Mutual Induction Questions in English

(B) According to the principle of mutual induction,the magnetic flux $\phi$ linked with a coil is proportional to the current $I$ flowing through the other coil,given by $\phi = MI$,where $M$ is the coefficient of mutual induction.
For the first case,when current $I_P = 3\, A$ flows through coil $P$,the flux through coil $Q$ is $\phi_Q = 10^{-3}\, Wb$.
Using the relation $\phi_Q = M I_P$:
$10^{-3} = M \times 3$
$M = \frac{1}{3} \times 10^{-3}\, H$
For the second case,when current $I_Q = 2\, A$ flows through coil $Q$,the flux through coil $P$ is $\phi_P = M I_Q$.
Since the coefficient of mutual induction $M$ is the same for both cases:
$\phi_P = (\frac{1}{3} \times 10^{-3}) \times 2$
$\phi_P = \frac{2}{3} \times 10^{-3}\, Wb$
$\phi_P = 0.666... \times 10^{-3}\, Wb = 6.67 \times 10^{-4}\, Wb$.

(B) Let a current $i$ flow through the large square loop of side $L$. The magnetic field $B$ produced by the large loop at its center is approximately uniform over the area of the small loop of side $l$.
The magnetic field at the center of a square loop of side $L$ is given by $B = \frac{2\sqrt{2}\mu_0 i}{\pi L}$.
The magnetic flux $\phi$ linked with the small loop is $\phi = B \times A$,where $A = l^2$ is the area of the small loop.
So,$\phi = \left(\frac{2\sqrt{2}\mu_0 i}{\pi L}\right) l^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{i}$.
Therefore,$M = \frac{2\sqrt{2}\mu_0 l^2}{\pi L}$.
Since $\mu_0$ and $\pi$ are constants,$M \propto \frac{l^2}{L}$.

(A) The magnetic flux $\phi$ linked with a coil is related to the current $I$ in the neighboring coil by the relation $\phi = MI$,where $M$ is the mutual inductance.
For a change in current $dI$ and a corresponding change in flux $d\phi$,the relation is $d\phi = M \cdot dI$.
Given: $dI = 2 \ A$ and $d\phi = 0.4 \ Wb$.
Substituting the values into the formula:
$0.4 = M \times 2$
$M = \frac{0.4}{2} = 0.2 \ H$.
Therefore,the mutual inductance of the coils is $0.2 \ H$.

(B) The induced $emf$ $e$ in the secondary coil is given by the formula:
$|e| = M \left| \frac{dI}{dt} \right|$
Given:
Mutual inductance $M = 5\, H$
Change in current $dI = 10\, A - 0\, A = 10\, A$
Time interval $dt = 5 \times 10^{-4}\, s$
Substituting the values into the formula:
$|e| = 5 \times \frac{10}{5 \times 10^{-4}}$
$|e| = \frac{50}{5 \times 10^{-4}}$
$|e| = 10 \times 10^4 = 1 \times 10^5\, V$
Therefore,the induced $emf$ is $1 \times 10^5\, V$.

(A) The mutual inductance $M$ of the solenoid and the coil is given by the formula: $M = \frac{\mu_{0} N_{1} N_{2} A}{l}$.
Substituting the given values: $N_{1} = 2000$,$N_{2} = 300$,$A = 1.2 \times 10^{-3} \, m^2$,$l = 0.3 \, m$,and $\mu_{0} = 4\pi \times 10^{-7} \, T \cdot m/A$.
$M = \frac{4 \pi \times 10^{-7} \times 2000 \times 300 \times 1.2 \times 10^{-3}}{0.3} = 3.016 \times 10^{-3} \, H \approx 3 \times 10^{-3} \, H$.
The induced emf $\varepsilon$ is given by $\varepsilon = -M \frac{dI}{dt}$.
The current changes from $1 \, A$ to $-1 \, A$,so $dI = -1 - 1 = -2 \, A$.
The time interval $dt = 0.25 \, s$.
$\varepsilon = - (3 \times 10^{-3}) \times \left( \frac{-2}{0.25} \right) = 3 \times 10^{-3} \times 8 = 24 \times 10^{-3} \, V = 24 \, mV$.

(C) The induced $emf$ $(e)$ in the secondary coil is given by the formula:
$e = M \left| \frac{dI}{dt} \right|$
Given:
Change in current $(dI)$ = $2\,A - 0\,A = 2\,A$
Time interval $(dt)$ = $0.01\,s$
Induced $emf$ $(e)$ = $1000\,V$
Substituting the values into the formula:
$1000 = M \times \left( \frac{2}{0.01} \right)$
$1000 = M \times 200$
$M = \frac{1000}{200} = 5\,H$.
Therefore,the mutual inductance of the two coils is $5\,H$.

(N/A) Let a current $I_2$ flow through the outer circular coil. The magnetic field at the center of the coil is $B_2 = \frac{\mu_0 I_2}{2 r_2}$.
Since the inner coil has a very small radius $(r_1 \ll r_2)$,the magnetic field $B_2$ can be considered uniform over the area of the inner coil.
The magnetic flux $\phi_1$ through the inner coil is given by $\phi_1 = B_2 \cdot A_1 = B_2 (\pi r_1^2)$.
Substituting the value of $B_2$,we get $\phi_1 = \left( \frac{\mu_0 I_2}{2 r_2} \right) (\pi r_1^2) = \left( \frac{\mu_0 \pi r_1^2}{2 r_2} \right) I_2$.
By definition,$\phi_1 = M_{12} I_2$,where $M_{12}$ is the mutual inductance.
Therefore,$M_{12} = \frac{\mu_0 \pi r_1^2}{2 r_2}$.
Since the mutual inductance is reciprocal,$M_{12} = M_{21} = M = \frac{\mu_0 \pi r_1^2}{2 r_2}$.

