Physics M.C.Q (1 Marks)

Key concept: In this problem, the Lenz’s law is applicable so let us introduce Lenz’s law first.
Lenz's law gives the direction of induced emf$/$induced current. According to this law, the direction of induced emf or current in a circuit is such as to oppose the cause that produces it. This law is based upon law of conservation of energy.
When the current in coil $B ($at $t = 0)$ is counter-clockwise and the coil $A$ is considered above it. The counter clockwise flow of the current in coil $B$ is equivalent to north pole of magnet and magnetic field lines are eliminating upward to coil $A$. When coil A starts rotating at $t = 0$, the current in $A$ is constant along clockwise direction by Lenz's rule.

$\text{L}=\frac{\mu_0\text{N}^2\pi\text{r}^2}{1}$
Length of wire $=\text{N}2\pi\text{r}=$ Constant$( = C,$ suppose$)$
$\therefore\text{L}=\mu_0\Big(\frac{\text{C}}{2\pi\text{r}}\Big)^2\frac{\pi\text{r}^2}{1}$
$\therefore\text{L}\propto\frac{1}{1}$
$\therefore\text{Self-inductance will become 2L.}$

emf $=$ rate of change of flux $\times$ number of turns $=\frac{12\times10^3\times10^{-8}}{0.2}\times100=0.06\text{V}$
emf $=$ rate of change of flux $\times$ number of turns.

Flux through a circular coil $\phi=\text{NBA}\cos\omega\text{t}$
Voltage required $\in=\frac{-\text{d}\phi}{\text{dt}}$
$\Rightarrow9=\text{NBA}\omega\sin\omega\text{t}$
$9=\frac{8\times10^{-5}\times\text{A}\times30\times2\pi\times2000}{60}$
$\text{A}=\frac{9\times10^5}{50\times10^3}=18\text{m}^2$

Weber $= ML ^2 T^{-2} I ^{-1}$
$= ML ^2 T^{-2} Q ^{-1} T= ML ^2 T^{-1} Q ^{-1}\left(I= QT ^{-1}\right)$
Henry $H$ is $SI$ unit of inductance.
$H = ML ^2 T^{-2} I ^{-2}$
also $I = QT ^{-1}$
So $H = ML ^2 T^{-2} Q ^{-2} T^2= ML ^2 Q ^{-2}$

The situation given in question is shown in the figure given below,

Clearly, loop $\text{ABCDA}$ lies in $x-y$ plane whose area vector $\vec{\text{A}_1}=\text{L}^2\hat{\text{K}}$ whereas lppo $\text{ADEFA}$ lies in $y-z$ plane whose area vector $\vec{\text{A}_2}=\text{L}^2\hat{\text{i}}$.
As, $\phi=\text{B.A}$
Now, $\text{B}=\text{B}_0(\hat{\text{i}}+\hat{\text{K}})\text{T}$ and $\text{A}=\text{A}_1+\text{A}_2=(\text{L}^2\hat{\text{K}}+\text{L}^2\hat{\text{i}})$
Therefore, $\phi=\text{B.A}=\big[\text{B}_0(\hat{\text{i}}+\hat{\text{K}})\big].\big[\text{L}^2\hat{\text{K}}+\text{L}^2\hat{\text{i}}\big]=2\text{B}_0\text{L}^2\text{Wb}$.

The magnetic flux is given by $\phi=\text{BA}\cos\theta$ where $\theta$ is angle between magnetic field and area of ring.
in this case the area vector is out of the plane,
Hence $\theta=90^\circ$
$\Rightarrow\phi=0$
The ring will not move from the magnet, as there is no change in flux through ring and current in it will be zero. Ring will not behave as a magnet.

Formula for self$-$inductance of a coil: $\text{L}=\frac{\mu_0\text{N}^2\text{A}}{1}$
Hence, it varies as $N^2$

An $AC$ generator converts mechanical energy into electrical energy, while a motor does the opposite $-$ it converts electrical energy into mechanical energy. Both devices work because of electromagnetic induction, which is when a voltage is induced by a changing magnetic field.

emf $=$ rate of change of flux
$=$ area $\times$ number of turns $\times$ rate of change of flux density
$=10 \times 100 \times 1000=106 V$

The angle between magnetic field and area vector is $90^{\circ}$, so the flux associated with coil is zero. Although magnetic field is changing but flux is remaining constant equal to zero, so emf induced and hence current in the loop is equal to zero.

As we know that flux $=$ inductance $\times$ current
Hence,
Flux $\phi= Li$
$=0.05 \times 0.02$
$=10^{-3} Wb$

$\text{emf}=\text{L}\frac{\text{di}}{\text{dt}}$
$\text{emf}=\text{L}\frac{\triangle\text{i}}{\triangle\text{i}}$
So, $\text{L}=\text{emf}\times\frac{\triangle\text{t}}{\triangle\text{i}}$
$=32\times\frac{0.1}{8}$
$=0.4\text{H}$

Here, induced emf in $B$ is due to the variation of magnetic flux associated with it.
If the current in $A$ would be variable, there must be an induced emf $($current$)$ in $B$ even if the $A$ stops moving.
If the current in $A$ is constant and $A$​​​​​​​ stops moving the current in $B$ becomes zero.

