InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a valocity v towards the arm RS. Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for (i) the current in the loop (ii) the force and (iii) the power required to move the arm PQ. |
|
Answer» Let length `PQ =l` and width of rod = dx and let area be A. Velocity = v, resistance = r. Current in the loop (i) `i = (epsilon)/(r)` `i=(Blv)/(r)` (ii) Force in the loop `F=ilB` `F=(B^(2)l^(2)v)/(r)` (iii) `P=Fv=(B^(2)l^(2)v^(2))/(r)` There is a decrease in area Hence emf is `epsilon=(-dphi)/(dt)` `epsilon =(-d)/(dt)(BA) , epsilon = -B(dA)/(dt)` `epsilon = +(Bldx)/(dt) , epsilon = Blv` |
|