Mutual Induction Questions in English

(D) The magnetic field at the center of a circular loop of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.
Since the rings are concentric and coplanar,the magnetic field $B$ is uniform over the area of the smaller ring (radius $r_2$).
The magnetic flux $\phi$ linked with the smaller ring is $\phi = B \times A_2$,where $A_2 = \pi r_2^2$ is the area of the smaller ring.
$\phi = \left( \frac{\mu_0 I}{2 r_1} \right) \times \pi r_2^2 = \frac{\mu_0 \pi r_2^2}{2 r_1} I$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

(C) The mutual inductance $M$ between two coils with self-inductances $L_1$ and $L_2$ is given by the formula $M = k \sqrt{L_1 L_2}$,where $k$ is the coefficient of coupling.
Since the flux in one coil is completely linked with the other,the coupling is perfect,meaning $k = 1$.
Given $L_1 = 25 \ mH$ and $L_2 = 9 \ mH$.
Substituting these values into the formula:
$M = \sqrt{25 \ mH \times 9 \ mH} = \sqrt{225 \ mH^2} = 15 \ mH$.

(B) The induced e.m.f. in the second coil is given by the formula: $|e_s| = M \left| \frac{dI_p}{dt} \right|$.
Given $I_p = I_0 \sin \omega t$, we differentiate with respect to time $t$:
$\frac{dI_p}{dt} = I_0 \omega \cos \omega t$.
Substituting this into the e.m.f. equation:
$|e_s| = M I_0 \omega \cos \omega t$.
The maximum value of e.m.f. occurs when $\cos \omega t = 1$:
$|e_s|_{\max} = M I_0 \omega$.
Substituting the given values $M = 0.003 \ H$, $I_0 = 8 \ A$, and $\omega = 100 \pi \ rad \ s^{-1}$:
$|e_s|_{\max} = 0.003 \times 8 \times 100 \pi = 2.4 \pi \ V$.

(D) The magnetic field at the center of a circular loop of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.
Since $r_2 \ll r_1$,the magnetic field $B$ is approximately uniform over the area of the smaller loop.
The magnetic flux $\phi$ linked with the smaller loop of radius $r_2$ is $\phi = B \times A_2$,where $A_2 = \pi r_2^2$ is the area of the smaller loop.
Substituting the values,we get $\phi = \left( \frac{\mu_0 I}{2 r_1} \right) \times (\pi r_2^2) = \frac{\mu_0 \pi r_2^2}{2 r_1} I$.
The mutual inductance $M$ is defined by the relation $\phi = M I$.
Therefore,$M = \frac{\phi}{I} = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

(A) The formula for induced e.m.f. in the secondary coil due to mutual induction is given by:
$e_s = M \cdot \frac{dI_p}{dt}$
Where:
$M = 2 \ H$ (coefficient of mutual induction)
$e_s = 2 \ kV = 2 \times 10^3 \ V$
$dI_p = 6 \ A - 3 \ A = 3 \ A$
Rearranging the formula to solve for time $(dt)$:
$dt = M \cdot \frac{dI_p}{e_s}$
Substituting the values:
$dt = 2 \times \frac{3}{2 \times 10^3} \ s$
$dt = 3 \times 10^{-3} \ s$

(C) The formula for mutual inductance $M$ between two coils is given by $M = K \sqrt{L_1 L_2}$,where $K$ is the coefficient of coupling.
Given values are $M = 45 \ mH$,$L_1 = 75 \ mH$,and $L_2 = 48 \ mH$.
Rearranging the formula to solve for $K$:
$K = \frac{M}{\sqrt{L_1 L_2}}$
Substituting the values:
$K = \frac{45}{\sqrt{75 \times 48}}$
$K = \frac{45}{\sqrt{3600}}$
$K = \frac{45}{60}$
$K = 0.75$

(C) The coefficient of coupling $(K)$ is defined by the formula:
$K = \frac{M}{\sqrt{L_1 L_2}}$
Given:
$M = 3 \ H$
$L_1 = 4 \ H$
$L_2 = 9 \ H$
Substituting the values into the formula:
$K = \frac{3}{\sqrt{4 \times 9}}$
$K = \frac{3}{\sqrt{36}}$
$K = \frac{3}{6}$
$K = 0.5$

