Physics M.C.Q (1 Marks)

$\text{LPT}=$ leading pole tip
$\text{TPT}=$ tailling pole top
Considering armature reaction.
where $\text{GNA}=$Geometrical axis
$\text{MNA}=$ Magnetic neutral axis
here clearly see that $\text{TPT}$ side flux increases due to armature in case of generator.

The moving rod develops an electric field $= Bvl$
and the $\text{EMF}$ induced is such that it opposes the change of magnetic field.
Due to shifting of electrons, one end of the rod becomes positive and the other end negative. This developes a electric field in the rod.

Henry $($symbol $H)$ is the $SI$ derived unit of self$-$inductance.

$\text{L}=\frac{\text{F}}{\text{I}},\text{f}=\text{NAB}$
$\text{B}=\text{M}_0\text{nI}$
Where $\text{n}=\frac{\text{N}}{2\pi\text{R}}$
$\therefore\text{F}=\text{N}\pi\text{r}^2=\bigg(\text{M}_0\frac{\text{N}}{2\pi\text{R}}\text{I}\bigg)$
$\text{F}=\frac{\text{M}_0\text{N}^2{\text{r}}^2\text{I}}{2\text{R}}$
$\text{L}=\frac{\text{F}}{\text{I}},=\frac{\text{M}_0\text{N}^2{\text{r}}^2}{2\text{R}}$

As per Faradays law, $\text{EMF}$ is always directly proportional to the rate of change of magnetic flux. Therefore, electromagnetic induction takes place whenever the magnetic flux passing through the object changes.

$\text{emf}=-\text{M}\frac{\text{dI}}{\text{dt}}$
$= − 1.25 \times-80 = 100\text{V}$

Induced emf is $AB$ is $Bvl$ and Induced emf is $DC$ is also Bvl.
Net emf in the closed circuit $($loop$)$ is zero.
So induced current in the loop is zero.

An anticlockwise current$-$pulse generated and then a clock$-$wise current pulse.

From Faradays law the induced emf ise $=\text{L}\frac{\text{di}}{\text{dt}}$
Given, $di = 10A, dt = 1s, e = 10$
$\therefore10=\text{L}\frac{10}{1}$
$\Rightarrow\text{L}=1\text{H}$

The magnetic field above the wire is out of the plane. This flux is decreasing and should be compensated by the current in the loop $F$ and so the current in loop $F$ will be anti$-$clockwise. For loop $G$ the situation is opposite

Current will be zero at $t = 0+$
if the current is zero magnetic field will be zero at $t = 0+$
power delivered will be zero at $t = 0+$
The $\text{EMF}$ induced will be equal to the applied voltage in the inductor to oppose the current.

Self$-$inductance of a coil is,
$\text{L}=\frac{\mu_0\mu_\text{r}\text{N}^2\text{A}}{1}$
Current flow through the circuit,
$\text{I}=\frac{\text{E}}{(\text{W}^2\text{L}^2+\text{R}^2)^\frac{1}{2}}$
Current in the circuit decrease with increasing self-inductance,
$\text{L}\alpha\mu_\text{r}.$
When an iron rod is inserted in the coil, self$-$inductance increases, and the current decreases.
The brightness of the bulb decreases.

$\text{e}=-\text{L}\frac{\text{di}}{\text{dt}}$
Current flow in the CKt is clock wise direction, due to Mutual Induction current flow in the loop anti clockwise direction. The net force applied on the loop in east direction. So we can say that the ring will move away from the solenoid.

Induced emf in the axle $= Blv$
$v -$ velocity of car
$l -$ length of car
$B -$ component of magnetic field perpendicular to both $l$ and $v.$
That is $B$ is the vertical component of magnetic field.
Vertical component of magnetic field is maximum at the poles.
Therefore emf induced in the axle will be maximum at the poles.

Take a small element dx at a distance of $'x'$ centre

$\text{Þ}\text{d}\in\int_{0}^{\frac{1}{2}}\text{B}\omega\text{x}\text{dx}=\frac{\text{B}\omega\text{x}^2}{2}\Big|_{0}^{\frac{1}{2}}$
$\in=\frac{1}{8}\omega\text{Bl}^2$

If we apply a high$-$frequency supply of the same peak voltage to the coil, the current still is being delayed by $90o$ but the time it requires to reach its maximum value has been reduced due to the increase in frequency. Because the frequency is inversely proportional to time $(T).$ Hence, the rate of change of the flux within the coil has also increased due to the increase in frequency. Hence, the induced $\text{EMF}$ is maximum in case of $1$ amp $100 \ Hz$ supply source in the coil.

Upon connecting to the $AC$ source, a current starts to flow in the coil $ P$. This current induces a current in the galvanometer coil $Q$ due to mutual inductance because of the flux linkages. This induced current causes a deflection in the galvanometer.

Induced emf is given by:
$\varepsilon= Bvl$
On putting the values we get
$=5 \times 10^{-5} \times 1.50 \times 2$
$=0.15 \ mV$

$\text{emf}=-\frac{\text{d}\phi}{\text{dt}}=-12\text{t}+5$
$\text{Current}=-\frac{\text{emf}}{10}=-1.2\text{t}+0.5$
$\text{Current at t}=0.25\text{s},$
$=1.2\times0.25+0.5=0.2\text{A}$