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Q. In an ac dynamo, the peak value of emf is $60\, V$. The induced emf (in $V$) in the position when the armature makes an angle of $30^{\circ}$ with the magnetic field perpendicular to the coil, will be

Electromagnetic Induction

Solution:

$\varepsilon=\varepsilon_{0} \sin \omega t=\varepsilon_{0} \sin \theta=60 \sin 30^{\circ}=30\, V$