question_answer

A metallic rod of length \[l\] is placed normal to the magnetic field B and revolved in a circular path about one of the ends with angular frequency co. The potential difference across the ends will be ;

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)  \[B\omega {{l}^{2}}\]

Correct Answer: B

Solution :

Suppose a conducting rod of length \[l\] rotates with a constant angular speed co 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 = no. 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. \[\therefore \] \[e=\int_{0}^{l}{de=\int_{0}^{l}{Br\omega dr=\frac{B\omega {{l}^{2}}}{2}}}\] \[\therefore \] \[e=\frac{B\omega {{l}^{2}}}{2}\] From right hand rule we can see that P is at higher potential than 0. Thus, \[{{V}_{P}}-{{V}_{0}}=\frac{B\omega {{l}^{2}}}{2}\]