(B) The induced e.m.f. in coil $B$ is given by the formula $e = M \frac{dI}{dt}$.
Given: $M = 0.008 \ H$, $I = I_{m} \sin \omega t$, $I_{m} = 5 \ A$, and $\omega = 200 \pi \ rad \ s^{-1}$.
Differentiating the current with respect to time: $\frac{dI}{dt} = I_{m} \omega \cos \omega t$.
Substituting this into the e.m.f. equation: $e = M I_{m} \omega \cos \omega t$.
The maximum value of e.m.f. $(e_{\max})$ occurs when $\cos \omega t = 1$.
Therefore, $e_{\max} = M I_{m} \omega$.
Substituting the values: $e_{\max} = 0.008 \times 5 \times 200 \pi$.
$e_{\max} = 0.04 \times 200 \pi = 8 \pi \ V$.

(A) Let $I_1$ be the current flowing through the coil of radius $r_1$.
The magnetic field produced at the center of this coil is given by $B_1 = \frac{\mu_0 I_1}{2 r_1}$.
Since $r_2 \ll r_1$,the magnetic field $B_1$ is approximately uniform over the area of the smaller coil.
The magnetic flux $\phi_2$ passing through the coil of radius $r_2$ is $\phi_2 = B_1 \cdot A_2 = B_1 \cdot \pi r_2^2$.
Substituting the value of $B_1$,we get $\phi_2 = \left( \frac{\mu_0 I_1}{2 r_1} \right) \pi r_2^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I_1}$.
Thus,$M = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

(D) The total induced e.m.f. in the secondary coil due to mutual inductance is given by the formula: $e_{total} = M \frac{dI}{dt}$.
Given that the current changes from $I$ to $0$ in time $t$,the magnitude of the induced e.m.f. is $e_{total} = M \frac{I}{t}$.
Since the coil has $N$ turns,the total e.m.f. is distributed across these $N$ turns.
Therefore,the e.m.f. induced per turn is given by: $e_{per\ turn} = \frac{e_{total}}{N} = \frac{MI}{Nt}$.
Thus,the correct option is $D$.

(B) The magnetic field at the center of a large coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 R}$.
Since $R \gg r$,the magnetic field $B$ is approximately uniform over the area of the smaller coil of radius $r$.
The magnetic flux $\phi$ passing through the smaller coil is $\phi = B \times A$,where $A = \pi r^2$ is the area of the smaller coil.
Substituting the values,we get $\phi = \left( \frac{\mu_0 I}{2 R} \right) \times \pi r^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{\frac{\mu_0 I}{2 R} \times \pi r^2}{I} = \frac{\mu_0 \pi r^2}{2 R}$.

(D) The induced e.m.f. in the second coil is given by the formula $|e_s| = M \frac{dI}{dt}$.
Given $I = I_0 \sin \omega t$,we differentiate with respect to time $t$:
$\frac{dI}{dt} = I_0 \omega \cos \omega t$.
Substituting this into the e.m.f. equation:
$|e_s| = M I_0 \omega \cos \omega t$.
The maximum value of the induced e.m.f. occurs when $\cos \omega t = 1$,so $|e_s|_{\max} = M I_0 \omega$.
Substituting the given values: $M = 0.004 \ H$,$I_0 = 10 \ A$,and $\omega = 50 \pi \ rad \ s^{-1}$:
$|e_s|_{\max} = 0.004 \times 10 \times 50 \pi = 2 \pi \ V$.

(D) The magnetic flux $\Phi$ linked with coil $B$ due to current $i_1$ in coil $A$ is given by the relation: $\Phi = M i_1$.
Here, $M$ is the coefficient of mutual inductance.
Given that the change in current $\Delta i_1 = 0.8 \,A$ results in a change in magnetic flux $\Delta \Phi = 0.16 \,Wb$.
The mutual inductance $M$ is calculated as:
$M = \frac{\Delta \Phi}{\Delta i_1} = \frac{0.16 \,Wb}{0.8 \,A} = 0.2 \,H$.

