question_answer

A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity \[\omega \]. The induced e.m.f. between the two ends is [MP PMT 1992; Orissa JEE 2003]

A)            \[\frac{1}{2}B\omega {{l}^{2}}\]                                      

B)            \[\frac{3}{4}B\omega {{l}^{2}}\]

C)            \[B\omega {{l}^{2}}\]       

D)            \[2B\omega {{l}^{2}}\]

Correct Answer: A

Solution :

                   If in time t. the rod turns by an angle q, the area generated by the rotation of rod will be \[=\frac{1}{2}l\times l\theta \] \[=\frac{1}{2}{{l}^{2}}\theta \] So the flux linked with the area generated by the rotation of rod                     \[\varphi =B\ \left( \frac{1}{2}{{l}^{2}}\theta  \right)\cos 0=\frac{1}{2}B{{l}^{2}}\theta =\frac{1}{2}B{{l}^{2}}\omega \,t\]                    And so \[e=\frac{d\varphi }{dt}=\frac{d}{dt}\left( \frac{1}{2}B{{l}^{2}}\omega t \right)=\frac{1}{2}B{{l}^{2}}\omega \]