Faraday's and Lenz's Law Questions in English

(D) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = (B_0 e^{-t})(\pi r^2)$.
According to Faraday's law of induction,the induced electromotive force $(emf)$ is $E = -\frac{d\phi}{dt}$.
$E = -\frac{d}{dt} (B_0 \pi r^2 e^{-t}) = -\pi r^2 B_0 \frac{d}{dt}(e^{-t}) = \pi r^2 B_0 e^{-t}$.
At the instant the key $(K)$ is closed,$t = 0$.
Therefore,the induced $emf$ at $t = 0$ is $E_0 = \pi r^2 B_0 e^0 = \pi r^2 B_0$.
The electrical power $P$ developed in the resistor $R$ is given by $P = \frac{E_0^2}{R}$.
Substituting the value of $E_0$,we get $P = \frac{(\pi r^2 B_0)^2}{R} = \frac{B_0^2 \pi^2 r^4}{R}$.

(D) According to Faraday's law of electromagnetic induction,the induced electromotive force $(Emf)$ is given by $E = -\frac{d\phi}{dt}$.
Given $\phi = 5t - 10t^2$.
$E = -\frac{d}{dt}(5t - 10t^2) = -(5 - 20t)$.
At $t = 0.25\, s = \frac{1}{4}\, s$,the $Emf$ is:
$E = -(5 - 20 \times 0.25) = -(5 - 5) = 0\, V$.
Since the induced $Emf$ is $0\, V$,the induced current $I = \frac{E}{R} = 0\, A$.

(C) According to Lenz's Law,the direction of the induced current in the ring $B$ will be such that it opposes the change in magnetic flux linked with it.
$1$. If the current $I$ in the electromagnet $A$ increases,the magnetic flux through the ring $B$ increases. To oppose this increase,the ring $B$ will develop an induced current such that it creates a magnetic field opposing the field of $A$. This results in a repulsive force between $A$ and $B$.
$2$. If the current $I$ in the electromagnet $A$ decreases,the magnetic flux through the ring $B$ decreases. To oppose this decrease,the ring $B$ will develop an induced current such that it creates a magnetic field that supports the field of $A$. This results in an attractive force between $A$ and $B$.
Therefore,if $I$ increases,$A$ will repel $B$.

(B) When the switch is closed in the primary circuit,the current in the primary circuit increases from $0$ to a maximum value.
This increase in current creates an increasing magnetic flux through the secondary circuit.
According to Faraday's law of electromagnetic induction,an induced electromotive force $(EMF)$ is generated in the secondary circuit.
According to Lenz's law,the direction of the induced current is such that it opposes the cause that produces it.
Since the magnetic flux is increasing in the direction of the primary circuit's magnetic field,the secondary circuit will induce a current to create a magnetic field in the opposite direction to oppose this increase.
Based on the right-hand thumb rule,for the given configuration,the induced current in the secondary circuit will flow in the $Anticlockwise$ direction.

(A) $1$. According to the right-hand thumb rule,the magnetic field produced by the current-carrying wire at the position of the loop is directed into the plane of the loop (inward).
$2$. As the loop is moved away from the wire,the distance from the wire increases,which causes the magnetic field strength at the loop to decrease.
$3$. Consequently,the magnetic flux linked with the loop decreases.
$4$. According to Lenz's Law,the induced current will flow in a direction that opposes this decrease in magnetic flux.
$5$. To oppose the decrease of the inward magnetic field,the induced current must create an additional inward magnetic field.
$6$. By the right-hand rule,an inward magnetic field is produced by a clockwise current.
$7$. Therefore,the induced current in the circular loop will be clockwise.

(B) According to Faraday's law of electromagnetic induction,the induced $emf$ is given by $e = -\frac{d\phi}{dt}$.
In the first case,the relative velocity between the magnet and the coil is $v$. Thus,the rate of change of magnetic flux is $\left(\frac{d\phi}{dt}\right)_1 = k \cdot v$,where $k$ is a constant depending on the magnetic field gradient and coil geometry. So,$e = k \cdot v$.
In the second case,both the magnet and the coil move away from each other with speed $v$. The relative velocity between them is $v_{rel} = v + v = 2v$.
The rate of change of magnetic flux is proportional to the relative velocity,so $\left(\frac{d\phi}{dt}\right)_2 = k \cdot (2v) = 2(k \cdot v) = 2e$.
Therefore,the induced $emf$ in the second case will be $2e$.

