Question

AB and CD are two parallel conductors kept 1 m apart and connected by a resistance R of 6 Ω as shown in Figure below. They are placed in a magnetic field B = 3 × 10^-2 T which is perpendicular to the plane of the conductors and directed into the paper. A Wire. MN is placed over AB and CD. and then made to slide with a velocity 2 ms^-1 (Neglect the resistance of AB, CD, and MN.)

Calculate the induced c.urrent flowing through the resistor R.

Answer 1

Let \[\overrightarrow{F}\] be the force on any free electron within the wire MN.
              \[\overrightarrow{|F|} = q  | \overrightarrow{V}|\ |\overrightarrow{B}|\] sin 90°
⇒              F = q x 2 x 3 x 10^-2

Now, W, the work done = force x displacement

               W = F x 1
              W = q x 2  x 3 x 10^-2 

Potential difference or emf induced
                  =`W/q`

⇒           `e = (qxx6xx10^-2)/q V`        

               e = 6 x 10^-2 V

∴ The induced current in R(6Ω)

                  =`V/R =E/R`

                  =`(6xx10^-2)/6 = 10^-2` A

∴   Current = 10^-2=.001 A