Faraday's and Lenz's Law Questions in English

(A) Given: Magnetic flux $\phi = 2t^2 - 5t + 1$ and resistance $R = 10 \ \Omega$.
According to Faraday's law of electromagnetic induction,the induced electromotive force (emf) is given by $\varepsilon = -\frac{d\phi}{dt}$.
Differentiating $\phi$ with respect to $t$: $\frac{d\phi}{dt} = \frac{d}{dt}(2t^2 - 5t + 1) = 4t - 5$.
Thus,$\varepsilon = -(4t - 5) = 5 - 4t$.
At $t = 0.25 \ s$,the induced emf is $\varepsilon = 5 - 4(0.25) = 5 - 1 = 4 \ V$.
The magnitude of the induced current $I$ is given by $I = \frac{|\varepsilon|}{R}$.
$I = \frac{4 \ V}{10 \ \Omega} = 0.4 \ A$.

(A) According to Lenz's law,the direction of the induced current is such that it opposes the change in magnetic flux that produced it.
$(i)$ The rectangular loop $abcd$ is moving into the magnetic field region. The magnetic flux through the loop increases. To oppose this increase,the induced current must create a magnetic field directed out of the page. By the right-hand rule,the induced current flows in the counter-clockwise direction,i.e.,along $adcba$.
(ii) The triangular loop $abc$ is moving out of the magnetic field region. The magnetic flux through the loop decreases. To oppose this decrease,the induced current must create a magnetic field directed into the page. By the right-hand rule,the induced current flows in the clockwise direction,i.e.,along $abc$.
(iii) The irregular loop $abcd$ is moving out of the magnetic field region. The magnetic flux through the loop decreases. To oppose this decrease,the induced current must create a magnetic field directed into the page. By the right-hand rule,the induced current flows in the clockwise direction,i.e.,along $abcda$.

(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Taking the magnitude of the induced $EMF$:
$|\varepsilon| = \left| \frac{d}{dt}(3t^2 + 2t + 5) \right|$
$|\varepsilon| = 6t + 2$
At time $t = 2 \text{ s}$,the induced $EMF$ is:
$|\varepsilon| = 6(2) + 2 = 14 \text{ V}$
The induced current $I$ is given by Ohm's law: $I = \frac{|\varepsilon|}{R}$.
Given $R = 14 \ \Omega$,we have:
$I = \frac{14 \text{ V}}{14 \ \Omega} = 1 \text{ A}$.

(B) According to Faraday's law of electromagnetic induction,the induced electromotive force (emf) $\varepsilon$ is given by the rate of change of magnetic flux: $\varepsilon = |\frac{d\phi}{dt}|$.
Given $\phi(t) = 2t^2 + 2t + 1$,we differentiate with respect to $t$:
$\varepsilon = |\frac{d}{dt}(2t^2 + 2t + 1)| = |4t + 2|$.
At $t = 2 \ s$,the induced emf is:
$\varepsilon = 4(2) + 2 = 10 \ V$.
Using Ohm's law,the current $i$ is given by $i = \frac{\varepsilon}{R}$.
Given resistance $R = 10 \ \Omega$,the current is:
$i = \frac{10 \ V}{10 \ \Omega} = 1 \ A$.

$(A)$ Given: Number of turns $n = 100$, Area $A = 0.1 \,m \times 0.05 \,m = 0.005 \,m^{2}$.
Initial magnetic field $B_{1} = 0.1 \,T$, final magnetic field $B_{2} = 0.05 \,T$, and time interval $dt = 0.05 \,s$.
The magnetic flux $\phi$ is given by $\phi = nBA \cos \theta$. Since the coil is perpendicular to the field, the angle $\theta = 0^{\circ}$, so $\cos 0^{\circ} = 1$.
The induced e.m.f. $e$ is given by Faraday's Law: $e = \left| -\frac{d\phi}{dt} \right| = nA \frac{|dB|}{dt}$.
Substituting the values: $e = 100 \times 0.005 \times \frac{(0.1 - 0.05)}{0.05}$.
$e = 0.5 \times \frac{0.05}{0.05} = 0.5 \,V$.

(A) According to Lenz's law,the induced current in the coil always opposes the change in magnetic flux that produces it.
When the north pole of the magnet is pushed towards the coil,the magnetic flux through the coil increases,and the induced current flows in a direction to oppose this increase.
When the magnet is pulled away from the coil,the magnetic flux through the coil decreases. To oppose this decrease,the induced current flows in the opposite direction compared to the first case.
Since the galvanometer pointer deflected towards $X$ when the magnet was pushed in,it will deflect in the opposite direction,i.e.,towards $X^1$,when the magnet is pulled away.