(A) Given: Number of turns $N = 1000$.
Face area $A = 4 \, cm^2 = 4 \times 10^{-4} \, m^2$.
Change in magnetic field $\Delta B = 10^{-2} \, Wb \, m^{-2}$.
Time interval $\Delta t = 0.01 \, s = 10^{-2} \, s$.
The induced $e.m.f.$ is given by Faraday's law: $e = N \frac{\Delta \phi}{\Delta t} = N A \frac{\Delta B}{\Delta t} \cos \theta$.
Since the axis is parallel to the magnetic field,the angle $\theta$ between the area vector and the magnetic field is $0^\circ$,so $\cos 0^\circ = 1$.
Substituting the values: $e = 1000 \times (4 \times 10^{-4} \, m^2) \times \frac{10^{-2} \, Wb \, m^{-2}}{10^{-2} \, s}$.
$e = 1000 \times 4 \times 10^{-4} = 0.4 \, V$.
Converting to $mV$: $0.4 \, V = 400 \, mV$.

(C) According to Faraday's law of electromagnetic induction,the induced electromotive force (emf) is given by $e = \frac{\Delta \phi}{\Delta t}$.
Since the coil has a resistance $R$,the induced current is $i = \frac{e}{R}$,which implies $e = iR$.
Equating the two expressions for $e$,we get $iR = \frac{\Delta \phi}{\Delta t}$.
Rearranging the terms,we find the change in magnetic flux: $\Delta \phi = R \times (i \cdot \Delta t)$.
The term $(i \cdot \Delta t)$ represents the area under the current-time $(i-t)$ graph.
From the given graph,the area is a right-angled triangle with base $0.1\,s$ and height $4\,A$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.1 \times 4 = 0.2$.
Substituting the values,we get $\Delta \phi = 10 \times 0.2 = 2\,Wb$.
Therefore,the magnitude $\left| \Delta \phi \right| = 2\,Wb$.

(B) When a wire of irregular shape turns into a circular loop,the area of the loop increases.
Since the magnetic field is directed into the paper,the magnetic flux linked with the loop increases as the area increases.
According to Lenz's law,the induced current will flow in a direction that opposes this increase in magnetic flux.
To oppose the inward magnetic flux,the induced magnetic field must be directed out of the paper.
Using the right-hand thumb rule,for the induced magnetic field to be directed out of the paper,the induced current must flow in the counter-clockwise direction,which is along $adcba$.

(B) According to Faraday's law of electromagnetic induction,an $emf$ is induced in a loop when the magnetic flux $\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta$ linked with the loop changes with time.
When the loop is rotated about its axis,the angle $\theta$ between the magnetic field $\vec{B}$ and the area vector $\vec{A}$ remains constant,so the flux does not change.
When the loop is rotated about a diameter,the angle $\theta$ changes with time,causing the magnetic flux to change,which induces an $emf$ in the loop.
Translational motion in a uniform magnetic field does not change the flux linked with the loop,so no $emf$ is induced.

(B) Given: Number of turns $N = 40$,Area $A = 4 \, cm^2 = 4 \times 10^{-4} \, m^2$,Charge $q = 2 \times 10^{-4} \, C$,Resistance $R = 80 \, \Omega$.
The induced charge $q$ is given by the formula $q = \frac{\Delta \phi}{R}$,where $\Delta \phi$ is the change in magnetic flux.
Since the coil is removed from the magnetic field,the final flux is $0$. Thus,$\Delta \phi = N \cdot A \cdot B$.
Substituting the values: $q = \frac{N \cdot A \cdot B}{R}$.
Rearranging for $B$: $B = \frac{q \cdot R}{N \cdot A}$.
$B = \frac{(2 \times 10^{-4} \, C) \times (80 \, \Omega)}{40 \times (4 \times 10^{-4} \, m^2)}$.
$B = \frac{160 \times 10^{-4}}{160 \times 10^{-4}} = 1 \, Wb \, m^{-2}$.