(D) The mutual inductance $M$ of a pair of coils is given as $1.5\; H$.
The change in current in the first coil is $\Delta I = I_2 - I_1 = 20\; A - 0\; A = 20\; A$.
The relationship between the magnetic flux linkage $\phi$ in the second coil and the current $I$ in the first coil is given by $\phi = M I$.
Therefore,the change in flux linkage $\Delta \phi$ is given by:
$\Delta \phi = M \Delta I$
Substituting the given values:
$\Delta \phi = 1.5\; H \times 20\; A$
$\Delta \phi = 30\; Wb$.
Thus,the change in the flux linkage with the other coil is $30\; Wb$.

(N/A) Mutual induction is the phenomenon where a change in current in one coil induces an $emf$ in a neighboring coil due to the change in magnetic flux linked with it.
As shown in the figure,when the current $I_{2}$ in coil $C_{2}$ changes,the magnetic flux $\Phi_{1}$ linked with coil $C_{1}$ also changes,inducing an $emf$ in coil $C_{1}$.
If the number of turns in coil $C_{1}$ is $N_{1}$,then the total flux linkage is proportional to the current in $C_{2}$:
$N_{1} \Phi_{1} \propto I_{2}$
$N_{1} \Phi_{1} = M_{12} I_{2}$
where $M_{12}$ is the coefficient of mutual induction.
According to Faraday's law of electromagnetic induction,the induced $emf$ $\varepsilon_{1}$ in coil $C_{1}$ is:
$\varepsilon_{1} = -\frac{d(N_{1} \Phi_{1})}{dt}$
Substituting the expression for flux linkage:
$\varepsilon_{1} = -\frac{d}{dt}(M_{12} I_{2})$
Assuming $M_{12}$ is constant:
$\varepsilon_{1} = -M_{12} \frac{dI_{2}}{dt}$
Similarly,for coil $C_{2}$,the induced $emf$ is:
$\varepsilon_{2} = -M_{21} \frac{dI_{1}}{dt}$

(N/A) The magnetic flux $\Phi_{2}$ linked with coil-$2$ when current $I_{1}$ flows through coil-$1$ is given by $\Phi_{2} = M_{21} I_{1}$.
Taking $I_{1} = 1 \text{ unit}$, we get $\Phi_{2} = M_{21}$. Thus, "The magnetic flux linked with one of the coils per unit current passing through the other coil is called the mutual inductance of the system."
The induced emf $\varepsilon_{2}$ in coil-$2$ is given by $\varepsilon_{2} = -M_{21} \frac{dI_{1}}{dt}$.
When $\frac{dI_{1}}{dt} = 1 \text{ unit}$, we get $\varepsilon_{2} = M_{21}$. Thus, "The mutual emf generated in one of the two coils due to a unit rate of change of current in the other coil is called the mutual inductance of the system."
The $SI$ unit of mutual inductance is the henry $(H)$, where $1 \text{ H} = 1 \text{ Wb A}^{-1} = 1 \text{ V s A}^{-1}$.
The value of mutual inductance depends on:
$(1)$ Shape and size of the coils.
$(2)$ Number of turns in the coils.
$(3)$ Distance between the coils.
$(4)$ Relative orientation (angle of inclination) of the coils.
$(5)$ Magnetic permeability of the core material on which the coils are wound.

(N/A) Consider two long coaxial solenoids,each of length $l$. Let the radius of the inner solenoid $S_{1}$ be $r_{1}$ and the number of turns per unit length be $n_{1}$. The corresponding quantities for the outer solenoid $S_{2}$ are $r_{2}$ and $n_{2}$ respectively. Let $N_{1}$ and $N_{2}$ be the total number of turns of coils $S_{1}$ and $S_{2}$ respectively.
When a current $I_{2}$ flows through $S_{2}$,it sets up a magnetic field $B_{2} = \mu_{0} n_{2} I_{2}$ inside it. This field passes through the inner solenoid $S_{1}$.
The magnetic flux $\Phi_{1}$ through each turn of $S_{1}$ is $\Phi_{1} = B_{2} A_{1} = (\mu_{0} n_{2} I_{2})(\pi r_{1}^{2})$.
The total flux linkage with $S_{1}$ is $N_{1} \Phi_{1} = (n_{1} l) (\mu_{0} n_{2} I_{2} \pi r_{1}^{2}) = (\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l) I_{2}$.
By definition,$N_{1} \Phi_{1} = M_{12} I_{2}$,where $M_{12}$ is the mutual inductance of $S_{1}$ with respect to $S_{2}$.
Thus,$M_{12} = \mu_{0} n_{1} n_{2} \pi r_{1}^{2} l$.
Similarly,if a current $I_{1}$ flows through $S_{1}$,the flux linkage with $S_{2}$ is $N_{2} \Phi_{2} = M_{21} I_{1}$,where $M_{21} = \mu_{0} n_{1} n_{2} \pi r_{1}^{2} l$.
Since $M_{12} = M_{21} = M$,this demonstrates the reciprocity theorem,which states that the mutual inductance of two coils is the same regardless of which coil is used as the primary or secondary.

