The arm `PQ` of the rectangular conductor is moved from `x=0`, outwards in the uniform magnetic field which extends from `x=0` to `x=b` and is zero for `x gt b` as shown. Only the arm `PQ` possess substantial resistance `r`. Consider the situation when the arm `PQ` is pulled outwards from `x = 0` to `x = 2b`, and is then moved back to `x = 0` with constant speed `v`. Obtain expression for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance.

Explanation

For forward motion from x=9 x=2b
The flux `phi_(B)` linkd with circtuit SPQR is

`{:(phi_(B)=Blx,0lexgltb),(BlB,blex2b):}`
The induced emf is ,
`{:(e=(-d phiE)/(JL),,),(e=-Blv,0lexltb,),(=0,,blexlt2v):}`
When induced emf is non-zero, the current, I in the magnitude,
`I=(theta)/(tau)=(Blv)/(r)`
The force requried to keep are PQ in constant motion is F=llB. Its is to hte left. In magnitude
`{:(F=l//B=(B^(z)l^(2)v)/(r),,,olexltb),(=0,,,blexlt2b):}`
The joule heating loss is
`{:(P_(l)=I^(2)r,,),(=(B^(2)l^(2)v^(2))/(r),,blexlt2b):}`
One obtains similar expressions for the inward motion fromx=2b to x=0