rails, the road and the magnetic field are in three mutually perpendicular directions as shown in the figure. If the ends A and C of the rails are short circuited, find the
(I) external force required to move the rod with uniform velocity v = 10 cm/s, and
(II) power required to do so.​

Answer 1

Answer:

The external force required to move the rod is 0.008 v, and the power is 6.410^-4w.

Explanation:

Given R=0.1l

         b=0.4T  

         l=0.2m

         v=0.1 m/s

and  uniform velocity v = 10 cm/s,

Force F=Ilsin

         F=0.0080.2

         F=0.016N

now identify the external force,

         E=BlV

          E=0.4**0.2**0.1

          E= 0.008V

External force  E= 0.008V

To Find power

P =V^2/R

   =I^2R

I=V/R

=0.008**0.1

I=0.08Amp

P=(0.008)^2/0.1

P=0.0064**0.1

P=0.0064

Then the power P=(0.008)^2/0.1

                         P=0.0064**0.1

                         P=6.410^-4w

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