(C) Given,the magnetic flux $\phi = (2t^2 + 4t + 6) \, mWb = (2t^2 + 4t + 6) \times 10^{-3} \, Wb$.
According to Faraday's law of electromagnetic induction,the induced $emf$ $(\varepsilon)$ is given by the magnitude of the rate of change of magnetic flux:
$\varepsilon = \left| \frac{d\phi}{dt} \right|$
$\varepsilon = \frac{d}{dt} [(2t^2 + 4t + 6) \times 10^{-3}] \, V$
$\varepsilon = (4t + 4) \times 10^{-3} \, V$
At $t = 4 \, s$:
$\varepsilon = (4 \times 4 + 4) \times 10^{-3} \, V$
$\varepsilon = (16 + 4) \times 10^{-3} \, V$
$\varepsilon = 20 \times 10^{-3} \, V = 0.02 \, V$.

(C) When the key $K$ is pressed,the current in the solenoid increases from zero to a steady value.
This change in current produces a changing magnetic flux through the conducting ring.
According to Faraday's law of electromagnetic induction,an induced electromotive force $(EMF)$ is produced in the ring,which drives an induced current.
According to Lenz's law,the direction of this induced current is such that it opposes the cause that produced it.
In this case,the increasing magnetic flux through the ring creates a magnetic pole on the face of the ring facing the solenoid,which is of the same polarity as the pole of the electromagnet.
Since like poles repel each other,the ring experiences an upward repulsive force and jumps out of the core.

(B) According to Lenz's law,the induced current in a loop opposes the change in magnetic flux that produces it.
When viewed from the side of the magnet,an anticlockwise current in the ring creates a magnetic field directed towards the magnet (acting as a North pole).
If the North pole of the magnet faces the ring and moves towards it,the magnetic flux through the ring increases. To oppose this,the ring induces a North pole to repel the magnet.
If the South pole of the magnet faces the ring and moves away from it,the magnetic flux through the ring decreases. To oppose this,the ring induces a North pole to attract the magnet.
Therefore,both the North pole facing the ring and moving towards it,or the South pole facing the ring and moving away from it,would result in an anticlockwise current. Among the given options,$B$ is a correct possibility.

(C) The induced $emf$ in a loop is given by Faraday's law: $emf = -\frac{d\phi}{dt}$,where $\phi = B A \cos\theta$.
$(a)$ If the loop is translated in a uniform magnetic field,the magnetic flux $\phi$ remains constant,so no $emf$ is induced.
$(b)$ If the loop is rotated about its axis,the angle $\theta$ between the area vector and the magnetic field remains constant,so $\phi$ does not change. No $emf$ is induced.
$(c)$ If the loop is rotated about a diameter,the angle $\theta$ changes with time,causing the flux $\phi$ to change. Thus,an $emf$ is induced.
$(d)$ If the loop is deformed,its area $A$ changes,which causes the flux $\phi$ to change. Thus,an $emf$ is induced.
Therefore,an $emf$ is induced in cases $(c)$ and $(d)$.

(A) Area of the square loop,$A = 10 \, cm \times 10 \, cm = 100 \, cm^2 = 10^{-2} \, m^2$.
Initial magnetic flux linked with the loop,$\phi_1 = B_1 A \cos \theta = 0.1 \times 10^{-2} \times \cos 45^{\circ} = 0.1 \times 10^{-2} \times \frac{1}{\sqrt{2}} = \frac{10^{-3}}{\sqrt{2}} \, Wb$.
Final magnetic flux linked with the loop,$\phi_2 = 0 \, Wb$ (since $B_2 = 0$).
The induced $EMF$ in the loop is $e = -\frac{d\phi}{dt} = -\frac{\phi_2 - \phi_1}{\Delta t} = -\frac{0 - \frac{10^{-3}}{\sqrt{2}}}{0.7} = \frac{10^{-3}}{0.7 \times 1.414} \approx \frac{10^{-3}}{0.99} \approx 10^{-3} \, V$.
The induced current in the loop is $I = \frac{e}{R} = \frac{10^{-3} \, V}{1 \, \Omega} = 10^{-3} \, A = 1.0 \, mA$.

(B) $1$. According to the right-hand thumb rule,the magnetic field produced by the current in the wire is directed into the page on the right side (near ring $B$) and out of the page on the left side (near ring $A$).
$2$. When the current in the wire decreases,the magnetic flux through both rings decreases.
$3$. According to Lenz's law,the induced current will oppose this change by creating its own magnetic field to support the decreasing flux.
$4$. For ring $A$ (left side),the magnetic field is outward and decreasing. To support this,the induced current will flow in an anticlockwise direction to produce an outward magnetic field.
$5$. For ring $B$ (right side),the magnetic field is inward and decreasing. To support this,the induced current will flow in a clockwise direction to produce an inward magnetic field.
$6$. Therefore,the induced current is anticlockwise in $A$ and clockwise in $B$.

