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Q.

A rod PQ of length L and resistance r rotates about OZ axis with angular velocity $ω$ . The ends of the rod slide on a circular conducting frame of negligible resistance. A resistance $\frac{r}{2}$ is connected across O and the frame. The voltage across $\frac{r}{2}$ resistor is:

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a

$\frac{{ωB}_{0} L^{2}}{4}$

b

$\frac{{ωB}_{0} L^{2}}{2}$

c

$\frac{{ωB}_{0} L^{2}}{12}$

d

$\frac{{ωB}_{0} L^{2}}{8}$

answer is B.

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Detailed Solution

$E = \frac{{ωB}_{0} {(\frac{L}{2})}^{2}}{2} = \frac{{ωB}_{0} L^{2}}{8} \Rightarrow$ Both segments act as source of
$emf = \frac{B_{0} {ωL}^{2}}{8}$ and internal resistance $\frac{r}{2}$
Question Image
$V = \frac{E}{(\frac{r}{2} + \frac{r}{4})} \times (\frac{r}{2}) = \frac{E}{(\frac{3}{4})} \times \frac{1}{2} = \frac{2 E}{3} = \frac{2}{3} \frac{B_{0} {ωL}^{2}}{8} = \frac{B_{0} {ωL}^{2}}{12}$

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