Solution:

Ans. (i) Labelled diagram of $a c$ generator $11 / 2$ Expression for instantaneous value of induced emf.

(ii) Calculation of maximum value of current

(i) Labelled diagram :

Try yourself, Similar to Q. 3(a), Long Answer Type Questions

(ii) Maximum value of emf

$$ \begin{aligned} e_{0}= & \text { NBA } \omega \ = & 20 \times 200 \times 10^{-4} \ & \times 3 \times 10^{-2} \times 50 \mathrm{~V} / \frac{1}{2} \ = & 600 \mathrm{mV} \end{aligned} $$

Maximum induced current,

[Note 1 : If the student calculates the value of the maximum induced emf and says that “since $R$ is not given, the value of maximum induced current cannot be calculated”, the $1 / 2$ mark, for the last part, of the question, can be given.]

[Note 2 : The direction of magnetic field has not been given. If the student takes this direction along the axis of rotation and hence obtains the value of induced emf and, therefore, maximum current, as zero, award full marks for this part.]

[CBSE Marking Scheme 2017]

Detailed Answer :

(i) Expression :

If at any moment, $t$ be the perpendicular vector to the plane of coil that makes an angle $\theta$ with direction of magnetic field $(B)$, so flux passing through the coil will be :

$$ \begin{aligned} & \phi=n B A \cos \theta \ & \theta=\omega t \end{aligned} $$

$[\omega=$ angular velocity of the coil] Now,

$$ \begin{equation*} \phi=n A B \cos \omega t \tag{1} \end{equation*} $$

If $e$ is instantaneous induced emf produced in the coil, then

$$ \begin{align*} e & =-\frac{d \phi}{d t} \ & =-\frac{d(n A B \cos \omega t)}{d t} \ & =-n A B \omega(-\sin \omega t) \end{align*} $$

Maximum value or peak value of instantaneous induced emf $e$ which is attained when $\sin \omega \mathrm{t}= \pm 1$

$\therefore \quad e_{\max }=n \omega A B$

So, $\quad e=e_{\text {max }} \sin \omega t$

As value of sine function varies from +1 to -1 , polarity of emf changes with time. Also the output voltage is sinusoidalin nature.

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