(C) The change in current in the primary coil is given by $di_1 = (8 - 4) \,A = 4 \,A$.
The time interval is $dt = 0.6 \,s$.
The induced e.m.f. in the secondary coil is $E_2 = 50 \,mV = 50 \times 10^{-3} \,V$.
The formula for mutual inductance $M$ is given by $E_2 = M \cdot \frac{di_1}{dt}$.
Rearranging for $M$, we get $M = \frac{E_2 \cdot dt}{di_1}$.
Substituting the values: $M = \frac{50 \times 10^{-3} \,V \times 0.6 \,s}{4 \,A}$.
$M = \frac{30 \times 10^{-3}}{4} \,H = 7.5 \times 10^{-3} \,H = 7.5 \,mH$.

(A) The induced emf $e$ in the neighboring coil is given by $e = M \frac{dI}{dt}$.
Given $I = 10 \sin(100 \pi t)$, we find the rate of change of current:
$\frac{dI}{dt} = 10 \times 100 \pi \cos(100 \pi t) = 1000 \pi \cos(100 \pi t)$.
Substituting this into the emf equation:
$e = M \times 1000 \pi \cos(100 \pi t)$.
The maximum value of the induced emf $e_0$ occurs when $\cos(100 \pi t) = 1$, so $e_0 = 1000 \pi M$.
Given $e_0 = 5 \pi \text{ V}$, we have $1000 \pi M = 5 \pi$.
Solving for $M$: $M = \frac{5 \pi}{1000 \pi} = 0.005 \text{ H} = 5 \text{ mH}$.

(D) The magnetic field $B$ at the center of the smaller coil $A$ due to the current $i$ in the larger coil $B$ is approximately uniform over the area of coil $A$.
The magnetic field produced by coil $B$ at its center is $B = \frac{\mu_{0} i}{2 R_{2}}$.
The magnetic flux $\phi$ linked with the smaller coil $A$ is $\phi = B \times A_{1} = \left( \frac{\mu_{0} i}{2 R_{2}} \right) (\pi R_{1}^{2})$.
By definition,the mutual inductance $M$ is given by $M = \frac{\phi}{i} = \frac{\mu_{0} \pi R_{1}^{2}}{2 R_{2}}$.
Therefore,$M \propto \frac{R_{1}^{2}}{R_{2}}$.

(D) If a current $I_{1}$ flows in the outer coil of radius $r_{1}$,the magnetic field at the center is given by $B_{1} = \frac{\mu_{0} I_{1}}{2 r_{1}}$.
Since $r_{2} \ll r_{1}$,we assume the magnetic field $B_{1}$ is uniform over the area of the smaller coil.
The magnetic flux $\phi_{2}$ passing through the inner coil of radius $r_{2}$ is $\phi_{2} = B_{1} \times A_{2} = B_{1} \times \pi r_{2}^{2}$.
Substituting the value of $B_{1}$,we get $\phi_{2} = \left( \frac{\mu_{0} I_{1}}{2 r_{1}} \right) \times \pi r_{2}^{2}$.
The mutual inductance $M$ is defined as $M = \frac{\phi_{2}}{I_{1}}$.
Therefore,$M = \frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}$.

(D) The magnetic field $B$ at the center of a large coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_{0} I}{2R}$.
Since $R \gg r$,we can assume the magnetic field is uniform over the area of the smaller coil.
The magnetic flux $\phi$ passing through the smaller coil of radius $r$ is $\phi = B \cdot A = B \cdot (\pi r^{2})$.
Substituting the value of $B$,we get $\phi = \left( \frac{\mu_{0} I}{2R} \right) \cdot (\pi r^{2})$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{\mu_{0} \pi r^{2}}{2R}$.
Since $\mu_{0}$,$\pi$,and $2$ are constants,we have $M \propto \frac{r^{2}}{R}$.

(C) The mutual inductance $M$ between two coils is related to their self-inductances $L_{1}$ and $L_{2}$ by the formula $M = k\sqrt{L_{1} L_{2}}$,where $k$ is the coefficient of coupling.
Given that the flux in one coil is completely linked with the other,the coupling is perfect,meaning $k = 1$.
Therefore,the expression simplifies to $M = \sqrt{L_{1} L_{2}}$.

