A) \[\frac{1}{2}{{B}^{2}}l\omega \]
B) \[\frac{1}{2}B\omega {{l}^{2}}\]
C) \[\frac{1}{8}B\omega {{l}^{3}}\]
D) \[B\omega {{l}^{3}}\]
Correct Answer: B
Solution :
Suppose a conducting rod of length \[l\]rotates with a constant angular speed \[\omega \] about a point at one end. A uniform magnetic field \[\vec{B}\]is directed perpendicular to the plane of rotation as shown in figure. Consider a segment of rod of length drat a distance r from O. This segment has a velocity \[v=r\omega .\] The induced emf in this segment is \[de=Bvdr=B(r\omega )dr\] Summing the emfs induced across all segments, which are in series, gives the total emf across the rod. \[\therefore \] \[e=\int_{0}^{1}{de}=\int_{0}^{1}{Br\omega dr}=\frac{B\omega {{l}^{2}}}{2}\] \[\therefore \] \[e=\frac{B\omega {{l}^{2}}}{2}\] rom right hand rule we can see that P is at higher potential than O. Thus, \[{{V}_{P}}-{{V}_{O}}=\frac{B\omega {{l}^{2}}}{2}\]You need to login to perform this action.
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