Relative velocity $= v + v = 2v$
Therefore, emf $= Bl(2v)$

Potential difference appears across the two ends $=\text{Bvl}$
$\text{v}^\hat{}\text{B}, \ \text{v}^\hat{}\text{I}, \ \text{I}^\hat{}\text{B}$

$\text{emf}=\frac{\text{BAn}}{\text{t}}$
$\text{i}=\frac{\text{emf}}{\text{R}}=\frac{\text{BAn}}{\text{tR}}$
$\text{q}=\text{it}=\frac{\text{emf}}{\text{tR}}=\frac{\text{BAn}}{\text{R}}$
$\therefore2\times10^{-6}=\frac{\text{B}\times10^{-4}\times20}{10}$
$B=10^{-2} T$
$B =0.01 T$

$AS$ the coil rotates by $\pi$ radians, $\triangle\phi=\text{NAB}-(-\text{NAB})=2\text{NAB}$
emf $=$ rate of change of flux
$=\frac{\triangle\phi}{\triangle\text{t}}=\frac{2\text{NAB}}{0.5}$
$=4\text{NAB}$

Let the inductance due to first loop alone be $L.$
Then inductance due to second loop alone is also $L.$
The effective self inductance can be calculate by: $L_{\text {eff }}=L_1+L_2-2 M$
$=\text{L}+\text{L}-2\sqrt{\text{L.L}}=0$

Force on positive charges in the rod will be in direction given by $qv \times B$, i.e., towards $A.$
Hence force on electron will be in opposite direction i.e., towards $B$. So negative charge will move towards $ B.$
Therefore $B$ will be negatively charged and $A$ will be positively charged.

$\text{emf(t)}=\frac{\text{d}\phi}{\text{dt}}=16\text{t}+3$
$\text{emf(4)}=16\times4+3=67\text{V}$

Initally the time constant can be representyed as:
$\frac{\text{L}}{\text{R}}=2\times10^{-3}...........(\text{i})$
When the resistance is increased, the new time constant is given as:
$\frac{\text{L}}{\text{R}+90}=0.5\times10^{-5}...........(\text{ii})$
From $(i)$ and $(ii) $, on solving , we get
$\text{L}=60\text{mH }\text{and } \text{R}=30\Omega$

The induced emf, $\text{E}=-\frac{\text{d}\phi}{\text{dt}}\text{where }\phi=\text{magnetic flux}$
Hence, emf will last as long as flux keeps changing.

When a magnet is moved closer to the current carrying coil it will generate electricity as the coil moves through the magnetic field. As the magnet is moved, there will be an induced electro-motive force $\text{(EMF)}$ which can cause a current in the coil. Once the magnet stops moving, the current will go to zero.
Hence, when a galvanometer is connected to the circuit, there will be deflection due to the flow of electricity. As the magnet is moved toward the coil of wire, the needle of the galvanometer moves one direction. As the magnet is moved away from the coil of wire, the needle of the galvanometer moves the opposite direction. If the magnet is moved faster, the magnitude of the deflection increases.

Use the formula of self$-$inductance of a solenoid.
Explanation for correct option:
Step $1:$ Find number of turns in the solenoid and area of cross section of solenoid
Let $x$ is the length of the wire, then the length of the solenoid is $\text{x}=2\pi\text{rN}$
$($Here $N ⇒$ number of turns$)$
$\Rightarrow\text{N}=\frac{\text{x}}{2\pi\text{r}}$
Since, the wire is in the shape of a cylinder, the area of cross section is
$\text{A}=\pi\text{r}^2$
Step $2:$ Find length of a wire required
The self$-$inductance of a solenoid is given by,
$\text{L}=\frac{\mu_0\text{N}^2\text{A}}{1}$ where $\mu_0$​ is the magnetic permeability, $l$ is the length of the solenoid and $A$ is the area of each turn in the solenoid.
$\Rightarrow\text{L}=\frac{\mu_0\big(\frac{\text{x}}{2\pi\text{r}}\big)^2\pi\text{r}^2}{\text{I}}$
$\Rightarrow\text{x}=\sqrt\frac{4\pi\text{LI}}{\mu_0}$

Energy in indicator $=\frac{1}{2}\text{Li}^2$
With Energy transfer, i becomes $\text{i}\sqrt{2}$
Now since $\phi=\frac{\text{Li}}{\sqrt{2}}=\frac{\phi}{\sqrt{2}}$
So, total flux will be $\frac{\phi}{\sqrt{2}}$

As per Faraday's law, the change in magnetic flux associated with coil produces an $\text{EMF}$ in the coil. Whenever there is a change in magnetic flux associated with a coil, an $\text{EMF}$ is induced in the coil. Magnetic flux is defined as the number of field lines passing normally through a given area. When magnetic flux changes, obviously, it means that, the number of magnetic field lines change.

$AC$ generators are made up of one or more magnetic pole pairs corresponding to their purposes.
The frequency of an $AC$ generator is dependent on the number of pairs of poles and the speed of the coil rotation. The frequency of the induced $AC$ current depends on therevolution
of the coil since, rate of revolution is Revolutions Per Minute $\text{(RPM)}$ and frequency is expressed as cycles per second, the frequency of the induced $AC$ current will be the rate of
revolution of the coil divided by 60 seconds.
That is, $F = RPM \times $ number of pole pairs $\div 60$
When the Revolutions per minute $\text{(RPM)}$ is doubled, then, obviously the frequency of the ac current will also be doubled as the $\text{RPM}$ is multiplied with the number of pole pairs.

On removal of load from the circuit, the circuit suddenly becomes an open circuit.
Thus $\frac{\text{di}}{\text{dt}}\rightarrow\infty$
For sparking, high voltage must appear across the open ends. This will happen only in case of an inductor as the voltage drop across the
inductor is $\text{L}\frac{\text{di}}{\text{dt}}$
Therefore, the circuit has high inductance.

$\text{EMF}$ is defined as the voltage developed by the electrical energy source. For example : Battery Thus $\text{EMF}$ has the unit of volt denoted by $V$