(D) The magnetic flux $\phi$ linked with the loop is given by $\phi = B \cdot A$,where $A$ is the area of the loop inside the magnetic field.
Since the magnetic field $B$ is uniform and constant,and the entire loop is moving within the region where the magnetic field exists,the area $A$ of the loop enclosed by the magnetic field remains constant.
As the magnetic flux $\phi = B \cdot A$ does not change with time,the rate of change of magnetic flux $d\phi/dt$ is zero.
According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is $\varepsilon = -d\phi/dt$.
Since $d\phi/dt = 0$,the induced $EMF$ $\varepsilon = 0$.
Consequently,the induced current $I = \varepsilon/R = 0$.
Therefore,no current is induced in the loop.

(B) According to Lenz's Law,the direction of the induced current in a closed loop is such that it opposes the change in magnetic flux that produced it.
As the solenoid approaches the loop,the magnetic flux linked with the loop increases.
To oppose this increase,the loop will develop a magnetic field in the opposite direction to the solenoid's magnetic field.
The solenoid acts like a bar magnet with its North pole facing the loop.
Therefore,the loop must behave like a North pole on the side facing the solenoid to repel it.
For the loop to act as a North pole,the current must flow in an anticlockwise direction when viewed from the side of the solenoid.
However,the observer is on the other side of the loop. When viewed from the observer's side,the current will appear to flow in the opposite direction,which is clockwise.

(A) The magnetic field inside the solenoid is given by $B = \mu_0 n I$,where $n = 100\, turns/cm = 10^4\, turns/m$ and $I = 5\, A$.
$B = (4\pi \times 10^{-7}) \times 10^4 \times 5 = 20\pi \times 10^{-3}\, T = 0.02\pi\, T$.
The magnetic flux through the coil is $\phi = N B A_{coil}$,where $N = 100$ and $A_{coil} = \pi r^2 = \pi (0.01)^2 = \pi \times 10^{-4}\, m^2$.
When the current is reversed,the change in flux is $\Delta \phi = \phi_f - \phi_i = (-NBA) - (NBA) = -2NBA$.
The induced charge $q$ is given by $q = \frac{|\Delta \phi|}{R} = \frac{2NBA}{R}$.
Substituting the values: $q = \frac{2 \times 100 \times (0.02\pi) \times (\pi \times 10^{-4})}{20} = \frac{4 \times 10^2 \times 0.01 \times \pi^2 \times 10^{-4}}{20} = \frac{4 \times 10^{-4} \times \pi^2}{20}$.
Using $\pi^2 \approx 10$,$q = \frac{4 \times 10^{-4} \times 10}{20} = 2 \times 10^{-4}\, C$.

(B) When the ring enters the magnetic field, the magnetic flux linked with the ring changes, which induces an electromotive force $(emf)$ and consequently an induced current. According to Lenz's law, this current opposes the cause of its production.
Once the ring is completely inside the uniform magnetic field, the magnetic flux linked with it remains constant. Since there is no change in magnetic flux $(\frac{d\phi}{dt} = 0)$, the induced $emf$ and the induced current become zero.
When the ring starts to emerge from the magnetic field, the magnetic flux linked with it changes again. This induces an $emf$ and a current in the opposite direction compared to the entry phase.
Therefore, the current is non-zero during entry, zero while inside, and non-zero (with opposite polarity) during exit. Graph $B$ correctly represents this variation.

(B) The magnetic flux linked with the coil is given by $\phi = 10t^2 - 50t + 250$.
According to Faraday's law of electromagnetic induction,the induced $emf$ $(e)$ is given by $e = -\frac{d\phi}{dt}$.
Differentiating the flux with respect to time $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(10t^2 - 50t + 250) = 20t - 50$.
Therefore,the induced $emf$ is $e = -(20t - 50) = 50 - 20t$.
At $t = 3 \ s$,the induced $emf$ is:
$e = 50 - 20(3) = 50 - 60 = -10 \ V$.