Consider two long coaxial solenoids of length $l$. Let the inner solenoid have $N_1$ turns and radius $r_1$, and the outer solenoid have $N_2$ turns and radius $r_2$.
Let $n_1 = N_1/l$ and $n_2 = N_2/l$ be the number of turns per unit length for the two solenoids respectively.
When a current $I_2$ flows through the outer solenoid, the magnetic field produced inside it is $B_2 = \mu_0 n_2 I_2$.
This magnetic field is uniform and confined to the volume of the inner solenoid.
The magnetic flux linked with each turn of the inner solenoid is $\phi_1 = B_2 A_1 = (\mu_0 n_2 I_2) (\pi r_1^2)$.
The total magnetic flux linked with the inner solenoid having $N_1$ turns is $\Phi_1 = N_1 \phi_1 = N_1 (\mu_0 n_2 I_2) (\pi r_1^2)$.
Substituting $N_1 = n_1 l$, we get $\Phi_1 = (n_1 l) (\mu_0 n_2 I_2) (\pi r_1^2) = \mu_0 n_1 n_2 l \pi r_1^2 I_2$.
By definition, the mutual inductance $M$ is given by $\Phi_1 = M I_2$.
Therefore, $M = \mu_0 n_1 n_2 l \pi r_1^2$ or $M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l}$.

(A) The mutual inductance $M$ of the system of two coils is defined by the relationship between the flux through one coil and the current in the other.
Given: When $I_A = 2 \ A$,$\phi_B = 10^{-2} \ Wb$.
The mutual inductance $M_{BA}$ is given by $M_{BA} = \frac{\phi_B}{I_A} = \frac{10^{-2}}{2} = 5 \times 10^{-3} \ H$.
According to the principle of reciprocity,$M_{AB} = M_{BA} = 5 \times 10^{-3} \ H$.
Now,when $I_A = 0$ and $I_B = 1 \ A$,the flux through coil $A$ is given by $\phi_A = M_{AB} \times I_B$.
Substituting the values: $\phi_A = (5 \times 10^{-3} \ H) \times (1 \ A) = 5 \times 10^{-3} \ Wb$.

(N/A) The magnetic field produced by a long solenoid is given by $B = \mu_0 n I$.
The magnetic flux $\phi$ linked with the smaller coil is $\phi = N A B$,where $A$ is the area of the smaller coil.
$\phi = N (\pi (b/2)^2) (\mu_0 n I) = \frac{\mu_0 N n \pi b^2 I}{4}$.
The induced emf $\varepsilon$ in the smaller coil is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt}$.
$\varepsilon = -\frac{d}{dt} \left( \frac{\mu_0 N n \pi b^2 I}{4} \right) = -\frac{\mu_0 N n \pi b^2}{4} \frac{dI}{dt}$.
Given the current varies as $I(t) = mt^2 + C$,we have $\frac{dI}{dt} = 2mt$.
Substituting this into the emf equation:
$\varepsilon = -\frac{\mu_0 N n \pi b^2}{4} (2mt) = -\left( \frac{\mu_0 N n \pi b^2 m}{2} \right) t$.
The magnitude of the induced emf is $|\varepsilon| = \left( \frac{\mu_0 N n \pi b^2 m}{2} \right) t$.
This represents a linear relationship between $|\varepsilon|$ and $t$,which is a straight line passing through the origin.

(A) The magnetic flux $\phi$ linked with the bigger loop due to the smaller loop is given by the formula for mutual induction between two loops where the smaller loop acts as a magnetic dipole:
$\phi = B \cdot A_{2} = \left( \frac{\mu_{0} I R_{1}^{2}}{2(R_{1}^{2} + x^{2})^{3/2}} \right) \cdot (\pi R_{2}^{2})$
Here,$R_{1} = 0.3 \, cm = 0.3 \times 10^{-2} \, m$ (radius of smaller loop),
$R_{2} = 20 \, cm = 0.2 \, m$ (radius of bigger loop),
$x = 15 \, cm = 0.15 \, m$ (distance between centers),
$I = 20 \, A$ (current in smaller loop).
Since $R_{1} \ll x$,we can approximate the field of the smaller loop as a dipole field,but using the exact formula for the flux through the larger loop due to the smaller one:
$\phi = \frac{\mu_{0} I \pi R_{1}^{2} R_{2}^{2}}{2(R_{1}^{2} + x^{2})^{3/2}}$
Substituting the values:
$\phi = \frac{(4\pi \times 10^{-7}) \times 20 \times \pi \times (0.3 \times 10^{-2})^{2} \times (0.2)^{2}}{2((0.3 \times 10^{-2})^{2} + (0.15)^{2})^{3/2}}$
$\phi \approx \frac{4\pi^{2} \times 10^{-7} \times 20 \times 9 \times 10^{-6} \times 0.04}{2(0.15)^{3}}$
$\phi \approx 9.1 \times 10^{-11} \, Wb$.

(A) The magnetic field $B$ at the center due to a square loop of side $b$ carrying current $I$ is given by the sum of the fields due to its four sides.
For one side,the field at the center is $B_{1} = \frac{\mu_{0} I}{4 \pi (b/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_{0} I}{2 \pi b} (2 \times \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} \mu_{0} I}{\pi b}$.
Since there are four sides,the total magnetic field at the center is $B = 4 \times B_{1} = \frac{4 \sqrt{2} \mu_{0} I}{\pi b}$.
Since $b \gg a$,we assume the magnetic field is uniform over the area of the smaller loop.
The magnetic flux $\phi$ linked with the smaller loop of area $A = a^{2}$ is $\phi = B \times A = \frac{4 \sqrt{2} \mu_{0} I}{\pi b} \times a^{2}$.
The coefficient of mutual inductance $M$ is defined as $M = \frac{\phi}{I} = \frac{4 \sqrt{2} \mu_{0} a^{2}}{\pi b}$.
To match the given options,we multiply and divide by $4$:
$M = \frac{\mu_{0}}{4 \pi} \times 16 \sqrt{2} \frac{a^{2}}{b}$.
Wait,re-evaluating the field calculation: $B_{side} = \frac{\mu_{0} I}{4 \pi d} (\sin \theta_1 + \sin \theta_2)$. Here $d = b/2$,$\theta_1 = \theta_2 = 45^{\circ}$.
$B_{side} = \frac{\mu_{0} I}{4 \pi (b/2)} (2 \sin 45^{\circ}) = \frac{\mu_{0} I}{2 \pi b} \sqrt{2} = \frac{\mu_{0} I}{\sqrt{2} \pi b}$.
Total $B = 4 \times \frac{\mu_{0} I}{\sqrt{2} \pi b} = \frac{2 \sqrt{2} \mu_{0} I}{\pi b}$.
Flux $\phi = B a^{2} = \frac{2 \sqrt{2} \mu_{0} I a^{2}}{\pi b}$.
$M = \frac{\phi}{I} = \frac{2 \sqrt{2} \mu_{0} a^{2}}{\pi b} = \frac{\mu_{0}}{4 \pi} \frac{8 \sqrt{2} a^{2}}{b}$.

