A metal bar with length `L` , mass `m` and resistance `R` is polaced on frictionless metal rails that are inclined at an angle `phi` above the horizontal. The rails have negiligible resistance. There is a uniform magnetic field of nagnitude `B` direction download in the figure. The bar is released from rest and slide down the rails.
(a) Is the direction of the current induced in the bar from `a` and `b` or `b` to `a` ?
(b) What is the terminal speed of the bar ?
What is the induced current in the bar when the terminal speed has been reached ?
(d) After the terminal speed has been reached, at what rate is electrical energy being converted to thermal energy in the resistance of the bar ?
(e) After the terminal speed has been reached, at what rate is work being done on the bar by gravity? Compare your answer to that in part (d).

Explanation

As bar slides down, flux through close area increases and hence magnetic field due to inducet current is upward therefore induced current is anticlockwise i.e. `a` to `b` .
(b) When speed of bar is `v`
Induced emf `=(Bcosphi)vL`
Induced current `=((Bcosphi)vL)/(R)`
Magnetic force due to
`Bcosphi : (Bcosphhi)IL` , along the plane upwards
`Bsinphi : (Bsinphi)IL` , along `mgcosphi`
When velocity is constant i.e. net force=0
`mgsinphi=(Bcosphi)IL=((B^(2)cos^(2)phi)vL^(2))/(R)`
Terminal speed `v=(mgRsinphi)/((B^(2)cos^(2)phi)L^(2))`
(c) `I=((Bcosphi)vL)/(R)=((Bcosphi)mgRsinphiL)/((B^(2)cos^(2)phi)L^(2)R)`
`=(mgtanphi)/(BL)`
(d) `P=I^(2)R=(m^(2)g^(2)tan^(2)phiR)/(B^(2)L^(2))`
Rate of work done by gravitational force
`P'=(d)/(dt)(mgsinphis)=mgsinphi(ds)/(dt)=mgsinphiv`
`=(m^(2)g^(2)tan^(2)phiR)/(B^(2)L^(2))=P`