TS EAMCET 03 May 2019 Shift 2 Solved Paper

The following shows a semicircle of length L, moving with velocity v in a uniform magnetic field. The expression for the induced emf in a conductor of length l is given by, The expression for the induced emf in a conductor of length $l$ is given by $e \;\; = B(v \times I)$ $\;\; = B v l \sin \theta \;\;\;\;\; \cdots \cdots \cdots \cdots (\mathrm{I})$ The effective length of arc for induced emf is given by, $l_{\text{eff}} = 2 R$ $= \; \frac{2L}{\pi} \;\;\;\;\; \cdots \cdots \cdots \cdots (\mathrm{II})$ From equations (1) and (2) $e = B v \; \frac{2L}{\pi} \sin 45^{\circ} ( \because \theta = 45^{\circ} )$ $e = Bv L \; \frac{\sqrt{2}}{\pi} \;\;\;\;\; \cdots \cdots \cdots \cdots (\mathrm{III})$ The induced emf is given by, $\phi = \alpha B v L \;\;\;\;\; \cdots \cdots \cdots \cdots (\mathrm{IV})$ Therefore, from equations (3) and (4) $\alpha = \; \frac{\sqrt{2}}{\pi}$