(D) The magnetic field $B_{1}$ at the center of a circular loop of radius $R_{1}$ carrying current $I_{1}$ is given by $B_{1} = \frac{\mu_{0} I_{1}}{2 R_{1}}$.
The magnetic flux $\phi_{2}$ linked with the smaller loop of radius $R_{2}$ is $\phi_{2} = B_{1} A_{2}$,where $A_{2} = \pi R_{2}^{2}$ is the area of the smaller loop.
Since $R_{1} >> R_{2}$,we can assume the magnetic field $B_{1}$ is uniform over the area of the smaller loop.
The mutual inductance $M$ is defined as $M = \frac{\phi_{2}}{I_{1}}$.
Substituting the expressions,we get $M = \frac{B_{1} A_{2}}{I_{1}} = \frac{(\frac{\mu_{0} I_{1}}{2 R_{1}}) (\pi R_{2}^{2})}{I_{1}} = \frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}$.
Thus,$M \propto \frac{R_{2}^{2}}{R_{1}}$.

(C) The induced electromotive force $(e_{2})$ in a secondary coil is given by the formula: $e_{2} = -M \frac{di_{1}}{dt}$.
Here,$M$ is the mutual inductance and $i_{1}$ is the current in the primary coil.
Rearranging for $M$,we get: $M = -\frac{e_{2}}{di_{1}/dt}$.
The dimensional formula for induced emf $(e_{2})$ is the same as potential difference,which is $[ML^{2}T^{-3}A^{-1}]$.
The dimensional formula for the rate of change of current $(di_{1}/dt)$ is $[AT^{-1}]$.
Therefore,the dimension of $M$ is: $[M] = \frac{[ML^{2}T^{-3}A^{-1}]}{[AT^{-1}]} = [ML^{2}T^{-2}A^{-2}]$.

(C) Let a current $I$ flow through the outer square loop of side $L$. The magnetic field $B$ produced by this loop at its centre $O$ is given by the sum of the fields due to its four sides:
$B = 4 \times \left( \frac{\mu_{0} I}{4 \pi (L/2)} \times (\sin 45^{\circ} + \sin 45^{\circ}) \right) = 4 \times \left( \frac{\mu_{0} I}{2 \pi L} \times 2 \times \frac{1}{\sqrt{2}} \right) = \frac{2 \sqrt{2} \mu_{0} I}{\pi L}$.
Since $L \gg l$,we can assume this magnetic field is uniform over the area of the small inner loop.
The magnetic flux $\phi$ linked with the inner loop of side $l$ is:
$\phi = B \times \text{Area} = \left( \frac{2 \sqrt{2} \mu_{0} I}{\pi L} \right) \times l^{2}$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{2 \sqrt{2} \mu_{0} l^{2}}{\pi L}$.

(B) To calculate the mutual inductance $M$ of the system,we use the principle of reciprocity,which states that $M_{12} = M_{21} = M$.
Let a current $I_{1}$ flow through the long outer solenoid (solenoid $1$). The magnetic field produced by this solenoid inside it is uniform and given by $B = \mu_{0} N I_{1}$.
The magnetic flux $\phi_{21}$ linked with the short inner solenoid (solenoid $2$) is the product of the number of turns in the inner solenoid,the magnetic field,and the cross-sectional area of the inner solenoid.
Number of turns in the inner solenoid = $n \times l$.
Area of the inner solenoid = $\pi r^{2}$.
Therefore,$\phi_{21} = (n l) \times B \times (\pi r^{2}) = (n l) \times (\mu_{0} N I_{1}) \times (\pi r^{2})$.
By definition,$\phi_{21} = M I_{1}$.
Comparing the two expressions,we get $M = \mu_{0} N n l \pi r^{2} = \pi \mu_{0} r^{2} n N l$.

(D) The mutual inductance $M$ is given as $5 \,mH = 5 \times 10^{-3} \,H$.
The current in the primary coil is $i = 50 \sin(500t) \,A$.
The induced e.m.f. $\varepsilon$ in the secondary coil is given by $\varepsilon = -M \frac{di}{dt}$.
Taking the magnitude,$\varepsilon = M \left| \frac{di}{dt} \right|$.
$\frac{di}{dt} = \frac{d}{dt} [50 \sin(500t)] = 50 \times 500 \cos(500t) = 25000 \cos(500t)$.
Therefore,$\varepsilon = (5 \times 10^{-3}) \times (25000 \cos(500t)) = 125 \cos(500t) \,V$.
The peak value of the induced e.m.f. is the coefficient of the cosine term,which is $125 \,V$.