(C) The magnetic field $B$ produced by a long straight current-carrying conductor at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.
As the ring falls towards the conductor,the distance $r$ decreases,causing the magnetic field $B$ passing through the ring to increase.
According to the right-hand rule,the magnetic field lines from the conductor point out of the plane of the paper in the region where the ring is located.
Since the magnetic flux linked with the ring is increasing in the outward direction,according to Lenz's law,the induced current in the ring will create a magnetic field pointing into the plane of the paper to oppose this increase.
Using the right-hand grip rule,a magnetic field directed into the plane of the paper corresponds to a clockwise induced current in the ring.

(B) Given: Induced current $I = 2 \text{ A}$,Resistance $R = 10 \text{ } \Omega$,Time interval $\Delta t = 1 \text{ ms} = 10^{-3} \text{ s}$.
According to Faraday's law of induction,the magnitude of induced emf is $|\varepsilon| = \frac{\Delta \phi}{\Delta t}$.
Since $|\varepsilon| = I R$,we have $I R = \frac{\Delta \phi}{\Delta t}$.
Therefore,the change in magnetic flux is $\Delta \phi = I R \Delta t$.
Substituting the values: $\Delta \phi = 2 \times 10 \times 10^{-3} \text{ Wb}$.
$\Delta \phi = 20 \times 10^{-3} \text{ Wb} = 2 \times 10^{-2} \text{ Wb}$.

(D) Given: Magnetic flux density, $B = 1.0 \,Wb \,m^{-2}$
Number of turns, $N = 80$
Area of coil, $A = 0.01 \,m^2$
Time interval, $\Delta t = 0.2 \,s$
According to Faraday's law of electromagnetic induction, the magnitude of induced emf is given by:
$|e| = N \frac{|\Delta \phi|}{\Delta t}$
Since the coil is removed from the field, the final flux is $0$.
Change in flux, $\Delta \phi = B \cdot A - 0 = 1.0 \times 0.01 = 0.01 \,Wb$
Substituting the values:
$|e| = \frac{80 \times 0.01}{0.2} = \frac{0.8}{0.2} = 4 \,V$

(B) Given,magnetic flux $\phi = 3t^{2} + 4t + 9$.
According to Faraday's law of electromagnetic induction,the magnitude of the induced emf is given by $\varepsilon = \left| \frac{d\phi}{dt} \right|$.
First,differentiate the flux expression with respect to time $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(3t^{2} + 4t + 9) = 6t + 4$.
Now,substitute the value $t = 2 \text{ s}$ into the derivative:
$\varepsilon = |6(2) + 4| = |12 + 4| = 16 \text{ V}$.
Thus,the magnitude of the induced emf at $t = 2 \text{ s}$ is $16 \text{ V}$.

(C) According to Lenz's law,the induced current in coil $B$ will oppose the cause that produces it.
As the current in coil $A$ increases,the magnetic flux linked with coil $B$ increases.
To oppose this increase in magnetic flux,an induced current flows in coil $B$ in a direction opposite to the current in coil $A$.
Since the currents in the adjacent sides of the two coils flow in opposite directions,they exert a repulsive force on each other.
Therefore,coil $B$ is repelled.

(A) For a given perimeter,the area of a circle is the maximum among all plane shapes. As the irregular loop changes into a circular loop,its area $A$ increases.
Since the magnetic field $B$ is uniform and directed into the plane,the magnetic flux $\phi = B \cdot A$ increases.
According to Lenz's law,the induced current will create a magnetic field to oppose this increase in flux.
Therefore,the induced magnetic field must be directed out of the plane.
Using the right-hand thumb rule,a magnetic field directed out of the plane corresponds to an anticlockwise induced current.

(C) As the bar magnet falls through the copper coil, the magnetic flux linked with the coil changes.
According to Faraday's law of electromagnetic induction, this change in magnetic flux induces an electromotive force (emf) and consequently an induced current in the coil.
According to Lenz's law, the direction of this induced current is such that it opposes the cause that produces it, which is the motion of the falling magnet.
This induced current creates a magnetic field that exerts an upward repulsive force on the falling magnet.
Therefore, the net downward force on the magnet is $F_{net} = mg - F_{repulsive}$.
Since there is an upward force opposing gravity, the net acceleration $a$ of the magnet is given by $a = \frac{F_{net}}{m} = g - \frac{F_{repulsive}}{m}$.
Thus, the net acceleration of the magnet is less than $g$.

(A) The magnetic flux $\phi$ through the loop is given by $\phi = \vec{B} \cdot \vec{A}$.
Given, area vector $\vec{A} = (10 \times 10^{-4} \,m^2) \hat{k}$.
The magnetic field vector $\vec{B} = B \hat{n}$, where $\hat{n}$ is the unit vector in the direction $\hat{i}+\hat{j}+\hat{k}$.
Thus, $\vec{B} = 1.73 \times \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \,T$.
Initial flux $\phi_i = \vec{B} \cdot \vec{A} = \left( \frac{1.73}{\sqrt{3}} (\hat{i}+\hat{j}+\hat{k}) \right) \cdot (10^{-3} \hat{k}) = \frac{1.73}{\sqrt{3}} \times 10^{-3} \,Wb$.
Since $\sqrt{3} \approx 1.732$, $\phi_i \approx 10^{-3} \,Wb$.
Final flux $\phi_f = 0$ as the field decreases to zero.
The induced emf is $|e| = \left| -\frac{\Delta \phi}{\Delta t} \right| = \frac{\phi_i - \phi_f}{\Delta t} = \frac{10^{-3} \,Wb - 0}{10 \,s} = 10^{-4} \,V = 0.1 \,mV$.

