1.

A conducting rod of length 2l is rotating with constant angular speed omega about is perpendicular bisector. A uniform magnetic field vec(B) exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is

Answer»

`(1)/(2)B omega l^(2)`
`(1)/(8)B omega l^(2)`
Zero
`B omega l^(2)`

SOLUTION :E.m.f. between CENTRE of ROD and one end `= e_(1)`
`e_(1)=(1)/(2)Bl^(2)omega` and between centre and other end
`e_(2)=-(1)/(2)Bl^(2)omega`. Therefore between two ENDS
`e=e_(1)+e_(2)=0`
`therefore` Between two ends of the rod, e.m.f. induced = 0.


Discussion

No Comment Found

Related InterviewSolutions