(B) When two coils are wound on the same iron core,the magnetic flux $\phi$ passing through each turn of both coils is the same because the iron core provides a path of high permeability for the magnetic field lines.
However,the total magnetic flux linkage for a coil with $N$ turns is given by $\lambda = N\phi$.
For coil $A$,the flux linkage is $\lambda_A = N_A \phi$.
For coil $B$,the flux linkage is $\lambda_B = N_B \phi$.
The ratio of the flux linkages is $\frac{\lambda_A}{\lambda_B} = \frac{N_A \phi}{N_B \phi} = \frac{N_A}{N_B}$.
According to Faraday's law of induction,the induced emf $E$ is given by $E = -\frac{d\lambda}{dt} = -N \frac{d\phi}{dt}$.
Therefore,the ratio of induced emf is $\frac{E_A}{E_B} = \frac{N_A (d\phi/dt)}{N_B (d\phi/dt)} = \frac{N_A}{N_B}$.
Comparing this with the given options,option $(b)$ correctly describes the relationship between the flux linkages (often referred to as the magnetic flux linked with the coils).

(A) The magnetic field $B$ at the center of a large circular loop of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.
Since the square loop is small and placed at the center,we can assume the magnetic field is uniform over the area of the square loop.
The magnetic flux $\phi$ linked with the square loop of side $l$ is $\phi = B \cdot A$,where $A = l^2$ is the area of the square loop.
Thus,$\phi = \left( \frac{\mu_0 i}{2r} \right) l^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{i}$.
Substituting the expression for $\phi$,we get $M = \frac{(\mu_0 i / 2r) l^2}{i} = \frac{\mu_0 l^2}{2r}$.
Therefore,the mutual inductance $M$ is proportional to $l^2/r$.

(A) The magnetic field $B$ produced by the smaller loop (radius $r = 0.3 \ cm = 0.003 \ m$) at a distance $x = 15 \ cm = 0.15 \ m$ along its axis is given by: $B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.
Since $r \ll x$,we can approximate $B \approx \frac{\mu_0 I r^2}{2x^3}$.
The magnetic flux $\phi$ linked with the bigger loop (radius $R = 20 \ cm = 0.2 \ m$) is $\phi = B \times A_{big} = B \times (\pi R^2)$.
Substituting the values: $B = \frac{(4\pi \times 10^{-7} \ T \cdot m/A) \times (2.0 \ A) \times (0.003 \ m)^2}{2 \times (0.15 \ m)^3} = \frac{4\pi \times 10^{-7} \times 2 \times 9 \times 10^{-6}}{2 \times 0.003375} = \frac{72\pi \times 10^{-13}}{0.00675} \approx 3.35 \times 10^{-9} \ T$.
However,using the reciprocity theorem,the flux through the bigger loop due to the smaller loop is the same as the flux through the smaller loop due to the bigger loop: $\phi = M I$.
Using $M = \frac{\mu_0 \pi r^2 R^2}{2(R^2 + x^2)^{3/2}} = \frac{4\pi \times 10^{-7} \times \pi \times (0.003)^2 \times (0.2)^2}{2 \times ((0.2)^2 + (0.15)^2)^{3/2}} = \frac{4\pi^2 \times 10^{-7} \times 9 \times 10^{-6} \times 0.04}{2 \times (0.0625)^{3/2}} = \frac{14.21 \times 10^{-12}}{0.03125} \approx 4.55 \times 10^{-10} \ H$.
Flux $\phi = M \times I = 4.55 \times 10^{-10} \times 2.0 = 9.1 \times 10^{-10} \ Wb = 91 \times 10^{-11} \ Wb$. Given the options,$9.1$ is the intended answer.

(C) The magnetic field $B$ at the center of a square loop of side $L$ carrying current $i$ is given by the sum of the fields due to its four sides.
For one side,the field at the center is $B_1 = \frac{\mu_0 i}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2 \pi L} (2 \cdot \frac{1}{\sqrt{2}}) = \frac{\sqrt{2} \mu_0 i}{2 \pi L} = \frac{\mu_0 i}{\sqrt{2} \pi L}$.
Since there are four sides,the total magnetic field at the center is $B = 4 \cdot \frac{\mu_0 i}{\sqrt{2} \pi L} = \frac{2 \sqrt{2} \mu_0 i}{\pi L}$.
Since $L \gg R$,we assume the magnetic field is uniform over the area of the small circular loop.
The magnetic flux $\phi$ through the circular loop of area $A = \pi R^2$ is $\phi = B \cdot A = \left( \frac{2 \sqrt{2} \mu_0 i}{\pi L} \right) (\pi R^2) = \frac{2 \sqrt{2} \mu_0 R^2 i}{L}$.
By definition,$\phi = Mi$,therefore $M = \frac{2 \sqrt{2} \mu_0 R^2}{L}$.

(C) Given:
Radius of inner coil,$r_1 = 1\,cm = 0.01\,m$
Number of turns in inner coil,$N_1 = 10$
Radius of outer coil,$r_2 = 1000\,cm = 10\,m$
Number of turns in outer coil,$N_2 = 200$
The magnetic field produced by the outer coil at its center is $B_2 = \frac{\mu_0 N_2 I_2}{2 r_2}$.
The magnetic flux linked with the inner coil due to the outer coil is $\phi_1 = N_1 B_2 A_1$,where $A_1 = \pi r_1^2$.
Thus,$\phi_1 = N_1 \left( \frac{\mu_0 N_2 I_2}{2 r_2} \right) (\pi r_1^2) = M I_2$.
Therefore,the mutual inductance $M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2}$.
Substituting the values:
$M = \frac{(4 \pi \times 10^{-7}) \times 10 \times 200 \times \pi \times (0.01)^2}{2 \times 10}$
$M = \frac{4 \pi^2 \times 10^{-7} \times 2000 \times 10^{-4}}{20}$
Using $\pi^2 = 10$:
$M = \frac{4 \times 10 \times 10^{-7} \times 2000 \times 10^{-4}}{20} = 4 \times 10^{-8}\,H$.
Thus,the value is $4$.