(A) The magnetic flux linkage $\phi$ in the second coil is related to the current $I$ in the first coil by the relation $\phi = M I$,where $M$ is the mutual inductance.
For a change in current $\Delta I$,the change in flux linkage $\Delta \phi$ is given by:
$\Delta \phi = M \Delta I$
Given:
$M = 1.5 \ H$
$\Delta I = I_f - I_i = 10 \ A - 0 \ A = 10 \ A$
Substituting the values:
$\Delta \phi = 1.5 \ H \times 10 \ A$
$\Delta \phi = 15 \ Wb$
Therefore,the change in flux linkage is $15 \ Wb$.

(B) The flux linkage $\phi$ in a coil due to current $I$ in an adjacent coil is given by $\phi = M I$,where $M$ is the mutual inductance.
The change in flux linkage $\Delta \phi$ is given by:
$\Delta \phi = M \Delta I$
Given:
Mutual inductance $M = 1.5 \ H$
Change in current $\Delta I = I_{final} - I_{initial} = 20 \ A - 0 \ A = 20 \ A$
Substituting the values:
$\Delta \phi = 1.5 \times 20$
$\Delta \phi = 30 \ Wb$
Therefore,the change in flux linkage is $30 \ Wb$.

(D) When we pass a current $I$ through the outer loop of radius $R$,the magnetic field at the centre is given by:
$B_{\text{center}} = \frac{\mu_0 I}{2 R}$
Since $r \ll R$,the magnetic field $B$ is approximately uniform over the area of the smaller loop $A$.
The magnetic flux $\phi$ associated with the smaller loop is:
$\phi = B \cdot A = \left( \frac{\mu_0 I}{2 R} \right) \times (\pi r^2)$
$\phi = \frac{\mu_0 I \pi r^2}{2 R}$
The mutual inductance $M$ is defined as the ratio of the magnetic flux $\phi$ through the secondary coil to the current $I$ in the primary coil:
$M = \frac{\phi}{I} = \frac{\mu_0 \pi r^2}{2 R}$

(B) The mutual inductance $M$ between two coils is defined by the relationship between the change in magnetic flux $\Delta \phi$ linked with one coil and the change in current $\Delta I$ in the other coil.
The formula is given by:
$M = \frac{\Delta \phi}{\Delta I}$
Given values:
Change in current in coil $X$,$\Delta I = 4 \ A$
Change in magnetic flux in coil $Y$,$\Delta \phi = 0.4 \ Wb$
Substituting these values into the formula:
$M = \frac{0.4 \ Wb}{4 \ A}$
$M = 0.1 \ H$
Therefore,the mutual inductance of the system is $0.1 \ H$.

(D) The correct answer is $D$.
Mutual inductance $(M)$ of a system of two coils depends on the geometric factors such as the number of turns $(N_1, N_2)$, the area of cross-section $(A)$, the distance between the coils, and the magnetic permeability $(\mu)$ of the medium between them.
It is defined by the relation $\phi_2 = M I_1$, where $\phi_2$ is the magnetic flux linked with the second coil and $I_1$ is the current in the first coil.
Since $M = \frac{\phi_2}{I_1}$, the value of $M$ is independent of the magnitude of the current $I_1$ flowing through the coils.

(A) The magnetic field $B$ produced by an infinitely long wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.
The magnetic flux $\phi$ through the square loop is $\phi = \int B \cdot dA = \int_{x}^{x+a} \frac{\mu_0 I}{2\pi r} (a \, dr) = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{x+a}{x}\right)$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I} = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x}\right)$.
Since the loop is moving at a constant speed $v$,the distance $x$ increases with time $t$ as $x = x_0 + vt$.
Substituting this into the expression for $M$,we get $M(t) = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x_0 + vt}\right)$.
As $t$ increases,the term $\frac{a}{x_0 + vt}$ decreases,and therefore $\ln(1 + \frac{a}{x_0 + vt})$ decreases.
This corresponds to a curve that starts at a maximum value and decreases asymptotically towards zero as $t \to \infty$,which matches the graph shown in option $A$.

