Physics M.C.Q (1 Marks)

$\text{Max.emf = nAB}\omega$
$=50\times0.008\times0.05\times2000\frac{2\pi}{60}=\frac{4\pi}{3}\text{V}$

Self-inductance of a solenoid is given by $\text{L}=\frac{\mu_0\text{N}^2\text{A}}{1}$
where, $N =$ number of turns in solenoid
$l =$ length of the solenoid
$A =$ area of one turn of solenoid.
The self-inductance of solenoid does not depend on current passing through it.

After performing a large number of experiments, Faraday and Henry made the following observations about electromagnetic induction

1. a current is induced in a coil when it is moved $($rotated$)$ relative to a fixed magnet.
2. a current is also induced in a fixed coil when a magnet is moved $($or rotated$)$ relative to the fixed coil.
3. no current is induced in a coil when the coil and magnet both are stationary relative to one another.
4. when the direction of motion of coil $($or magnet$)$ is reversed , the direction of current induced in the coil also get reversed.
5. the magnitude of current induced in the coil can be increased.

1. by winding the coil on a soft iron core.
2. by increasing the number of turns in the coil.
3. by increasing the strength of magnet.
4. by increasing the speed of rotation of coil $($magnet$).$

$\oint\overrightarrow{\text{B}}\cdot=\mu_0\text{I}$
$\Rightarrow\text{B}.2\text{b}=\mu_0\text{I}$
$\Rightarrow\text{B}=\frac{\mu_0\text{I}}{2\text{b}}\rightarrow$ due to one tape
Net field $= 2B$
Magnetic flux passing through this double tape
$\phi=\text{2BA}=2\text{B(Ih)}$
$\Rightarrow\phi=\frac{\mu_0\text{I}}{\text{b}}\text{Ih}$
$\text{L}=\frac{\phi}{\text{I}}=\frac{\mu_0\text{Ih}}{\text{b}}$
$\Rightarrow\frac{\text{L}}{\text{I}}=\frac{\mu_0\text{h}}{\text{b}}$

emf between the wing tips $= Blv$
$=0.02\times50\times360\times\frac{1000}{3600}$
$=100\text{V}$

$I=t^2 e^{-t}$
$\text{e}=\frac{\text{LdI}}{\text{dt}}$
$\frac{\text{dI}}{\text{dt}}=2\text{te}^{\text{-t}}\text{-t}^2\text{e}^{\text{-t}}=0$
$\text{t}=2\sec$

Fleming’s Right Hand Rule gives the direction of current in the conductor whenever it is moved in any magnetic field. It states that if we stretch our right hand fingers perpendicular to each other as shown in fig, then if fore finger is in the direction of Magnetic field and thumb is in the direction of motion of conductor then middle finger gives the direction of current in the conductor.
Explanation for the correct answer:
For induction of emf in the conductor, the motion of conductor should be perpendicular to the direction of magnetic field. Thus, Option $B$ and $C$ are eliminated.
Now if we move conductor in $P$ direction, the motion will be perpendicular to the direction of magnetic field but then the charge separation in the conductor will take place perpendicular to the length of the conductor. Thus, due to less space for charge separation emf can not be induced in the conductor. Thus, Option A is eliminated.
If conductor is moved in direction $M$ then the charge separation in the conductor takes place across its length, there will be enough space for charge separation. And thus emf will be induced across the conductor.

$\text{L}=\frac{\mu\text{n}^2\text{A}}{\text{L}}\cdot$
$\text{L}\propto\text{n}^2$
Thus, for $500$ turns, self inductance would be $\frac{108\times(500)^2}{(600)^2}=\frac{108\times25}{36}=75\text{mH}$

Let $M$ be the mutual inductance,
We can write $E = MI′$
We know that $\text{E}=-\frac{\text{d}\phi_\text{B}}{\text{dt}}$
$\phi_\text{B}=-\int\text{Edt}=-\text{M}\int\text{I dt}=-\text{MI}$
whereI is the current in the circuit.
Given $I = I0​.$
So $\phi_\text{B}=-\text{MI}_0=-\big(\frac{\text{E}}{\text{I}}\big)\text{I}_0$
Sign is indicative of the direction. So for magnitude only
$\phi_\text{B}=\Big(\frac{\text{E}}{\text{I}}\Big)\text{I}_0$

