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# A closed coil having 100 turns is rotated in a uniform magnetic field B=0.0004 T about a diameter which is perpendicular to the field.The angular velocity of rotation=300 revolutions/minute.The area of the coil is 0.025m^2 and its resistance=4.0.Find the average emf developed in a half a turn from the position where the coil is perpendicular to the magnetic field

Arun Kumar IIT Delhi
6 years ago
$\phi =B.A=B.Acos(\theta)=BAcos(\omega t) \\=>E=-\d(\phi)/dt=-B.\omega.Acos(\omega t)$

$\omega=5*2\pi r$
$E_{avg}=\int_{0}^{\pi}Ed\theta/\pi$
Notice its theta in place of t.

Thanks & Regards
Arun Kumar
IIT Delhi