Home » Uncategorized » GATE 2015 EE - SET 2 - Question 10

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

Translate »