(C) The magnetic flux linked with the second coil is given by $\phi = Mi = M i_0 \sin \omega t$.
The induced $emf$ in the second coil is given by Faraday's law: $emf = -\frac{d\phi}{dt} = -M \frac{di}{dt}$.
Substituting the given expression for current: $emf = -M \frac{d}{dt}(i_0 \sin \omega t) = -M i_0 \omega \cos \omega t$.
The maximum value of the induced $emf$ is $|emf|_{max} = M i_0 \omega$.
Given $M = 0.002 \ H$,$i_0 = 5 \ A$,and $\omega = 50 \pi \ rad/s$:
$|emf|_{max} = (0.002) \times (5) \times (50 \pi) = 0.01 \times 50 \pi = 0.5 \pi = \frac{\pi}{2} \ V$.
Comparing this with $\frac{\pi}{\alpha} \ V$,we get $\alpha = 2$.

(C) The magnetic field $B$ at the center of a large square loop of side $L$ carrying current $i$ is given by:
$B = 4 \times \frac{\mu_0 i}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{\pi L/2} \times \sqrt{2} = \frac{2\sqrt{2} \mu_0 i}{\pi L}$.
Since $L \gg \ell$, we assume the magnetic field is uniform over the area of the small loop.
The magnetic flux $\phi$ through the small loop of area $\ell^2$ is:
$\phi = B \times \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi L} \ell^2$.
Given $L = \ell^2$, we substitute this into the expression:
$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi \ell^2} \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi}$.
The mutual inductance $M$ is defined as $M = \phi / i$:
$M = \frac{2\sqrt{2} \mu_0}{\pi} = \frac{2\sqrt{2} (4\pi \times 10^{-7})}{\pi} = 8\sqrt{2} \times 10^{-7} \text{ H}$.
$M = \sqrt{64 \times 2} \times 10^{-7} \text{ H} = \sqrt{128} \times 10^{-7} \text{ H}$.
Thus, $x = 128$.

(A) Let a current $i$ flow through the outer loop $B$ of radius $b$.
The magnetic field at the center $O$ due to the current $i$ in loop $B$ is given by $B_{center} = \frac{\mu_0 i}{2b}$.
Since $b >> a$,we can assume this magnetic field is uniform over the area of the inner loop $A$ of radius $a$.
The magnetic flux $\phi$ linked with the inner loop $A$ is $\phi = B_{center} \cdot A_{area} = \left( \frac{\mu_0 i}{2b} \right) (\pi a^2)$.
By definition of mutual inductance,$\phi = Mi$.
Equating the two expressions for flux: $Mi = \frac{\mu_0 i \pi a^2}{2b}$.
Therefore,the mutual inductance $M$ is $M = \frac{\mu_0 \pi a^2}{2b}$.

(B) The magnetic field $B$ on the axis of a circular loop of radius $R$ at a distance $X$ from the center is given by:
$B = \frac{\mu_0 i R^2}{2(R^2 + X^2)^{3/2}}$
Given $X = \sqrt{3} R$,the magnetic field at the center of the square loop is:
$B = \frac{\mu_0 i R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 i R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 i R^2}{2 \cdot 8 R^3} = \frac{\mu_0 i}{16 R}$
Since the square loop has $N = 2$ turns,area $A = a^2$,and the angle between the area vector and the magnetic field (which is along the $z$-axis) is $\theta = 45^{\circ}$,the magnetic flux $\phi$ is:
$\phi = N B A \cos(45^{\circ}) = 2 \cdot \left(\frac{\mu_0 i}{16 R}\right) \cdot a^2 \cdot \frac{1}{\sqrt{2}} = \frac{\mu_0 i a^2}{8 \sqrt{2} R}$
Since $\sqrt{2} = 2^{1/2}$ and $8 = 2^3$,the denominator is $2^3 \cdot 2^{1/2} = 2^{7/2}$.
Thus,$\phi = \frac{\mu_0 i a^2}{2^{7/2} R}$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{i} = \frac{\mu_0 a^2}{2^{7/2} R}$.
Comparing this with the given expression $\frac{\mu_0 a^2}{2^{p/2} R}$,we get $p = 7$.

(C) In the steady state,the currents in the two circuits are:
$I_1 = \frac{V_1}{R_1} = \frac{5 \text{ V}}{5 \text{ }\Omega} = 1 \text{ A}$
$I_2 = \frac{V_2}{R_2} = \frac{20 \text{ V}}{10 \text{ }\Omega} = 2 \text{ A}$
The total energy stored in the system of two coupled inductors is given by:
$U = \frac{1}{2} L_1 I_1^2 + \frac{1}{2} L_2 I_2^2 + M I_1 I_2$
Substituting the given values $(L_1 = 10 \text{ mH}, L_2 = 20 \text{ mH}, M = 5 \text{ mH}, I_1 = 1 \text{ A}, I_2 = 2 \text{ A})$:
$U = \frac{1}{2} \times (10 \times 10^{-3}) \times (1)^2 + \frac{1}{2} \times (20 \times 10^{-3}) \times (2)^2 + (5 \times 10^{-3}) \times 1 \times 2$
$U = 5 \times 10^{-3} + 40 \times 10^{-3} + 10 \times 10^{-3}$
$U = 55 \times 10^{-3} \text{ J} = 55 \text{ mJ}$

(C) The total magnetic flux $\phi_1$ linked with coil $1$ is due to its own current $I_1$ and the current $I_2$ in the nearby coil $2$.
This is given by the expression: $\phi_1 = L_1 I_1 + M_{12} I_2$.
According to Faraday's law of electromagnetic induction,the induced emf $\varepsilon_1$ in coil $1$ is the negative rate of change of magnetic flux:
$\varepsilon_1 = -\frac{d\phi_1}{dt}$.
Substituting the expression for $\phi_1$:
$\varepsilon_1 = -\frac{d}{dt}(L_1 I_1 + M_{12} I_2)$.
Assuming $L_1$ and $M_{12}$ are constants,we get:
$\varepsilon_1 = -L_1 \frac{dI_1}{dt} - M_{12} \frac{dI_2}{dt}$.

