Show a rectangular loop MNOP being pullled out of a magnetic field with a uniform velocity v by applying an external force F. The length MN is equal to l and the total resistance of the loop is R. Find(a) the current in the loop,(b) the magnetic force on the loop, (c) the external force F needed to magnetic the velocity, (d) the power delivered by the external force and (e) the thermal power developed in the loop.

Explanation

No motion emf in rod `OP` , it is outside magnetic field.
No motional emf in rods `NO` and `MP` , because velocity is parallel to length. The rod `NM` can be repalced by a battery of emf `e=Bvl` , with positine terminal at `N` .
(a) Induced current `I=(e)/(R)=(BVl)/(R)` , clockwise
(b) Magnetic force: No force on `OP` , it is outside magnetic field, the force on `NO` is equal and opposite to force on `MP` . The magnetic force om `MN` equals net magnetic force on loop
`F_(m)=Bll=(B^(2)vl^(2))/(R)=F_(loop)` , towards left
(c) Since loop is moving with constant velocity,
Net force=0 implies Applied force `F=F_(m)=(B^(2)vl^(2))/(R)`
(d) power delivered by external force `P=vec(F).vec(v)=Fv=(B^(2)v^(2)l^(2))/(R)`
(e) Thermal power `p'=I^(2)R=(B^(2)v^(2)l)/(R)`
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