(C) The magnetic flux $\phi_B$ linked with a coil is given by $\phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$.
When the plane of the coil is parallel to the magnetic field,the area vector $\vec{A}$ (which is perpendicular to the plane) is perpendicular to the magnetic field $\vec{B}$. Thus,$\theta = 90^\circ$ and $\phi_B = BA \cos 90^\circ = 0$.
Since the magnetic field is uniform and the coil expands radially while remaining in the same plane parallel to the field,the flux remains zero at all times. Therefore,the induced emf $\varepsilon = -\frac{d\phi_B}{dt} = 0$.
Thus,Assertion $(A)$ is true.
The Reason $(R)$ states that there is a constant magnetic field in the perpendicular direction to the plane of the coil. If this were true,the flux would be $\phi_B = BA$,and expanding the coil would change the area $A$,thereby inducing an emf. Since the problem states the plane is parallel to the field,the Reason $(R)$ is false.

(B) When a copper rod is moved in a magnetic field,an electromotive force $(emf)$ is induced.
According to Faraday's law of electromagnetic induction,the induced $emf$ $(e)$ is given by $e = -\frac{d\phi}{dt}$.
Since the rod has a resistance $(R)$,the induced current $(i)$ is given by $i = \frac{e}{R} = -\frac{1}{R} \frac{d\phi}{dt}$.
We know that current is the rate of flow of charge,so $i = \frac{dQ}{dt}$.
Substituting this into the equation,we get $\frac{dQ}{dt} = -\frac{1}{R} \frac{d\phi}{dt}$.
Integrating both sides,we find that the total charge $Q$ developed is proportional to the change in magnetic flux $\Delta\phi$.
However,the instantaneous rate of charge flow is proportional to the rate of change of magnetic flux,$\frac{d\phi}{dt}$.

(D) According to Faraday's law of electromagnetic induction,the magnitude of induced emf $e$ is given by $e = \frac{\Delta \phi}{\Delta t}$.
Since the circuit has a resistance $R$,the induced current $i$ is given by $i = \frac{e}{R} = \frac{\Delta \phi}{R \Delta t}$.
The total charge $Q$ passing through any point in the circuit in time $\Delta t$ is given by $Q = i \Delta t$.
Substituting the value of $i$,we get $Q = \left( \frac{\Delta \phi}{R \Delta t} \right) \Delta t = \frac{\Delta \phi}{R}$.

(C) Given:
Number of turns $N = 45$
Radius $r = 4 \ cm = 0.04 \ m$
Area $A = \pi r^2 = \pi \times (0.04)^2 = 16\pi \times 10^{-4} \ m^2$
Initial magnetic field $B_1 = 0 \ T$
Final magnetic field $B_2 = 0.70 \ T$
Time interval $\Delta t = 220 \ s$
The magnetic flux $\phi = B \cdot A \cdot \cos(\theta)$. Since the plane is perpendicular to the field,the angle $\theta = 0^\circ$,so $\cos(0^\circ) = 1$.
Change in flux $\Delta \phi = A(B_2 - B_1) = 16\pi \times 10^{-4} \times (0.70 - 0) = 11.2\pi \times 10^{-4} \ Wb$.
According to Faraday's law,the induced emf $\varepsilon = -N \frac{\Delta \phi}{\Delta t}$.
Magnitude of emf $|\varepsilon| = \frac{45 \times 11.2\pi \times 10^{-4}}{220}$.
$|\varepsilon| = \frac{504\pi \times 10^{-4}}{220} \approx \frac{1583.36 \times 10^{-4}}{220} \approx 7.2 \times 10^{-4} \ V = 0.72 \ mV$.

(B) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ $\varepsilon$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Given the magnetic flux $\phi = 50t^2 + 4$.
Differentiating $\phi$ with respect to time $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(50t^2 + 4) = 100t$.
The magnitude of the induced $EMF$ is $|\varepsilon| = |-\frac{d\phi}{dt}| = 100t$.
At time $t = 2 \ s$,the induced $EMF$ is $|\varepsilon| = 100(2) = 200 \ V$.
The induced current $I$ is given by Ohm's law: $I = \frac{|\varepsilon|}{R}$.
Given resistance $R = 200 \ \Omega$.
Therefore,$I = \frac{200 \ V}{200 \ \Omega} = 1 \ A$.