(A) The formula for magnetic flux linkage in terms of mutual inductance is given by $\Delta \phi = M \Delta I$.
Given values are:
Change in magnetic flux,$\Delta \phi = 2 \times 10^{-2} \ Wb$
Change in current,$\Delta I = 0.01 \ A$
Rearranging the formula to solve for the coefficient of mutual inductance $M$:
$M = \frac{\Delta \phi}{\Delta I}$
Substituting the values:
$M = \frac{2 \times 10^{-2}}{0.01} = \frac{0.02}{0.01} = 2 \ H$
Therefore,the coefficient of mutual inductance is $2 \ H$.

(B) Given: Peak current $I_{0} = \frac{2}{\pi} \ A$,frequency $f = 50 \ Hz$,and mutual inductance $M = 1 \ H$.
The angular frequency is $\omega = 2 \pi f = 2 \pi (50) = 100 \pi \ rad/s$.
The current in the primary coil is $I = I_{0} \sin(\omega t)$.
The induced e.m.f. in the secondary coil is $\mathcal{E} = M \frac{dI}{dt}$.
Differentiating the current with respect to time,$\frac{dI}{dt} = I_{0} \omega \cos(\omega t)$.
The peak value of the induced e.m.f. is $\mathcal{E}_{0} = M \cdot I_{0} \cdot \omega$.
Substituting the values: $\mathcal{E}_{0} = 1 \times (\frac{2}{\pi}) \times (100 \pi) = 200 \ V$.

(D) transformer is an electrical device that transfers electrical energy between two or more circuits through electromagnetic induction.
It consists of two coils,the primary coil and the secondary coil,which are magnetically coupled.
When an alternating current flows through the primary coil,it produces a changing magnetic flux.
This changing magnetic flux is linked to the secondary coil,inducing an electromotive force $(EMF)$ in it.
This phenomenon,where a change in current in one coil induces an $EMF$ in a nearby coil,is known as mutual induction.
Therefore,a transformer works on the principle of mutual induction.

(D) The induced e.m.f. in the first coil is given by $e_1 = M \frac{dI_2}{dt}$.
Given $e_1 = 20 \,mV = 20 \times 10^{-3} \,V$ and $\frac{dI_2}{dt} = 10 \,A/s$.
Substituting these values, we get $M = \frac{e_1}{dI_2/dt} = \frac{20 \times 10^{-3}}{10} = 2 \times 10^{-3} \,H$.
The flux linkage in the second coil when current $I_1$ flows through the first coil is given by $\phi_2 = M I_1$.
Given $I_1 = 3.6 \,A$, we have $\phi_2 = (2 \times 10^{-3} \,H) \times (3.6 \,A) = 7.2 \times 10^{-3} \,Wb$.

(D) The induced e.m.f. in the second coil is given by the formula: $e = M \frac{dI_1}{dt} \dots (i)$
Given $I_1 = I_0 \sin \omega t$, we differentiate with respect to time $t$:
$\frac{dI_1}{dt} = I_0 \omega \cos \omega t$
Substituting this into equation $(i)$, we get:
$e = M I_0 \omega \cos \omega t$
The maximum value of e.m.f. $(e_{max})$ occurs when $\cos \omega t = 1$:
$e_{max} = M I_0 \omega$
Given values: $M = 5 \times 10^{-3} \text{ H}$, $I_0 = 10 \text{ A}$, and $\omega = 100 \pi \text{ rad/s}$.
Substituting these values:
$e_{max} = (5 \times 10^{-3}) \times 10 \times 100 \pi$
$e_{max} = 5 \times 10^{-3} \times 10^3 \pi$
$e_{max} = 5 \pi \text{ V}$

(D) The instantaneous current in the primary coil is given by $I = I_0 \sin(\omega t)$.
Given peak current $I_0 = \frac{2}{\pi} \text{ A}$ and frequency $v = 50 \text{ Hz}$.
The angular frequency is $\omega = 2 \pi v = 2 \pi(50) = 100 \pi \text{ rad/s}$.
The rate of change of current is $\frac{dI}{dt} = I_0 \omega \cos(\omega t)$.
The maximum value of the rate of change of current is $\left(\frac{dI}{dt}\right)_{\text{max}} = I_0 \omega$.
Substituting the values: $\left(\frac{dI}{dt}\right)_{\text{max}} = \left(\frac{2}{\pi}\right) \times (100 \pi) = 200 \text{ A/s}$.
The induced e.m.f. in the secondary coil is $E = M \left|\frac{dI}{dt}\right|$,where $M = 1 \text{ H}$.
Therefore,the peak induced e.m.f. is $E_0 = 1 \times 200 = 200 \text{ V}$.

(B) The induced e.m.f. in coil $S$ due to a change in current in coil $P$ is given by Faraday's law of induction: $E = -M \frac{dI}{dt}$.
Given,mutual inductance $M = 3 \times 10^{-3} \ H$ and current $I = 20 \sin(50 \pi t) \ A$.
Differentiating the current with respect to time $t$: $\frac{dI}{dt} = 20 \times 50 \pi \cos(50 \pi t) = 1000 \pi \cos(50 \pi t) \ A/s$.
Substituting these values into the e.m.f. equation: $E = -(3 \times 10^{-3}) \times (1000 \pi \cos(50 \pi t)) = -3 \pi \cos(50 \pi t) \ V$.
The maximum value of the induced e.m.f. is $|E_{max}| = 3 \pi \ V$.
Using $\pi \approx 3.14$,we get $|E_{max}| = 3 \times 3.14 = 9.42 \ V$.