(A) According to Faraday's law of electromagnetic induction,the induced $emf$ in a closed loop is given by $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi}{dt}$.
Since the magnetic flux $\phi$ is varying,the line integral $\oint \vec{E} \cdot d\vec{l}$ is non-zero.
$A$ field is defined as conservative if the line integral of the field around any closed loop is zero. Since the line integral here is non-zero,the induced electric field $\vec{E}$ is a non-conservative field.
Therefore,both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(A) The Assertion describes Lenz's Law,which states that the direction of induced current is such that it opposes the change in magnetic flux that produced it.
Lenz's Law is a direct consequence of the Law of Conservation of Energy. If the induced current were to aid the change in magnetic flux,it would lead to an infinite increase in energy,which violates the principle of conservation of energy.
Therefore,the Reason correctly explains the Assertion.

(D) Lenz's law states that the direction of the induced $emf$ is always such that it opposes the change in magnetic flux that produces it.
This law is a direct consequence of the law of conservation of energy.
If the induced $emf$ were to assist the change in flux,it would lead to an increase in energy without any external work,which is impossible.
Therefore,the Assertion is incorrect,and the Reason is correct.

(C) Faraday's law,specifically Lenz's law,is a direct consequence of the law of conservation of energy. If this were not true,we could create energy out of nothing,which violates physical laws.
In a purely resistive $AC$ circuit,the voltage $(emf)$ and current are in the same phase. The phase difference between them is $0$. Therefore,the statement in the Reason is incorrect.
Thus,the Assertion is correct,but the Reason is incorrect.

(C) Faraday's laws of electromagnetic induction are a direct consequence of the law of conservation of energy. If they were not,we could create energy out of nothing,which violates the fundamental laws of physics.
In a purely resistive $A.C.$ circuit,the current and the $e.m.f.$ (voltage) are in the same phase. The phase difference between them is $0$. The statement that current lags behind the $e.m.f.$ is incorrect,as this only happens in an inductive circuit.
Therefore,the Assertion is correct,but the Reason is incorrect.

(N/A) To obtain a large deflection,one or more of the following steps can be taken:
$(i)$ Use a rod made of soft iron inside the coil $C_2$.
$(ii)$ Connect the coil to a more powerful battery.
$(iii)$ Move the arrangement more rapidly towards or away from the test coil $C_1$.
$(b)$ Replace the galvanometer with a small bulb,such as one found in a small torch light. The relative motion between the two coils will cause the bulb to glow,thereby demonstrating the presence of an induced current.
In experimental physics,one must learn to innovate. Michael Faraday,who is ranked as one of the best experimentalists ever,was legendary for his innovative skills.

(A) The area of the square loop is $A = (10 \; cm)^2 = 100 \; cm^2 = 10^{-2} \; m^2$.
The angle $\theta$ between the area vector (perpendicular to the east-west plane) and the magnetic field (in the north-east direction) is $45^{\circ}$.
The initial magnetic flux is $\phi_i = B A \cos \theta = 0.10 \times 10^{-2} \times \cos(45^{\circ}) = \frac{10^{-3}}{\sqrt{2}} \; Wb$.
The final magnetic flux is $\phi_f = 0 \; Wb$.
The change in flux is $\Delta \phi = |\phi_f - \phi_i| = \frac{10^{-3}}{\sqrt{2}} \; Wb$.
The magnitude of the induced $emf$ is $\varepsilon = \frac{|\Delta \phi|}{\Delta t} = \frac{10^{-3}}{\sqrt{2} \times 0.70} \approx 1.01 \times 10^{-3} \; V \approx 1.0 \; mV$.
The magnitude of the induced current is $I = \frac{\varepsilon}{R} = \frac{1.01 \times 10^{-3} \; V}{0.5 \; \Omega} \approx 2.02 \times 10^{-3} \; A \approx 2.0 \; mA$.

(N/A) $(i)$ The magnetic flux through the rectangular loop $abcd$ increases due to the motion of the loop into the region of the magnetic field. The induced current must flow along the path $bcdab$ so that it opposes the increasing flux.
$(ii)$ Due to the outward motion,the magnetic flux through the triangular loop $abc$ decreases,due to which the induced current flows along $bacb$ to oppose the change in flux.
$(iii)$ As the magnetic flux decreases due to the motion of the irregular-shaped loop $abcd$ out of the region of the magnetic field,the induced current flows along $cdabc$ to oppose the change in flux.
Note that there is no induced current as long as the loops are completely inside or outside the region of the magnetic field.