(C) The magnetic flux $\phi$ through a single loop is given by $\phi = B \cdot A \cdot \cos(\theta)$.
Since the plane of the coil is normal to the magnetic field,the angle $\theta$ between the area vector and the magnetic field is $0^\circ$,so $\cos(0^\circ) = 1$.
Thus,$\phi = B \cdot A$.
The induced emf $\epsilon$ in a coil with $N$ turns is given by Faraday's Law: $\epsilon = -N \frac{d\phi}{dt}$.
Here,$N = 100$,the area $A = (10 \ cm)^2 = (0.1 \ m)^2 = 0.01 \ m^2$,and the rate of change of magnetic field $\frac{dB}{dt} = 0.7 \ T \ s^{-1}$.
Substituting these values: $\epsilon = N \cdot A \cdot \frac{dB}{dt} = 100 \times 0.01 \times 0.7$.
$\epsilon = 1 \times 0.7 = 0.7 \ V$.

(B) When two coaxial coils carrying current in the same direction are brought closer to each other,the magnetic flux linked with each coil increases.
According to Lenz's law,the induced electromotive force $(EMF)$ will oppose this change in magnetic flux.
To oppose the increase in flux,the induced current flows in a direction opposite to the original current in each coil.
Consequently,the net current in both coils decreases.

(D) Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produced it.
This law is a direct consequence of the law of conservation of energy.
If the induced current were to assist the change in magnetic flux,it would lead to a perpetual motion machine,which violates the principle of conservation of energy.
Therefore,the work done in moving a magnet against the induced magnetic force is converted into electrical energy,satisfying the law of conservation of energy.

(A) According to Faraday's law of electromagnetic induction,the induced emf in a single loop is given by $\varepsilon = -\frac{d \phi_{B}}{dt}$.
For a coil consisting of $N$ closely wound turns,the total magnetic flux linked with the coil is $N \phi_{B}$.
Therefore,the total induced emf $\varepsilon$ is given by the rate of change of the total flux linkage:
$\varepsilon = -\frac{d}{dt} (N \phi_{B}) = -N \frac{d \phi_{B}}{dt}$.

(A) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.
Here,$n = 200 \ \text{turns/cm} = 20000 \ \text{turns/m}$.
The magnetic flux $\phi$ linked with the circular coil is $\phi = N B A$,where $N = 100$ is the number of turns in the coil and $A = \pi r^2$ is the area of the coil with $r = 0.03 \ m$.
Thus,$\phi = N (\mu_0 n i) (\pi r^2)$.
According to Faraday's law,the induced $EMF$ is $\varepsilon = -\frac{\Delta \phi}{\Delta t}$.
The induced current $I_{ind} = \frac{|\varepsilon|}{R} = \frac{N \mu_0 n \pi r^2}{R} \times \frac{\Delta i}{\Delta t}$.
Substituting the values: $I_{ind} = \frac{100 \times (4 \pi \times 10^{-7}) \times 20000 \times \pi \times (0.03)^2}{4} \times \frac{2 - 0}{0.04}$.
$I_{ind} = \frac{100 \times 4 \pi \times 10^{-7} \times 20000 \times \pi \times 0.0009}{4} \times 50$.
$I_{ind} = 9 \pi^2 \times 10^{-3} \ A = 9 \pi^2 \ mA$.

(B) Given: Radius of the circular loop $r = 14 \ cm = 0.14 \ m$.
Rate of change of magnetic field $\frac{dB}{dt} = 0.05 \ Ts^{-1}$.
According to Faraday's law of electromagnetic induction,the magnitude of induced emf $|e|$ is given by $|e| = \frac{d\phi}{dt}$.
Since the magnetic field is perpendicular to the plane,the magnetic flux $\phi = B \cdot A$.
Thus,$|e| = \frac{d}{dt}(BA) = A \frac{dB}{dt}$.
The area of the loop $A = \pi r^2 = \frac{22}{7} \times (0.14)^2 = \frac{22}{7} \times 0.0196 = 0.0616 \ m^2$.
Substituting the values: $|e| = 0.0616 \times 0.05 = 0.00308 \ V$.
Converting to millivolts: $|e| = 3.08 \times 10^{-3} \ V = 3.08 \ mV$.

(A) The induced current $i$ is given by $i = \frac{e}{R} = \frac{1}{R} \frac{d\phi}{dt} = \frac{A}{R} \frac{dB}{dt}$.
Here, $A$ is the area of the loop: $A = \pi r^2 = \pi \times (0.05 \text{ m})^2 = 25\pi \times 10^{-4} \text{ m}^2$.
The resistance $R$ of the wire is $R = \frac{\rho l}{a}$, where $l = 2\pi r$ and $a = \pi (r_{wire})^2$.
$l = 2 \times \pi \times 0.05 = 0.1\pi \text{ m}$.
$a = \pi \times (10^{-3} \text{ m})^2 = \pi \times 10^{-6} \text{ m}^2$.
$R = \frac{2 \times 10^{-8} \times 0.1\pi}{\pi \times 10^{-6}} = 2 \times 10^{-3} \Omega$.
Now, $\frac{dB}{dt} = \frac{iR}{A} = \frac{11 \times 2 \times 10^{-3}}{25\pi \times 10^{-4}} = \frac{22 \times 10^{-3}}{25 \times 3.14 \times 10^{-4}} \approx 2.8 \text{ T s}^{-1}$.

