Physics M.C.Q (1 Marks)

$\text{Reading of voltmeter = emf induced}=\text{B1v}=0.2\times10^{-4}\times1\times180\times\frac{1000}{3600}=\text{1mV}$

It is because after every $\frac{1}{2}$ revolution the current becomes zero and mode of change in flux changes thereafter $($If before the current becomes zero, the mode of flux change was from left to right then after the current becomes zero the mode of flux change becomes right to left$).$

emf $=$ Blv
$=0.2 \times 10^{-4} \times 1 \times 20$
$=0.4 \ mV$

Total resistance of the circuit $=4000+400=4400$ ohms
Current flowing $i =\frac{ v }{ R }=\frac{440}{4400}=0.1 . A$
Voltage across load $V _{ L }= Ri =4000 \times 0.1=400 V$

The property of induction of $e.m.f.$ in the same coil when there is a change in current in it is called selfinduction.

As shown in the image $\text{L}\propto\text{N}^2$
and transformer has coils which cause impedance and impedance of a inductor is $2\pi\text{fL}$
so impedance of each coil $\propto\text{N}^2$
so impedance of secondary coil $=\text{Z}_\text{P}\Big(\frac{\text{N}_S}{\text{N}_P}\Big)^2$
$=250\times2^2=1000\Omega$

Let at time $t = 0,$ the coil is vertical and at time $t$, plane of coil makes an
angle $\theta$ with the vertical, then
$\theta=\omega\text{t},(\omega=\text{uniform angular velocity}),$
in this position, magnetic flux linked with coil will be,
$\phi=\text{NBA}\cos\theta, ($where, $A =$ area of coil,$)$
$\phi=\text{NBA}\cos\omega\text{t},$
now, differentiating this equation $w.r.t.$ time, we get
$\frac{\text{d}\phi}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{NBA}\cos\omega\text{t}),$
$\frac{\text{d}\phi}{\text{dt}}=-\text{NBA}\omega\sin\omega\text{t},$
if $e$ is the emf induced in coil then by Faraday's law,
$\text{e}=-\frac{\text{d}\phi}{\text{dt}}=\text{NBA}\omega\sin\omega\text{t},$
Now if, $\sin\omega\text{t}=1 ($maximum$),$
then $\text{e}_\text{max}=\text{NBA}\omega,$
If $\omega=\frac{1\text{rad}}{\text{s,}}$
then $\text{e}_\text{max}=\text{NBA}$