(B) Given: Mutual inductance $ M = 0.005 \ H $,current $ i = i_{m} \sin \omega t $,peak current $ i_{m} = 10 \ A $,and angular frequency $ \omega = 100 \pi \ rad \ s^{-1} $.
The induced emf $ \varepsilon $ in the second coil is given by $ \varepsilon = M \frac{di}{dt} $.
Substituting the expression for current: $ \varepsilon = M \frac{d}{dt} (i_{m} \sin \omega t) = M i_{m} \omega \cos \omega t $.
The maximum value of the induced emf $ \varepsilon_{\max} $ occurs when $ \cos \omega t = 1 $.
Therefore,$ \varepsilon_{\max} = M \omega i_{m} $.
Substituting the values: $ \varepsilon_{\max} = 0.005 \times 100 \pi \times 10 = 5 \pi \ V $.
Thus,the maximum value of the emf induced in the second coil is $ 5 \pi \ V $.

(B) Given: Mutual inductance $M = 3 \text{ mH} = 3 \times 10^{-3} \text{ H}$,Resistance of circuit $Y$ $R_2 = 4 \text{ } \Omega$,Induced current in $Y$ $I_2 = 60 \times 10^{-4} \text{ A}$,Time interval $\Delta t = 0.02 \text{ s}$.
The induced electromotive force $(e_2)$ in circuit $Y$ due to change in current in circuit $X$ is given by $e_2 = M \frac{\Delta I_1}{\Delta t}$.
The induced current in circuit $Y$ is $I_2 = \frac{e_2}{R_2}$.
Substituting the expression for $e_2$,we get $I_2 = \frac{M \cdot \Delta I_1}{R_2 \cdot \Delta t}$.
Rearranging to solve for the change in current $\Delta I_1$ in circuit $X$:
$\Delta I_1 = \frac{I_2 \cdot R_2 \cdot \Delta t}{M}$.
Substituting the values:
$\Delta I_1 = \frac{60 \times 10^{-4} \times 4 \times 0.02}{3 \times 10^{-3}}$.
$\Delta I_1 = \frac{60 \times 10^{-4} \times 0.08}{3 \times 10^{-3}} = \frac{4.8 \times 10^{-4}}{3 \times 10^{-3}} = 1.6 \times 10^{-1} = 0.16 \text{ A}$.

(C) The coefficient of mutual induction $M$ between two coils is defined by the relation between the change in magnetic flux linkage in the secondary coil and the change in current in the primary coil.
$M = \frac{\Delta \phi_2}{\Delta I_1}$
Given:
Change in flux linkage in the second coil, $\Delta \phi_2 = 40 \,Wb$
Change in current in the first coil, $\Delta I_1 = 16 \,A - 0 \,A = 16 \,A$
Substituting these values into the formula:
$M = \frac{40}{16} \,H$
$M = 2.5 \,H$
Therefore, the value of $M$ is $2.5 \,H$.

(A) The mutual inductance $M$ of two co-axial solenoids is given by the formula: $M = \mu_0 n_1 n_2 A l$,where $n_1$ and $n_2$ are the number of turns per unit length,$A$ is the cross-sectional area of the inner solenoid,and $l$ is the length of the solenoids.
Given:
Length $l = 60 \ cm = 0.6 \ m$
Turns per unit length of outer solenoid $n_1 = 15 \ \text{turns/cm} = 1500 \ \text{turns/m}$
Turns per unit length of inner solenoid $n_2 = 40 \ \text{turns/cm} = 4000 \ \text{turns/m}$
Area of inner solenoid $A = 2 \times 10^{-3} \ m^2$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$
Substituting the values:
$M = (4\pi \times 10^{-7}) \times 1500 \times 4000 \times (2 \times 10^{-3}) \times 0.6$
$M = (4 \times 3.14 \times 10^{-7}) \times (6 \times 10^6) \times (1.2 \times 10^{-3})$
$M = 12.56 \times 10^{-7} \times 7.2 \times 10^3$
$M \approx 9.04 \times 10^{-3} \ H = 9 \ mH$.