(D) Faraday's law states that an induced $emf$ is produced whenever the magnetic flux linked with a circuit changes with time.
$1$. Moving a magnet near a circuit changes the magnetic flux,thus inducing an $emf$.
$2$. Moving a circuit near a magnet also changes the magnetic flux,thus inducing an $emf$.
$3$. Changing the current in one circuit placed near another changes the magnetic field and thus the magnetic flux in the second circuit,inducing an $emf$.
$4$. Maintaining a large but constant current in a circuit produces a constant magnetic field. Since the magnetic flux remains constant over time,no induced $emf$ is produced.
Therefore,the correct answer is option $D$.

(D) Given that,number of loops,$N = 500$.
Side of square,$a = 10 \ cm = 0.1 \ m$.
Rate of increase of magnetic field,$\frac{dB}{dt} = 1 \ T/s$.
Since the coil is placed normal to the magnetic field,the magnetic flux $\phi = B \cdot A$.
The induced emf is given by Faraday's law: $\varepsilon = -N \frac{d\phi}{dt} = -N \frac{d}{dt}(BA)$.
Since the area $A$ is constant,$\varepsilon = -NA \frac{dB}{dt} = -N a^2 \frac{dB}{dt}$.
Substituting the values: $\varepsilon = -500 \times (0.1)^2 \times 1 = -5 \ V$.
The magnitude of the induced emf is $|\varepsilon| = 5 \ V$.

(A) Lenz's law states that the induced current always flows in a direction that opposes the change in magnetic flux that produced it.
To overcome this opposing force,external mechanical work must be performed.
This mechanical work is converted into electrical energy in the circuit.
Since energy is neither created nor destroyed but only transformed from one form to another,Lenz's law is a direct consequence of the law of conservation of energy.

(C) According to Faraday's law of electromagnetic induction,the magnitude of the induced electromotive force $(emf)$ in a circuit is equal to the time rate of change of magnetic flux through the circuit. Mathematically,it is expressed as $|\varepsilon| = |\frac{d\Phi_B}{dt}|$. Therefore,the correct statement corresponds to Faraday's law.

(A) According to Lenz's law,the direction of the induced current is such that it opposes the change in magnetic flux that produced it.
Therefore,Lenz's law provides the direction of the induced current in a circuit.

(D) Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produced it. If the induced current were to assist the change,it would violate the law of conservation of energy by creating energy out of nothing. Therefore,Lenz's law is a direct consequence of the law of conservation of energy.

(B) Let $r$ be the radius of the loop and $A_w$ be the cross-sectional area of the wire.
The mass of the loop is $m = (2 \pi r) A_w d$,so $A_w = \frac{m}{2 \pi r d}$.
The resistance of the loop is $R = \rho \frac{l}{A_w} = \rho \frac{2 \pi r}{A_w} = \rho \frac{2 \pi r}{m / (2 \pi r d)} = \frac{4 \pi^2 r^2 \rho d}{m}$.
The magnetic flux through the loop is $\phi = B \cdot \pi r^2$.
The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = -\pi r^2 \frac{dB}{dt}$.
The induced current is $I = \frac{|\varepsilon|}{R} = \frac{\pi r^2 (dB/dt)}{4 \pi^2 r^2 \rho d / m} = \frac{m}{4 \pi \rho d} \left(\frac{dB}{dt}\right)$.

(B) According to Faraday's law,the induced $emf$ $(\varepsilon)$ in a coil is given by $\varepsilon = -N \frac{d\phi}{dt}$,where $N$ is the number of turns.
As the number of turns $(N)$ increases,the induced $emf$ and consequently the induced current increase for a given rate of change of magnetic flux.
According to Lenz's law,the direction of the induced current is such that it opposes the cause that produces it,which is the motion of the magnet.
Since a larger $N$ results in a stronger induced current,the opposing magnetic force becomes greater,making it more difficult to push the magnet into the coil.
Thus,both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.