$\phi=\text{BA}\phi=\text{BA}$

The emf induced is independent of the number of spoke and only depends on the lenght of spoke and the angular speed.
Therefore emf induced is same as before $= E$

$\text{flux}(\phi)=\text{nBA}$
$=\text{nB}\pi\text{r}^2$
Let total length of coil is $l$
$\text{I}=(\text{n}\pi\text{r})$
Now, since same wire length has double radius $80,$
$\text{I}=\text{n}_1(2\pi2\text{r})$
$\text{I}=\text{n}_1=\frac{\text{n}}{2}$
Now, flux $=\frac{\text{n}}{2}\text{BA}$
$=\frac{\text{n}}{2}\text{B}\pi4\text{r}^2$
$=2\text{nB}\pi\text{r}^2$
$=2\phi$
So, new flux $=2\phi$

Force on charge q due to the motion of rod in the field $= F = qVB$
This force on the charge is attributed to the induced electric field $E$
Therefore,
$Eq = F = qVB$
$E = VB$
therefore, an electric field $E = VB$ is said to be induced in the rod due to the motion in the magnetic field.
Potential difference due to the field $=$ emf induced $= E \times $ length $= El = V$ Blvolts

Displacement current is $\text{i}_\text{D}=\in\frac{\text{d}\phi_\text{E}}{\text{dt}}$
or $\in=\frac{\text{i}_\text{D}}{(\frac{\text{d}\phi_\text{E}}{\text{dt}})}$
$\in=\frac{12.9\times10^{-9}}{4(8.76)\times10^3\times(26.1\times10^{-3})^3}$
$\simeq2\times10^{-8}$

Magnetic field due to wire at the ring is
$\text{B}=\frac{\mu0\text{i}}{4\pi\text{R}}$
$\therefore\text{then}\phi$
$=\phi\overrightarrow{\text{B}\cdot}\overrightarrow{\text{A}}$
$\Rightarrow\mid\text{B}\mid\mid\text{A}\mid\cos\theta$
$\phi=\text{O},\text{as}\cos90^\circ=0$
$\therefore\text{mutual induction = 0}$

Key concept: The phenomenon of electromagnetic induction is used in this problem. Whenever the number of magnetic lines of force $($magnetic flux$)$ passing through a circuit changes $($or a moving conductor cuts the magnetic flux$)$ an emf is produced in the circuit $($or emf induces across the ends of the conductor$)$ is called induced emf. The induced emf persists only as long as there is a change or cutting of flux.
When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A.
Hence the ammeter shows no deflection.

When the switch is closed than a clock wise current pulse generated $($Because initially current flow the terminal to negative terminal$).$
Due to Mutual Induction, current is generated in the loop. If circuit is open after some time. Dut to loop an anticlock wise current pulse generated in the circuit.

$Þ$ An emf con be induced by moving a condcutor in a magnetic field.
$\hat{\text{I}}=\text{Bvl}$
$Þ$ An emf can be induced by charging the magnetic field.
$\in=\frac{-\text{d}\phi}{\text{dt}} \phi\rightarrow\text{flux}$

Force on $a$ charge $q$ in the rod $= qvB$
electric field inside the rod due to displacements of charges due to force by relative motion in magnetic field $=\frac{\text{emf}}{1}$
As the rod moves in constant velocity, net force of constituent charge $q$ in the rod $= 0$
Therefore,
$\text{q}\frac{\text{emf}}{1}=\text{qvB}$
$\Rightarrow\text{emf}=\text{BIv}$

The deflection in the galvanometer $(G)$ occurs, when the magnet is pushed into the coil. This is because the relative motion between the two induces an emf into the coil.

$\in=\text{L}\times(\text{rate of change of current})$
$=\frac{(5\times10^{-3})}{(\frac{1}{0.1})}$
$=0.05\text{V}$

$\text{M}=\mu\text{N}_1\text{N}_2\frac{\text{Area}_{12}}{\text{length}_{12}}$
Area$_{12}​ -$ area in common to both the coils where the flux links both of them together.
length$_{12}​ -$ length in common to both the coils where the flux links both of them together.
Therefore to increase the mutual inductance, the number of turns can be increased.

