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(B) Given: Length of the rod,$l = 1.0 \; m$. Angular frequency,$\omega = 400 \; rad \; s^{-1}$. Magnetic field strength,$B = 0.5 \; T$. When a rod of

Vedclass · Accepted Answer

$100$

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(B) Given: Length of the rod,$l = 1.0 \; m$. Angular frequency,$\omega = 400 \; rad \; s^{-1}$. Magnetic field strength,$B = 0.5 \; T$. When a rod of length $l$ rotates about one of its ends in a uniform magnetic field $B$ with angular velocity $\omega$,the induced emf $(e)$ across its ends is given by the formula: $e = \int_{0}^{l} B v \; dr = \int_{0}^{l} B (\omega r) \; dr$ $e = B \omega \left[ \frac{r^2}{2} ight]_{0}^{l} = \frac{1}{2} B \omega l^2$ Substituting the given values: $e = \frac{1}{2} 	imes 0.5 	imes 400 	imes (1.0)^2$ $e = 0.25 	imes 400 = 100 \; V$. Thus,the emf developed between the centre and the ring is $100 \; V$.

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