(B) Given:
$N_1 = 2000$,$L = 0.30 \ m$,$N_2 = 300$,$A = 1.2 \times 10^{-3} \ m^2$.
The rate of change of current is $\frac{di}{dt} = \frac{I_f - I_i}{\Delta t} = \frac{-2 - 2}{0.25} = -16 \ A/s$. The magnitude is $16 \ A/s$.
The mutual inductance $M$ of the solenoid-coil system is given by $M = \frac{\mu_0 N_1 N_2 A}{L}$.
Substituting the values:
$M = \frac{4\pi \times 10^{-7} \times 2000 \times 300 \times 1.2 \times 10^{-3}}{0.30} = 3.016 \times 10^{-3} \ H$.
The induced emf $e$ is given by $e = M \left| \frac{di}{dt} \right|$.
$e = (3.016 \times 10^{-3}) \times 16 = 4.825 \times 10^{-2} \ V \approx 4.8 \times 10^{-2} \ V$.

(B) Given: The coefficient of mutual inductance $M = 0.2 H$.
The rate of change of current in the primary coil is $\frac{dI}{dt} = 5 A s^{-1}$.
The induced emf $(e)$ in the secondary coil is given by the formula:
$e = M \frac{dI}{dt}$
Substituting the given values:
$e = 0.2 H \times 5 A s^{-1}$
$e = 1 V$
Therefore,the induced emf in the secondary coil is $1 V$.

(C) The induced emf $(e_s)$ in the secondary coil is given by the relation $e_s = -M \frac{di_p}{dt}$,where $M$ is the coefficient of mutual induction and $\frac{di_p}{dt}$ is the rate of change of current in the primary coil.
If the rate of change of current in the primary coil is unit,i.e.,$\frac{di_p}{dt} = 1 \ A/s$,then the magnitude of induced emf is equal to the coefficient of mutual induction $(|e_s| = M)$.
Therefore,the emf induced in the secondary coil due to a unit rate of change of current in the primary coil is defined as the mutual induction (or mutual inductance) of the two coils.

(C) Consider two concentric coplanar circular loops. Let the outer loop have radius $R$ and the inner loop have radius $r$.
When a current $I$ flows through the outer loop,the magnetic field $B$ at its center is given by $B = \frac{\mu_0 I}{2R}$.
Since $R \gg r$,we can assume the magnetic field is uniform over the area of the smaller loop.
The magnetic flux $\phi$ linked with the smaller loop is $\phi = B \cdot A$,where $A = \pi r^2$ is the area of the smaller loop.
Thus,$\phi = \left( \frac{\mu_0 I}{2R} \right) (\pi r^2) = \left( \frac{\mu_0 \pi r^2}{2R} \right) I$.
By definition,the mutual inductance $M$ is given by $\phi = MI$.
Comparing the two expressions,we get $M = \frac{\mu_0 \pi r^2}{2R}$.
Therefore,the mutual inductance $M$ is proportional to $\frac{r^2}{R}$.

(C) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.
Consider a small rectangular strip of width $dr$ at a distance $r$ from the wire within the frame. The area of this strip is $dA = b \cdot dr$.
The magnetic flux $d\phi$ through this strip is $d\phi = B \cdot dA = \left( \frac{\mu_0 I}{2 \pi r} \right) (b \cdot dr) = \frac{\mu_0 I b}{2 \pi} \frac{dr}{r}$.
To find the total magnetic flux $\phi$ through the frame,we integrate from $r = a$ to $r = 2a$:
$\phi = \int_a^{2a} \frac{\mu_0 I b}{2 \pi} \frac{dr}{r} = \frac{\mu_0 I b}{2 \pi} [\ln r]_a^{2a} = \frac{\mu_0 I b}{2 \pi} \ln \left( \frac{2a}{a} \right) = \frac{\mu_0 I b}{2 \pi} \ln 2$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Therefore,$M = \frac{\mu_0 b}{2 \pi} \ln 2$.

(C) The magnetic field at the center of a large circular coil of radius $r_2$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_2}$.
Since $r_1 \ll r_2$,we assume the magnetic field $B$ is uniform over the area of the smaller coil of radius $r_1$.
The magnetic flux $\phi$ linked with the smaller coil is $\phi = B \cdot A = \left(\frac{\mu_0 I}{2 r_2}\right) \cdot (\pi r_1^2)$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Substituting the expression for $\phi$,we get $M = \frac{\mu_0 \pi r_1^2}{2 r_2}$.