(C) The magnetic field lines produced by a circular coil are perpendicular to the plane of the coil at its center.
Since the two coils $A$ and $B$ are placed mutually perpendicular,the magnetic field lines produced by coil $A$ will lie in the plane of coil $B$,and vice-versa.
The magnetic flux $\Phi$ is given by $\Phi = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is perpendicular to the plane of the coil).
Since the magnetic field lines are parallel to the plane of the other coil,they are perpendicular to the area vector of that coil. Thus,$\theta = 90^\circ$ and $\cos 90^\circ = 0$.
Therefore,the magnetic flux linked with the other coil is zero.
According to Faraday's law of electromagnetic induction,the induced $EMF$ is $\varepsilon = -d\Phi/dt$. Since $\Phi = 0$ at all times,the induced $EMF$ and consequently the induced current will be zero.

(A) Given: Mutual inductance $M = 0.01 \ H$,Current $I = 5 \sin(200 \pi t)$.
The induced e.m.f. in the second coil is given by Faraday's law of induction: $\varepsilon = -M \frac{dI}{dt}$.
Substituting the expression for current:
$\varepsilon = -0.01 \cdot \frac{d}{dt} [5 \sin(200 \pi t)]$
$\varepsilon = -0.01 \cdot 5 \cdot 200 \pi \cdot \cos(200 \pi t)$
$\varepsilon = -10 \pi \cos(200 \pi t)$.
The maximum value of the induced e.m.f. is the magnitude of the coefficient of the cosine term:
$\varepsilon_{max} = | -10 \pi | = 10 \pi \ V$.

(B) The magnetic field $B$ at the center of a circular coil of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.
Since $r_1 > r_2$,the magnetic field produced by the larger coil is approximately uniform over the area of the smaller coil.
The magnetic flux $\phi$ linked with the smaller coil of radius $r_2$ is $\phi = B \cdot A = \left( \frac{\mu_0 I}{2 r_1} \right) (\pi r_2^2)$.
The mutual inductance $M$ is defined by the relation $\phi = M I$.
Therefore,$M = \frac{\phi}{I} = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

(D) The formula for induced e.m.f. in the secondary coil due to mutual induction is given by:
$|\varepsilon| = M \cdot \left| \frac{dI}{dt} \right|$
Given:
Mutual inductance $M = 2 \text{ H}$
Induced e.m.f. $\varepsilon = 2 \text{ kV} = 2000 \text{ V}$
Change in current $\Delta I = 6 \text{ A} - 3 \text{ A} = 3 \text{ A}$
Substituting the values into the formula:
$2000 = 2 \cdot \left( \frac{3}{\Delta t} \right)$
$2000 = \frac{6}{\Delta t}$
$\Delta t = \frac{6}{2000} \text{ s}$
$\Delta t = 3 \times 10^{-3} \text{ s}$
Therefore,the time required is $3 \times 10^{-3} \text{ s}$.

(A) The induced e.m.f. in coil $P$ due to the changing current in coil $Q$ is given by $\varepsilon_P = M \frac{dI_Q}{dt}$,where $M$ is the mutual inductance.
Given $\varepsilon_P = 15 \ mV = 15 \times 10^{-3} \ V$ and $\frac{dI_Q}{dt} = 10 \ A/s$.
$15 \times 10^{-3} = M \times 10 \implies M = 1.5 \times 10^{-3} \ H = 1.5 \ mH$.
The magnetic flux $\phi_Q$ linked with coil $Q$ when current $I_P$ flows through coil $P$ is given by $\phi_Q = M I_P$.
Given $I_P = 1.8 \ A$ and $M = 1.5 \ mH$.
$\phi_Q = 1.5 \ mH \times 1.8 \ A = 2.7 \ mWb$.

(C) The coefficient of mutual inductance $M$ is defined by the relation $\Delta \phi = M \Delta I$,where $\Delta \phi$ is the change in magnetic flux and $\Delta I$ is the change in current.
Given:
Initial flux $\phi_1 = 6.5 \times 10^{-2} \ Wb$
Final flux $\phi_2 = 11 \times 10^{-2} \ Wb$
Change in flux $\Delta \phi = \phi_2 - \phi_1 = (11 - 6.5) \times 10^{-2} \ Wb = 4.5 \times 10^{-2} \ Wb$
Change in current $\Delta I = 0.03 \ A = 3 \times 10^{-2} \ A$
Using the formula $M = \frac{\Delta \phi}{\Delta I}$:
$M = \frac{4.5 \times 10^{-2}}{3 \times 10^{-2}} = \frac{4.5}{3} = 1.5 \ H$
Therefore,the coefficient of mutual inductance is $1.5 \ H$.

(D) The induced emf in coil $P$ due to the changing current in coil $Q$ is given by $\epsilon_P = M \frac{dI_Q}{dt}$,where $M$ is the mutual inductance.
Given $\epsilon_P = 12 \ mV = 12 \times 10^{-3} \ V$ and $\frac{dI_Q}{dt} = 10 \ A/s$.
$12 \times 10^{-3} = M \times 10 \implies M = 1.2 \times 10^{-3} \ H = 1.2 \ mH$.
Now,the magnetic flux $\phi_Q$ linked with coil $Q$ when current $I_P = 1.5 \ A$ flows through coil $P$ is given by $\phi_Q = M \times I_P$.
$\phi_Q = 1.2 \ mH \times 1.5 \ A = 1.8 \ mWb$.
Thus,the correct option is $D$.