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(B) The magnetic flux $\phi$ through a coil rotating in a magnetic field is given by $\phi = NAB \cos(\omega t)$,where $	heta = \omega t$ is the angl

Vedclass · Accepted Answer

$\varphi=0, \varepsilon=NAB \omega$

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(B) The magnetic flux $\phi$ through a coil rotating in a magnetic field is given by $\phi = NAB \cos(\omega t)$,where $	heta = \omega t$ is the angle between the area vector and the magnetic field vector $\vec{B}$.The induced electromotive force (emf) $\varepsilon$ is given by Faraday's law of induction: $\varepsilon = -\frac{d\phi}{dt}$.Calculating the derivative: $\varepsilon = -\frac{d}{dt}(NAB \cos(\omega t)) = NAB \omega \sin(\omega t)$.When the magnetic field $\vec{B}$ is parallel to the plane of the coil,the angle between the area vector (which is perpendicular to the plane) and the magnetic field $\vec{B}$ is $90^\circ$ or $\frac{\pi}{2}$ radians.At this instant,$	heta = \omega t = \frac{\pi}{2}$.Substituting this into the expressions:$\phi = NAB \cos(\frac{\pi}{2}) = NAB(0) = 0$.$\varepsilon = NAB \omega \sin(\frac{\pi}{2}) = NAB \omega(1) = NAB \omega$.Therefore,the magnetic flux is $0$ and the induced emf is $NAB \omega$.

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