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# GATE 2015 EE - SET 2 - Question 10

Solution :

Given , r = 1m , N = 600rpm , $\mu _{0}$ = 4$\pi * 10^{-7}$ ,$\vec{H}$ =$10^{7}\hat {Z}$A/m

Area of  the ring perpendicular to the magnetic field  = $(\lambda r^{2}sin\omega t)$

Magnetic field density $\Rightarrow$ B = $\mu _{0}$H , B = 4 * $\pi * 10^{-7}* 10^{7}$= 4$\pi$

Flux $\phi$= B.A = 4$\pi \left [ \pi r^{2}sin\omega t \right ]$ = 4$\pi r^{2}Sin\omega t$

$V_{turn}$ = $\frac{-d\phi }{dt}$=$\frac{-d}{dt}(4\pi r^{2}Sin\omega t )$ = - 4$\pi^{2} r^{2}(cos\omega t)\omega$

Peak value = $V_{turn}$ = 4$\pi ^{2}r^{2}\omega$  = 4$\pi ^{2}r^{2}\left [ \frac{2\piN }{60} \right ]$

$V_{turn}$ = 4$\pi ^{2}(1)\left [ \frac{2\pi(600) }{60} \right ]$= 247.67 V

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