(N/A) The direction of the induced current in a closed loop is given by Lenz's law,which states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
$(a)$ As the North pole of the magnet moves towards the coil,the magnetic flux through the coil increases. To oppose this,the coil develops a North pole on the face towards the magnet. Thus,the induced current flows along $qrpq$.
$(b)$ As the North pole of the magnet moves away from the coil,the magnetic flux decreases. To oppose this,the coil develops a South pole on the face towards the magnet. Thus,the induced current flows along $prqp$ in the left coil and $zyxz$ in the right coil.
$(c)$ When the tapping key is just closed,the current in the left loop increases,creating an increasing magnetic field. By Lenz's law,the right loop develops an induced current to oppose this increase. The direction is along $zyxz$.
$(d)$ When the rheostat setting is changed to increase the resistance,the current in the right loop decreases,leading to a decrease in magnetic flux. The left loop develops an induced current to oppose this decrease. The direction is along $yzxy$.
$(e)$ When the tapping key is just released,the current in the primary coil drops to zero,causing a rapid decrease in magnetic flux. The secondary coil develops an induced current to oppose this change. The direction is along $xryx$.
$(f)$ No current is induced because the magnetic field lines are parallel to the plane of the loop,resulting in zero magnetic flux through the loop at all times.

(N/A) According to Lenz's law,the direction of the induced emf is such that it opposes the change in magnetic flux that produced it.
$(a)$ As the wire expands to form a circle,the area enclosed by the loop increases. Since the magnetic field is directed into the page (represented by crosses),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,this corresponds to an anticlockwise direction of current,i.e.,along $adcba$.
$(b)$ When the circular loop is deformed into a narrow straight wire,the area enclosed by the loop decreases,leading to a decrease in the magnetic flux through the loop. To oppose this decrease,the induced current must create a magnetic field directed into the page. By the right-hand rule,this corresponds to a clockwise direction of current,i.e.,along $abcda$.

(B) The number of turns per unit length $n = 15 \; \text{turns/cm} = 1500 \; \text{turns/m}$.
The area of the small loop is $A = 2.0 \; \text{cm}^2 = 2.0 \times 10^{-4} \; \text{m}^2$.
The change in current $\Delta I = 4.0 \; \text{A} - 2.0 \; \text{A} = 2.0 \; \text{A}$.
The time interval $\Delta t = 0.1 \; \text{s}$.
The magnetic field inside a long solenoid is $B = \mu_0 n I$.
The magnetic flux through the loop is $\phi = BA = \mu_0 n I A$.
According to Faraday's law, the induced $emf$ is $e = \left| \frac{d\phi}{dt} \right| = \mu_0 n A \left( \frac{dI}{dt} \right)$.
Substituting the values:
$e = (4\pi \times 10^{-7} \; \text{T m/A}) \times (1500 \; \text{m}^{-1}) \times (2.0 \times 10^{-4} \; \text{m}^2) \times \left( \frac{2.0 \; \text{A}}{0.1 \; \text{s}} \right)$.
$e = (4 \times 3.14159 \times 10^{-7}) \times 1500 \times 2.0 \times 10^{-4} \times 20$.
$e = 7.5398 \times 10^{-6} \; \text{V} \approx 7.54 \times 10^{-6} \; \text{V}$.

(0.75 T) Given:
Area of the coil,$A = 2 \; cm^{2} = 2 \times 10^{-4} \; m^{2}$
Number of turns,$N = 25$
Total charge,$Q = 7.5 \; mC = 7.5 \times 10^{-3} \; C$
Total resistance,$R = 0.50 \; \Omega$
The induced emf is given by Faraday's law: $e = -N \frac{d\phi}{dt}$.
The induced current is $I = \frac{e}{R} = -\frac{N}{R} \frac{d\phi}{dt}$.
The total charge $Q$ is the integral of current over time: $Q = \int I \; dt = -\frac{N}{R} \int_{\phi_i}^{\phi_f} d\phi = -\frac{N}{R} (\phi_f - \phi_i)$.
Since the coil is removed from the field,the final flux $\phi_f = 0$ and the initial flux $\phi_i = BA$.
Thus,$Q = \frac{N \phi_i}{R} = \frac{NBA}{R}$.
Rearranging for the magnetic field $B$: $B = \frac{QR}{NA}$.
Substituting the values: $B = \frac{7.5 \times 10^{-3} \times 0.50}{25 \times 2 \times 10^{-4}} = \frac{3.75 \times 10^{-3}}{50 \times 10^{-4}} = \frac{3.75 \times 10^{-3}}{5 \times 10^{-3}} = 0.75 \; T$.
Therefore,the field strength of the magnet is $0.75 \; T$.

