A rectangular coil of a single turn, having area A, rotates in a uniform magnetic field B with an angular velocity ω about an axis perpendicular to the field. If initially, the plane of the coil is perpendicular to the field, then the average induced emf when it has rotated through \[{{90}^{0}}\] is:
(A) \[\dfrac{\omega BA}{\pi }\]
(B) \[\dfrac{\omega BA}{2\pi }\]
(C) \[\dfrac{\omega BA}{4\pi }\]
(D) \[\dfrac{2\omega BA}{\pi }\]

Hint: In this problem, we have to use Faraday’s law of electromagnetic induction. It says a changing magnetic field induces potential difference inside a conductor placed in it. Faraday’s law states that a current will be induced in a conductor which is exposed to a changing magnetic field.


Complete step by step answer: Given the coil of single turn so, n=1
Area of the coil=A
It is rotating in an external magnetic field, B with an angular velocity ω. Given initially the plane of the coil is perpendicular to the field. It means the angle between the area vector and the magnetic field is \[{{0}^{0}}\]
Magnetic flux is given as \[{{\phi }_{1}}=\overrightarrow{B}.\overrightarrow{A}=BA\cos \alpha =BA\cos {{0}^{0}}=BA\]
Now the coil is rotated by an angle \[{{90}^{0}}\]. So final flux is given as,
\[{{\phi }_{2}}=\overrightarrow{B}.\overrightarrow{A}=BA\cos \alpha =BA\cos {{90}^{0}}=0\]
Now, using faraday’s law we get,
\[\begin{align}
  & e=\dfrac{\Delta \varphi }{\Delta t} \\
 & =\dfrac{{{\varphi }_{2}}-{{\phi }_{1}}}{\Delta t} \\
\end{align}\]
Now since it is rotated by one fourth, the time must be one fourth of time period.
\[\begin{align}
  & =\dfrac{BA}{\dfrac{T}{4}} \\
 & =\dfrac{4BA}{T} \\
\end{align}\]
Now using the relation, \[T=\dfrac{2\pi }{\omega }\]
We get, \[e=\dfrac{4BA}{T}=\dfrac{4BA\times \omega }{2\pi }=\dfrac{2BA\omega }{\pi }\]
Thus, the correct option is (D)


Note: while solving such problems we have to keep in mind that while finding out the angle between the field and the plane we have to take the area vector and area vector is always perpendicular to the plane. Also, in application of Faraday’s law, emf will be produced only and only if the flux changes and if there is no change of flux then there will no induced emf.