By energy conservation...
ω=√3gsinθ/l
dε=ωxBdx
ε=ωB∫xdx
=ωBl2/2
Put value of ω
ε= B√3gsinθ
2
______________
A conducting rod of length l is hinged at a point.It is free to rotate in a vertical plane.There exists a uniform mag. field B perpendicular to the plane.The rod is released from horizontal position.What is the potential difference between the ends of the rod as a function of angle θ the rod makes with the starting posn?
By energy conservation...
ω=√3gsinθ/l
dε=ωxBdx
ε=ωB∫xdx
=ωBl2/2
Put value of ω
ε= B√3gsinθ
2
______________
i understood the second part but can u explain how u got the value of ω or the exact expression you used