(A) Given: Mutual inductance $M = 8 \ mH = 8 \times 10^{-3} \ H$. Current $I = 12 \sin 100t$.
The induced emf in the second coil is given by the formula $\varepsilon = M \frac{dI}{dt}$.
Substituting the expression for $I$:
$\varepsilon = M \frac{d}{dt} (12 \sin 100t)$
$\varepsilon = M \times 12 \times 100 \cos 100t$
$\varepsilon = 1200 M \cos 100t$.
The maximum value of induced emf occurs when $\cos 100t = 1$:
$\varepsilon_{\max} = 1200 \times M$
$\varepsilon_{\max} = 1200 \times 8 \times 10^{-3} \ V$
$\varepsilon_{\max} = 9.6 \ V$.

(B) The induced emf $\varepsilon$ in a coil due to the change in current in an adjacent coil is given by the formula:
$\varepsilon = M \left| \frac{dI}{dt} \right|$
Given:
Change in current,$dI = 10 \,A - 2 \,A = 8 \,A$
Time interval,$dt = 0.2 \,s$
Induced emf,$\varepsilon = 120 \,V$
Substituting the values into the formula:
$120 = M \times \frac{8}{0.2}$
$120 = M \times 40$
$M = \frac{120}{40} = 3 \,H$
Therefore,the mutual inductance of the pair of coils is $3 \,H$.

(C) The mutual inductance $M$ of two concentric coils where $r \ll R$ is given by the formula: $M = \frac{\mu_0 \pi N_1 N_2 r^2}{2 R}$.
Here,$N_1$ and $N_2$ are the number of turns in the inner and outer coils respectively,$r$ is the radius of the inner coil,and $R$ is the radius of the outer coil.
From the formula,we observe that $M \propto r^2$.
Using the concept of relative error,we have $\frac{\Delta M}{M} = 2 \frac{\Delta r}{r}$.
Given that the percentage change in radius is $\frac{\Delta r}{r} \times 100 \% = 2 \%$.
Therefore,the percentage change in mutual inductance is $\frac{\Delta M}{M} \times 100 \% = 2 \times (2 \%) = 4 \%$.

(B) Given: Mutual inductance $M = 0.005 \ H$,current $I = I_0 \sin \omega t$,$I_0 = 10 \ A$,and $\omega = 100 \pi \ rad/s$.
The induced emf $(e)$ in the second coil is given by the formula:
$e = M \frac{dI}{dt}$
Substituting the expression for current:
$e = M \frac{d}{dt} (I_0 \sin \omega t) = M I_0 \omega \cos \omega t$
The maximum value of the induced emf $(e_{\max})$ occurs when $\cos \omega t = 1$:
$e_{\max} = M I_0 \omega$
Substituting the given values:
$e_{\max} = 0.005 \times 10 \times 100 \pi$
$e_{\max} = 0.05 \times 100 \pi = 5 \pi \ V$
Therefore,the maximum induced emf is $5 \pi \ V$.

(A) The magnetic field $B$ produced by a large square loop of side $L$ at its center is given by $B = \frac{2\sqrt{2}\mu_0 I}{\pi L}$.
Since the small loop of side $l$ is placed at the center,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.
Thus,$\phi = \left( \frac{2\sqrt{2}\mu_0 I}{\pi L} \right) l^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I}$.
Substituting the expression for $\phi$,we get $M = \frac{2\sqrt{2}\mu_0 l^2}{\pi L}$.
Therefore,$M \propto \frac{l^2}{L}$.

(B) The induced emf $(e)$ in the secondary coil is given by the formula: $e = M \frac{di}{dt}$.
Given:
Mutual inductance $(M) = 92 \times 10^{-6} \, H$.
Rate of change of current $(\frac{di}{dt}) = \frac{25 \, A}{1 \, ms} = \frac{25 \, A}{1 \times 10^{-3} \, s} = 25,000 \, A/s$.
Substituting these values into the formula:
$e = 92 \times 10^{-6} \times 25,000$
$e = 92 \times 10^{-6} \times 25 \times 10^3$
$e = 92 \times 25 \times 10^{-3}$
$e = 2300 \times 10^{-3} = 2.3 \, V$.
Therefore, the induced emf in the secondary coil is $2.3 \, V$.