(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.
Given: $n = 100 \text{ turns/cm} = 10^4 \text{ turns/m}$, $I = \frac{4}{\pi} \,A$, and $\mu_0 = 4\pi \times 10^{-7} \,T \cdot m/A$.
$B = (4\pi \times 10^{-7}) \times (10^4) \times (\frac{4}{\pi}) = 16 \times 10^{-3} \,T$.
The magnetic flux $\phi_B$ linked with the coil of $N$ turns and area $A$ is $\phi_B = N B A$.
Given: $N = 200$, $A = 25 \,cm^2 = 25 \times 10^{-4} \,m^2$.
When the current is reversed, the magnetic field changes from $B$ to $-B$. The change in flux is $\Delta \phi_B = N(B - (-B))A = 2NBA$.
The induced emf $e$ is given by Faraday's law: $e = \left| \frac{\Delta \phi_B}{\Delta t} \right| = \frac{2NBA}{\Delta t}$.
Substituting the values: $e = \frac{2 \times 200 \times (16 \times 10^{-3}) \times (25 \times 10^{-4})}{0.04} = \frac{400 \times 16 \times 10^{-3} \times 25 \times 10^{-4}}{0.04} = \frac{160000 \times 10^{-7}}{0.04} = \frac{0.016}{0.04} = 0.4 \,V$.

(A) According to Faraday's law of electromagnetic induction,the induced emf in a coil is given by $e = -N \frac{d\phi}{dt}$.
Here,$N$ represents the number of turns in the coil.
As the number of turns $N$ increases,the magnitude of the induced emf increases.
According to Lenz's law,the induced current creates a magnetic field that opposes the change in magnetic flux that produced it.
When a magnet is moved into a coil,the induced emf in each loop creates a current that opposes the motion of the magnet.
Since more loops result in a larger total induced emf,the opposing force becomes greater,making it more difficult to move the magnet.
Therefore,both $A$ and $R$ are true,and $R$ is the correct explanation for $A$.

(A) An electric generator is a device that converts mechanical energy into electrical energy.
When a coil rotates in a magnetic field,the magnetic flux linked with the coil changes continuously.
According to Faraday's laws of electromagnetic induction,this change in magnetic flux induces an electromotive force $(emf)$ across the coil.
This induced $emf$ is responsible for the flow of induced current in the circuit.
Therefore,the working principle of an electric generator is based on Faraday's laws of electromagnetic induction.

(A) According to Faraday's law of electromagnetic induction, an induced electromotive force $(EMF)$ and current are produced only when there is a change in the magnetic flux $(\Phi_B)$ linked with a circuit over time.
Magnetic flux is defined as $\Phi_B = \int \vec{B} \cdot d\vec{A}$.
In this scenario, the cylinder is placed parallel to a uniform magnetic field $\vec{B}$. Since the magnetic field is uniform and the cylinder is stationary, the magnetic flux passing through any cross-section of the cylinder remains constant over time.
Since $\frac{d\Phi_B}{dt} = 0$, there is no induced $EMF$ and consequently no induced current.
Therefore, the induced current is zero.

(B) The magnetic flux $\phi$ through a coil is given by $\phi = NBA \cos \theta$,where $\theta$ is the angle between the area vector and the magnetic field.
Given that the plane of the coil makes an angle of $\frac{\pi}{6}$ with the magnetic field,the angle $\theta_1$ between the area vector and the magnetic field is $\theta_1 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} = 60^\circ$.
Initial flux $\phi_1 = NBA \cos(60^\circ) = 250 \times 2 \times (120 \times 10^{-4}) \times 0.5 = 3 \ Wb$.
When the plane of the coil is parallel to the magnetic field,the area vector is perpendicular to the magnetic field,so $\theta_2 = 90^\circ$.
Final flux $\phi_2 = NBA \cos(90^\circ) = 0 \ Wb$.
The change in flux is $\Delta \phi = |\phi_2 - \phi_1| = 3 \ Wb$.
The induced $EMF$ is $\varepsilon = \frac{\Delta \phi}{\Delta t} = \frac{3}{100 \times 10^{-3}} = 30 \ V$.
The induced current is $I = \frac{\varepsilon}{R} = \frac{30}{8} = 3.75 \ A$.

(A) According to Faraday's law of electromagnetic induction, the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Given $\phi = (5t^2 + 4t + 2) \times 10^{-3} \text{ Wb}$.
Differentiating with respect to $t$:
$\frac{d\phi}{dt} = (10t + 4) \times 10^{-3} \text{ Wb/s}$.
At $t = 6 \text{ s}$, the magnitude of induced $EMF$ is:
$|\varepsilon| = |10(6) + 4| \times 10^{-3} = 64 \times 10^{-3} \text{ V}$.
The induced current $I$ is given by $I = \frac{|\varepsilon|}{R}$.
Given $R = 16 \Omega$, we have:
$I = \frac{64 \times 10^{-3}}{16} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$.
Thus, the correct option is $A$.

