Question

A copper rod ab of length l pivoted at one end a

Answer 1

Explanation:

A copper rod ab of length l pivoted at one end a rotates at a constant angular velocity w at right angles to a uniform magnetic field of induction b the emf developed between the mid point c of the rod and end b.

hope you get it now...

Answer 2

$\huge\boxed{\fcolorbox{cyan}{Yellow}{HeLLo MaTe..!!}}$

Motional EMF induced in a conductor of length L moving in magnetic field is given by equation

EMF = vBL

now lets take a small part of length "dx" which is at distance"x" from the hinge point

the speed of the point is given by

v = x\omega

now the motional emf in that small part of the rod will be

EMF =_{x=L/2}^{x=L}\int vBdx

EMF = _{x=L/2}^{x=L}\int x\omega Bdx

EMF = \frac{(L^2 - (L/2)^2)}{2} \omega B

EMF = \frac{3B \omega L^2}{8}

so above is the induced EMF frm centre of rod to end of rod