$\text{N}\phi=\text{Li}$
$\text{L}=\frac{\text{N}\phi}{\text{i}}$
$=\frac{100\times1\times10^{-6}}{5}$
$=20\times10^{-6}\text{H}$
$=20\mu\text{H}$

Field inside solenoid $=\text{B}=\mu\text{Ni}$
$N -$ number of turns per length in solenoid $=\frac{800}{\text{m}}$
Flux linking the smaller coil $=\lambda=100\times\pi\text{r}^2\text{B}=100\times\pi\text{r}^2\mu\text{Ni}$
mutual inductance $=\text{M}=\frac{\lambda}{\text{i}}=100\times\pi\text{r}^2\mu\text{N}=3.15\text{mH}$

lf e is the induced $e.m.f.$ in the coil, then $\text{e}=\text{-L}\frac{\text{di}}{\text{dt}}$
$\therefore\text{L}=-\frac{\text{e}}{\frac{\text{di}}{\text{dt}}}$
Substituting values, we get $\text{L}=\frac{-8\times0.05}{-4}=0.1\text{H}$

$emf = BlV$
$=0.2\times10^{-4}\times1.5\times180\times\frac{1000}{3600}$
$=1.5\times10^{-3}\text{V}$.
$=1.5\text{mV}$

Using the formulas$-$
$\text{I}=\frac{\text{e}}{\text{T}}=\frac{\text{e}\omega}{2\pi}$
$B$ at center $=\frac{\mu_0}{2\text{R}}=\frac{\mu_0\text{e}\omega}{4\pi\text{R}}$
$\phi=\text{B}.\pi\text{r}^2=\frac{\mu_0}{4\pi\text{R}}\text{e}\omega\pi\text{r}^2$
$=\frac{\mu_0\text{e}\omega\text{r}^2}{4\text{R}}$
$\text{e}=-\frac{\text{d}\phi}{\text{dt}}=\frac{\mu\text{e}\text{r}^2\alpha}{4\text{R}}$

An electromagnetic field exists only when there is current. electric current flowing through a wire caused a nearby compass to deflect. $...$ In most conductors, the magnetic field exists only as long as the current is flowing $($i.e. an electrical charge is in motion$).$

Key concept: The self inductance $L$ of $a$ solenoid depends on various factor like geometry and magnetic permeability of the core material.
$\text{L}=\mu_\text{r}\mu_0\text{n}^2\text{Al}$
where, $\text{n}=\frac{\text{N}}{\text{l}} ($no. of turns per unit length$)$

1. No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2. Area of cross section: Larger the area of cross section of the solenoid, larger is its self inductance.
3. Permeability of the core material. The self inductance of a solenoid increases $μr$ times if it is wound over an iron core of relative permeability $μr$.

The long solenoid of cross$-$sectional area $A$ and length $l$, having $A$ turns, filled inside of the solenoid with a material of relative permeability $($e.g., soft iron, which has a high value of relative permeability$)$ then its self inductance is $\text{L}=\mu_\text{r}\mu_0\text{n}^2\frac{\text{A}}{\text{l}}$.
So, the self inductance $L$ of a solenoid increases as $l$ decreases and $A$ increases because $L$ is directly proportional to area and inversely proportional to length.
Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/permeability of the medium.

$\mid\text{M}\mid=\frac{\text{e}_1}{\big(\frac{\text{di}_2}{\text{dt}}\big)}=\frac{\phi_2}{\text{i}_1}$
$\therefore\phi_2=\frac{\text{e}_1\text{i}_1}{(\frac{\text{di}^2}{\text{dt}})}=\frac{(25.0\times10^{-3})(3.6)}{(15)}$
$=6\times10^{-3}=6\text{mWb}$

Maximum power is transferred at resonance.
$\therefore\text{f}_0=\frac{1}{2\pi\sqrt{\text{LC}}}$
$\text{or C}=\frac{1}{4\pi^2\text{f}^2_0\text{L}}=\frac{1}{4\times10\times(50)^2\times10}$
$=10^{-6}\text{F}=1\mu\text{F}$

A. Series Cells in Series connection.
In series, cells are joined end to end so that the same current flows through each cell. In case if the cells are connected in series the emf of the battery connected to the sum of the emf of the individual cell, If $E$ is the overall emf of the battery combined with n number cells and $E1, E2,......Em$ is the $\text{EMFs}$ of individual cell Then $E = E1 + E2+.....+Em.$