(B) The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length.
Given $n = 20 \text{ turns/cm} = 2000 \text{ turns/m}$.
$B = (4\pi \times 10^{-7} \text{ T m/A}) \times (2000 \text{ m}^{-1}) \times I = 8\pi \times 10^{-4} I \text{ T}$.
The magnetic flux $\phi$ through the loop of area $A = \frac{4}{\pi} \text{ cm}^2 = \frac{4}{\pi} \times 10^{-4} \text{ m}^2$ is $\phi = B \cdot A$.
The induced emf $\varepsilon$ is given by Faraday's Law: $\varepsilon = \left| \frac{\Delta \phi}{\Delta t} \right| = A \left| \frac{\Delta B}{\Delta t} \right| = A \mu_0 n \left| \frac{\Delta I}{\Delta t} \right|$.
Substituting the values: $\Delta I = 3.0 \text{ A} - 1.0 \text{ A} = 2.0 \text{ A}$,$\Delta t = 0.2 \text{ s}$.
$\varepsilon = \left( \frac{4}{\pi} \times 10^{-4} \text{ m}^2 \right) \times (4\pi \times 10^{-7} \text{ T m/A}) \times (2000 \text{ m}^{-1}) \times \left( \frac{2.0 \text{ A}}{0.2 \text{ s}} \right)$.
$\varepsilon = (4 \times 10^{-4}) \times (4 \times 10^{-7}) \times (2000) \times (10) \text{ V}$.
$\varepsilon = 32 \times 10^{-7} \text{ V} = 3.2 \times 10^{-6} \text{ V} = 3.2 \mu \text{V}$.

(C) According to Faraday's Law of Electromagnetic Induction,the induced electromotive force $(EMF)$ in a coil is given by the rate of change of magnetic flux linkage.
$|\varepsilon| = N \left| \frac{d\phi}{dt} \right| = N A \left| \frac{dB}{dt} \right|$
Given:
Number of turns $N = 100$
Radius $r = 10 \ cm = 0.1 \ m$
Area $A = \pi r^2 = \pi \times (0.1)^2 = 0.01 \pi \ m^2$
Rate of change of magnetic field $\frac{dB}{dt} = 0.1 \ T \ s^{-1}$
Substituting the values:
$|\varepsilon| = 100 \times (0.01 \pi) \times 0.1$
$|\varepsilon| = 1 \times 0.1 \pi = 0.1 \pi \ V = \frac{\pi}{10} \ V$

(B) Given: Magnetic field $B = 10^{-12} \sin(5 \times 10^6 t) \text{ T}$,number of turns $N = 300$,and area $A = 20 \text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2$.
According to Faraday's law of induction,the induced emf $e$ is given by $e = -N \frac{d\phi}{dt}$,where $\phi = BA$.
Since the coil is perpendicular to the field,$\phi = BA \cos(0^\circ) = BA$.
Substituting the values:
$e = -N \frac{d}{dt}(BA) = -NA \frac{dB}{dt}$
$e = -300 \times (20 \times 10^{-4}) \times \frac{d}{dt} [10^{-12} \sin(5 \times 10^6 t)]$
$e = -300 \times 20 \times 10^{-4} \times 10^{-12} \times \cos(5 \times 10^6 t) \times (5 \times 10^6)$
$e = -6000 \times 10^{-16} \times 5 \times 10^6 \times \cos(5 \times 10^6 t)$
$e = -30000 \times 10^{-10} \times \cos(5 \times 10^6 t)$
$e = -3 \times 10^4 \times 10^{-10} \times \cos(5 \times 10^6 t)$
$e = -3 \times 10^{-6} \cos(5 \times 10^6 t) \text{ V}$.

(B) Given: Number of turns in the coil $= n$. Resistance of the coil $= R$. Resistance of the galvanometer $= 4R$. Total resistance of the circuit $R_{total} = R + 4R = 5R$.
According to Faraday's Law of electromagnetic induction,the induced electromotive force (emf) is given by $e = -n \frac{\Delta \phi}{\Delta t}$.
Here,the change in flux is $\Delta \phi = \phi_2 - \phi_1$ and the time taken is $\Delta t = t$.
Therefore,$e = -n \frac{(\phi_2 - \phi_1)}{t}$.
The induced current $i$ is given by $i = \frac{e}{R_{total}}$.
Substituting the values,we get $i = \frac{-n(\phi_2 - \phi_1)}{5Rt}$.

(D) The magnetic flux through the coil is given by $\Phi = BA \cos(\theta)$,where $\theta$ is the angle between the magnetic field and the area vector. Initially,the field is perpendicular to the plane,so $\theta_0 = 0^{\circ}$.
In step $A$,the coil rotates by $60^{\circ}$,so the final angle is $\theta_A = 60^{\circ}$. The change in flux is $\Delta \Phi_A = BA(\cos 60^{\circ} - \cos 0^{\circ}) = BA(0.5 - 1) = -0.5 BA$. The induced emf is $\varepsilon_A = -\frac{\Delta \Phi_A}{t} = \frac{0.5 BA}{t}$.
In step $B$,the coil rotates by another $120^{\circ}$ in the same sense,so the final angle is $\theta_B = 60^{\circ} + 120^{\circ} = 180^{\circ}$. The change in flux is $\Delta \Phi_B = BA(\cos 180^{\circ} - \cos 60^{\circ}) = BA(-1 - 0.5) = -1.5 BA$. The induced emf is $\varepsilon_B = -\frac{\Delta \Phi_B}{2t} = \frac{1.5 BA}{2t} = \frac{0.75 BA}{t}$.
The ratio of the induced emf is $\frac{\varepsilon_A}{\varepsilon_B} = \frac{0.5 BA / t}{0.75 BA / t} = \frac{0.5}{0.75} = \frac{2}{3}$.