(D) Let a current $I$ flow through the large square loop of side $L$. The magnetic field $B$ produced by one side of the large loop at its centre is given by the formula for a finite wire: $B_{side} = \frac{\mu_0 I}{4 \pi d} (\sin \alpha + \sin \beta)$,where $d = L/2$ and $\alpha = \beta = 45^\circ$.
Since there are four such sides,the total magnetic field at the centre is $B = 4 \times \frac{\mu_0 I}{4 \pi (L/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{\pi L} \times 2 \times \frac{2}{\sqrt{2}} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$.
Since $L \gg l$,we can assume the magnetic field $B$ is uniform over the area of the smaller loop $S_2 = l^2$.
The magnetic flux linked with the smaller loop is $\phi_2 = B \times S_2 = \left( \frac{2 \sqrt{2} \mu_0 I}{\pi L} \right) l^2$.
The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I} = \frac{2 \sqrt{2} \mu_0 l^2}{\pi L}$.
Therefore,$M \propto \frac{l^2}{L}$.

(C) The 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:
$\Delta i = 0.01 \,A = 10^{-2} \,A$
$\Delta \phi = 2 \times 10^{-2} \,Wb$
Using the formula $M = \frac{\Delta \phi}{\Delta i}$:
$M = \frac{2 \times 10^{-2} \,Wb}{10^{-2} \,A} = 2 \,H$.
Therefore,the mutual inductance of the two coils is $2 \,H$.

(B) $1$. For the circular loop: The circumference is $2 \pi R = L$, so $R = \frac{L}{2 \pi}$.
$2$. For the square loop: The perimeter is $4s = a$, where $s$ is the side length of the square. Thus, $s = \frac{a}{4}$.
$3$. The magnetic field $B$ at the centre of the circular loop due to a current $I$ flowing through it is $B = \frac{\mu_{0} I}{2R}$.
$4$. Since $a \ll R$, the magnetic field $B$ is approximately uniform over the area of the square loop. The magnetic flux $\phi$ linked with the square loop is $\phi = B \times \text{Area of square} = B \times s^2$.
$5$. Substituting the values: $\phi = \left( \frac{\mu_{0} I}{2R} \right) \times s^2 = \left( \frac{\mu_{0} I}{2(L / 2 \pi)} \right) \times \left( \frac{a}{4} \right)^2 = \left( \frac{\mu_{0} I \pi}{L} \right) \times \frac{a^2}{16} = \frac{\mu_{0} \pi a^2}{16 L} I$.
$6$. The mutual inductance $M$ is given by $M = \frac{\phi}{I} = \frac{\mu_{0} \pi a^2}{16 L}$.

(C) The magnetic flux linkage $\phi$ in the secondary coil is related to the current $I$ in the primary coil by the formula $\phi = M I$,where $M$ is the mutual inductance.
To find the change in flux linkage $\Delta \phi$,we use the formula $\Delta \phi = M \Delta I$.
Given:
Mutual inductance $M = 2 \text{H}$.
Change in current $\Delta I = I_f - I_i = 30 \text{A} - 0 \text{A} = 30 \text{A}$.
Substituting these values into the formula:
$\Delta \phi = 2 \text{H} \times 30 \text{A} = 60 \text{ Wb}$.
Thus,the change in flux linkage with the other coil is $60 \text{ Wb}$.

(C) The magnetic field $B$ at the center of a large circular coil of radius $r_2$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r_2}$.
Since $r_1 \ll r_2$,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 $r_1$ is $\phi = B \times A = \frac{\mu_0 I}{2r_2} \times (\pi r_1^2)$.
By definition,the mutual inductance $M$ is given by $M = \frac{\phi}{I}$.
Substituting the expression for $\phi$,we get $M = \frac{\mu_0 \pi r_1^2}{2r_2}$.
Therefore,the mutual inductance $M$ is proportional to $\frac{r_1^2}{r_2}$.