(N/A) For a long time,electricity and magnetism were considered separate and unrelated phenomena. In the early $19^{th}$ century,experiments by Oersted,Ampere,and others established that electricity and magnetism are interrelated,showing that moving electric charges produce magnetic fields.
Michael Faraday in England and Joseph Henry in the $USA$,around $1830$,demonstrated that electric currents are induced in closed coils when subjected to changing magnetic fields.
Electromagnetic induction is the phenomenon in which an electric current is generated in a conductor by a varying magnetic field.

(N/A) Faraday made public his discovery that relative motion between a bar magnet and a wire loop produces an electric current in the loop.
The phenomenon of electromagnetic induction is not merely of theoretical interest but is of immense practical utility. Imagine a world without electricity—no electric lights,no trains,no telephones,and no computers. Electromagnetic induction is the fundamental principle behind the generation of electricity.
The pioneering experiments of Faraday and Henry led directly to the development of modern-day generators,motors,and transformers. Today's civilization owes its progress to a great extent to the discovery of electromagnetic induction.

(A) The phenomenon of electromagnetic induction was discovered by $Michael \ Faraday$ in $1831$. He observed that a changing magnetic flux through a coil induces an electromotive force $(EMF)$ in it. Joseph Henry also discovered this phenomenon independently around the same time, but $Michael \ Faraday$ is primarily credited for the discovery and the subsequent laws.

(N/A) Electromagnetic induction is the phenomenon of generating an electromotive force $(EMF)$ or current in a conductor by changing the magnetic flux linked with it.
According to Faraday's law,the magnitude of the induced $EMF$ is equal to the rate of change of magnetic flux through the circuit,given by $\varepsilon = -\frac{d\Phi_B}{dt}$.
This phenomenon is the fundamental principle behind the operation of electric generators,transformers,and induction motors.

(N/A) The figure shows a coil $C_{1}$ connected to a galvanometer $G$.
When the North-pole of a bar magnet is pushed towards the coil,the pointer in the galvanometer deflects,indicating the presence of electric current in the coil. The deflection lasts as long as the bar magnet is in motion. When the magnet is pulled away from the coil,the galvanometer shows deflection in the opposite direction,which indicates a reversal of the current's direction. The galvanometer does not show any deflection when the magnet is held stationary. Moreover,when the South-pole of the bar magnet is moved towards or away from the coil,the deflections in the galvanometer are opposite to those observed with the North-pole for similar movements.
Further,the deflection (and hence current) is found to be larger when the magnet is pushed towards or pulled away from the coil faster.
Instead,when the bar magnet is held fixed and the coil $C_{1}$ is moved towards or away from the magnet,the same effects are observed.
It shows that it is the relative motion between the magnet and the coil that is responsible for the generation (induction) of electric current in the coil.

(N/A) In the figure,two coils $C_{1}$ and $C_{2}$ are shown. Coil $C_{1}$ is connected to a galvanometer $G$,and coil $C_{2}$ is connected to a battery.
The steady current in coil $C_{2}$ produces a steady magnetic field. As coil $C_{2}$ is moved towards coil $C_{1}$,the galvanometer shows a deflection. This indicates that an electric current is induced in coil $C_{1}$.
When $C_{2}$ is moved away,the galvanometer shows a deflection again,but this time in the opposite direction.
The deflection lasts as long as coil $C_{2}$ is in motion.
When coil $C_{2}$ is held fixed and $C_{1}$ is moved,the same effects are observed. Again,it is the relative motion between the coils that induces the electric current.

(N/A) The figure shows two coils $C_{1}$ and $C_{2}$ held stationary. Coil $C_{1}$ is connected to a galvanometer $G$,while the second coil $C_{2}$ is connected to a battery through a tapping key $K$.
It is observed that the galvanometer shows a momentary deflection when the tapping key $K$ is pressed. The pointer in the galvanometer returns to zero immediately. If the key is held pressed continuously,there is no deflection in the galvanometer.
When the key is released,a momentary deflection is observed again,but in the opposite direction. It is also observed that the deflection increases dramatically when an iron rod is inserted into the coils along their axis.
Thus,it can be concluded from this experiment that relative motion is not an absolute requirement to induce current.