(B) The induced emf $\varepsilon$ is given by Faraday's law of induction: $|\varepsilon| = |\frac{d\phi}{dt}| = |\frac{d(B \cdot A)}{dt}|$.
Since the area $A$ is constant,$|\varepsilon| = A \cdot |\frac{dB}{dt}|$.
Given $A = 0.2 \, m^2$,initial $B = 0.25 \, T$,final $B = 0 \, T$,and $\Delta t = 10^{-4} \, s$.
$|\varepsilon| = 0.2 \cdot \frac{0.25 - 0}{10^{-4}} = 0.2 \cdot 2500 = 500 \, V$.
The induced current $I$ is given by Ohm's law: $I = \frac{\varepsilon}{R}$.
Given $R = 20 \, \Omega$,$I = \frac{500}{20} = 25 \, A$.

(D) Given that:
Area of coil,$A = 100 \,cm^2 = 100 \times 10^{-4} \,m^2 = 10^{-2} \,m^2$
Number of turns,$N = 20$
Magnetic field,$B = 2 \,Wb/m^2$
Time interval,$\Delta t = 0.2 \,s$
The magnetic flux $\phi$ is given by $\phi = N B A \cos \theta$,where $\theta$ is the angle between the area vector $\hat{n}$ and the magnetic field $B$.
Initial position: The plane of the coil makes an angle of $30^{\circ}$ with the field. Therefore,the angle between the area vector $\hat{n}$ and the field $B$ is $\theta_1 = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
Initial flux,$\phi_1 = N B A \cos 60^{\circ} = 20 \times 2 \times 10^{-2} \times 0.5 = 0.2 \,Wb$.
Final position: The coil is perpendicular to the field. Therefore,the area vector $\hat{n}$ is parallel to the field $B$,so $\theta_2 = 0^{\circ}$.
Final flux,$\phi_2 = N B A \cos 0^{\circ} = 20 \times 2 \times 10^{-2} \times 1 = 0.4 \,Wb$.
Induced emf,$e = -\frac{\Delta \phi}{\Delta t} = -\frac{\phi_2 - \phi_1}{\Delta t} = -\frac{0.4 - 0.2}{0.2} = -\frac{0.2}{0.2} = -1 \,V$.
The magnitude of the induced emf is $|e| = 1 \,V$.

(C) Given: Resistance of the coil, $R = 10 \Omega$. Number of turns, $N = 180$. Diameter of the coil, $d = 4 \text{ cm} = 4 \times 10^{-2} \text{ m}$. Radius, $r = 2 \times 10^{-2} \text{ m}$. Resistance of the galvanometer, $R_g = 618 \Omega$. Total resistance, $R_{eq} = R + R_g = 10 + 618 = 628 \Omega$. Charge, $q = 360 \mu \text{C} = 360 \times 10^{-6} \text{ C}$.
The magnetic flux is maximum, so $\phi = BA$.
The induced charge is given by $q = \frac{N \Delta \phi}{R_{eq}}$.
Since the field is removed, $\Delta \phi = BA - 0 = BA$.
Area $A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4 \pi \times 10^{-4} \text{ m}^2$.
Substituting the values: $360 \times 10^{-6} = \frac{180 \times B \times 4 \pi \times 10^{-4}}{628}$.
$B = \frac{360 \times 10^{-6} \times 628}{180 \times 4 \times 3.14 \times 10^{-4}} = \frac{2 \times 10^{-6} \times 628}{12.56 \times 10^{-4}} = \frac{1256 \times 10^{-6}}{1256 \times 10^{-4}} = 1 \text{ T}$.

(B) According to Faraday's Law of electromagnetic induction,the induced electromotive force (Emf) is given by: $\epsilon = -\frac{d\phi}{dt}$.
Since the circuit has a resistance $R$,the induced current $i$ is: $i = \frac{\epsilon}{R} = -\frac{1}{R} \frac{d\phi}{dt}$.
We know that current $i = \frac{dq}{dt}$,where $dq$ is the small amount of charge flowing in time $dt$.
Substituting this,we get: $\frac{dq}{dt} = -\frac{1}{R} \frac{d\phi}{dt}$.
Integrating both sides,we find the total charge $q$ that flows:
$q = \int dq = -\frac{1}{R} \int d\phi = \frac{\Delta\phi}{R}$.
Given $\Delta\phi = 5 \text{ Wb}$ and $R = 100 \Omega$,the charge $q$ is:
$q = \frac{5}{100} = 0.05 \text{ C}$.