Question

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

Answer 1

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

Answer 2

Answer:

the ans is 3/8Bwl^2

Explanation:

Hope this ans helps you.....

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