(B) No,relative motion is not an absolute condition for inducing an $emf$. According to Faraday's Law of Induction,an $emf$ is induced whenever there is a change in the magnetic flux linked with a circuit. This change in magnetic flux can be achieved in several ways:
$1$. By changing the magnetic field $B$ with time (e.g.,using an alternating current in a nearby coil).
$2$. By changing the area $A$ of the loop in a magnetic field.
$3$. By changing the orientation (angle $\theta$) of the loop relative to the magnetic field.
Therefore,even if there is no relative motion between the source of the magnetic field and the conductor,an $emf$ can be induced if the magnetic field itself is time-varying.

(A) According to Faraday's Law of Electromagnetic Induction, the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$.
When a magnet is moved towards a coil with a greater velocity, the rate of change of magnetic flux $(\frac{d\phi}{dt})$ linked with the coil increases.
Since the induced electromotive force is directly proportional to the rate of change of magnetic flux, the induced electromotive force $(e)$ increases.
According to Ohm's Law, $I = \frac{e}{R}$, where $R$ is the resistance of the coil.
Since $e$ increases and $R$ remains constant, the induced current $(I)$ will increase.

(N/A) Faraday's law states: "The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit."
Mathematically, the induced emf $\varepsilon$ is given by:
$\varepsilon = -\frac{d \phi_{B}}{d t}$ ... $(1)$
The negative sign indicates the direction of $\varepsilon$ (Lenz's Law), which opposes the change in magnetic flux.
In the case of a closely wound coil of $N$ turns, the change of flux associated with each turn is the same. Therefore, the expression for the total induced emf is given by:
$\varepsilon = -N \frac{d \phi_{B}}{d t}$ ... $(2)$
The induced emf can be increased by increasing the number of turns $N$ of the coil or by increasing the rate of change of magnetic flux $\frac{d \phi_{B}}{d t}$.

(B) Faraday's law of electromagnetic induction is given by the formula $\varepsilon = -\frac{d\Phi_B}{dt}$.
The negative sign in this equation is a mathematical representation of Lenz's Law.
Lenz's Law states that the direction of the induced current (and thus the induced $EMF$) is such that it always opposes the change in magnetic flux that produced it.
Therefore,the negative sign signifies the opposition to the change in magnetic flux.

(A) Michael Faraday discovered the phenomenon of electromagnetic induction in $1831$. He demonstrated that a changing magnetic field through a closed loop induces an electromotive force $(EMF)$,which in turn creates an electric field. This is summarized by Faraday's law of induction,which states that the induced electric field is proportional to the rate of change of the magnetic flux.

(10 V) According to Faraday's law of electromagnetic induction,the induced emf $\varepsilon$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Given $\phi = 3t^2 - 2t + 5$.
Differentiating with respect to $t$,we get $\frac{d\phi}{dt} = \frac{d}{dt}(3t^2 - 2t + 5) = 6t - 2$.
The magnitude of induced emf is $|\varepsilon| = |6t - 2|$.
At $t = 2 \ s$,$|\varepsilon| = |6(2) - 2| = |12 - 2| = 10 \ V$.

(N/A) Lenz's law states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
Proof of Conservation of Energy:
Consider a bar magnet being pushed towards a closed coil. As the North-pole of the magnet moves towards the coil,the magnetic flux through the coil increases. According to Lenz's law,the induced current in the coil will flow in a direction that creates a North-pole on the face of the coil towards the magnet. This creates a repulsive force between the magnet and the coil.
To continue moving the magnet towards the coil,we must do work against this repulsive force. This mechanical work done by the external agent is converted into electrical energy,which manifests as the induced current in the coil. If the current were to flow in the opposite direction (attracting the magnet),the magnet would accelerate towards the coil,increasing the flux and the current further,leading to an infinite increase in energy without any work done,which violates the law of conservation of energy. Thus,Lenz's law is a direct consequence of the law of conservation of energy.

(C) 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 opposition means that work must be done against the magnetic force to move a magnet or change the current in a circuit.
This work done is converted into electrical energy in the circuit.
Therefore,Lenz's law is a direct consequence of the law of conservation of energy.