mg(l/2)sinθ=(1/2)m(l2/3)ω2
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?
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4 Answers
Abhishek Priyam
·2009-02-17 21:48:29
By energy conservation...
ω=√3gsinθ/l
dε=ωxBdx
ε=ωB∫xdx
=ωBl2/2
Put value of ω
ε= B√3gsinθ
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Vivek
·2009-02-18 07:19:59
i understood the second part but can u explain how u got the value of ω or the exact expression you used