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