A copper rod AB of length L,pivoted at one en | vedclass

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(C) Consider a small element of length $dx$ at a distance $x$ from the pivot $A$.The motional e.m.f. $de$ induced across this small element is given b

Vedclass · Accepted Answer

$\frac{3B\omega L^2}{8}$

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(C) Consider a small element of length $dx$ at a distance $x$ from the pivot $A$.The motional e.m.f. $de$ induced across this small element is given by $de = Bv dx$,where $v = \omega x$.Substituting $v$,we get $de = B(\omega x) dx$.To find the total e.m.f. $e$ between the midpoint $C$ (at distance $L/2$ from $A$) and the end $B$ (at distance $L$ from $A$),we integrate the expression:$e = \int_{L/2}^{L} B\omega x dx = B\omega \left[ \frac{x^2}{2} \right]_{L/2}^{L}$$e = \frac{B\omega}{2} \left( L^2 - \left(\frac{L}{2}\right)^2 \right) = \frac{B\omega}{2} \left( L^2 - \frac{L^2}{4} \right)$$e = \frac{B\omega}{2} \left( \frac{3L^2}{4} \right) = \frac{3B\omega L^2}{8}$

$A$ copper rod $AB$ of length $L$,pivoted at one end $A$,rotates at a constant angular velocity $\omega$,at right angles to a uniform magnetic field of induction $B$. The e.m.f. developed between the midpoint $C$ of the rod and end $B$ is

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