A conducting wire bent in the shape of a semi | vedclass

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(D) The induced emf in a moving conductor is given by $e = B l_{eff} v \sin(	heta)$,where $l_{eff}$ is the displacement vector between the two ends o

Vedclass · Accepted Answer

$\frac{\sqrt{2}}{\pi}$

Vedclass · Answer

(D) The induced emf in a moving conductor is given by $e = B l_{eff} v \sin(	heta)$,where $l_{eff}$ is the displacement vector between the two ends of the conductor.For a semicircular wire of length $L$,the radius $r$ is given by $L = \pi r$,so $r = \frac{L}{\pi}$.The effective length $l_{eff}$ (the straight-line distance between the two ends) is the diameter of the semicircle,$l_{eff} = 2r = \frac{2L}{\pi}$.The velocity $v$ makes an angle of $45^{\circ}$ with the diameter (the effective length vector).Substituting these values into the emf formula:$e = B 	imes (\frac{2L}{\pi}) 	imes v 	imes \sin(45^{\circ})$$e = B 	imes \frac{2L}{\pi} 	imes v 	imes \frac{1}{\sqrt{2}}$$e = \frac{\sqrt{2}}{\pi} B v L$Comparing this with the given expression $\Phi = \alpha(B v L)$,we find $\alpha = \frac{\sqrt{2}}{\pi}$.

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