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# In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf. generated in the coil is:A. NABRωB. NABC. NABRD. NABω

Last updated date: 11th Aug 2024
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Hint: In the question, magnetic field, frequency, area of the coil is given so we will first find the flux through the coil and then by using Faraday’s law we will find the instantaneous emf induced in a circuit.

Complete step by step answer:Numbers of turns is N; Magnetic field is B, Area of the coil is A and, Frequency is f.
We know the magnetic flux through a coil rotating in a magnetic field is given as
$\phi = NBA\cos \left( {\omega t} \right) - - (i)$
As we know according to Faraday’s law the instantaneous emf induced in a circuit is directly proportional to the time rate of change of magnetic flux through the circuit, is given as
$\varepsilon = - \dfrac{{d\phi }}{{dt}} - - (ii)$
Now by substituting the value of (i) in equation (ii), we can write
$\varepsilon = - N\dfrac{{d\left( {BA\cos \left( {\omega t} \right)} \right)}}{{dt}}$
Hence by solving we get
$\varepsilon = - NBA\dfrac{{d\left( {\cos \left( {\omega t} \right)} \right)}}{{dt}} \\ = NBA\omega \sin \left( {\omega t} \right) \\$[$\dfrac{{d}}{{dx}}\left( {\cos x} \right) = - \sin x$]
Now we know that the generated emf in the coil will be maximum when the coil is perpendicular to the magnetic line of forces, i.e. $\omega t = {90^ \circ }$
Therefore, $\sin \left( {\omega t} \right) = \sin {90^ \circ } = 1$
Hence by substituting $\sin \left( {\omega t} \right) = 1$, we get maximum generated emf as
$\varepsilon = NBA\omega \sin \left( {\omega t} \right) \\ = NBA\omega \\$
Option D is the correct answer.

Note:EMF generated is maximum when the angle between the magnetic field and the coil is maximum. Here, since we are asked to find the maximum emf value so, we will put $\omega t = {90^ \circ }$since maximum flux is cut by the coil when it is perpendicular to the field.