Self Induction Questions in English

(B) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 n^2 A l$, where $n$ is the number of turns per unit length, $A$ is the cross-sectional area, and $l$ is the length of the coil.
Given that the linear dimensions are increased by a factor of $2$, the new length $l' = 2l$ and the new area $A' = (2)^2 A = 4A$.
The number of turns per unit length $n$ remains constant.
Substituting these values into the formula for the new inductance $L'$:
$L' = \mu_0 n^2 (4A) (2l) = 8 (\mu_0 n^2 A l) = 8L$.
Therefore, the self-inductance increases by a factor of $8$.

(C) The self-inductance $L$ of a long solenoid is given by the formula: $L = \frac{\mu_0 N^2 A}{l}$,where $\mu_0$ is the permeability of free space,$N$ is the total number of turns,$A$ is the area of cross-section,and $l$ is the length of the solenoid.
From this formula,it is clear that $L \propto A$.
Therefore,the self-inductance is directly proportional to its area of cross-section.

(A) The relationship between magnetic flux $(\phi)$ and current $(I)$ for an inductor is given by $\phi = LI$,where $L$ is the self-inductance.
Comparing this equation with the equation of a straight line passing through the origin,$y = mx$,we get $L = \phi / I$.
This implies that the self-inductance $L$ is equal to the slope of the $\phi$ versus $I$ graph.
Since the line $A$ has the steepest slope among the four lines,it indicates the largest value of self-inductance.

(D) Given: Number of turns $N = 100$, Current $I = 1 \,A$, Flux per turn $\phi = 2.5 \times 10^{-5} \,Wb/\text{turn}$.
The total flux linkage is given by $N\phi$.
The formula for self-inductance $L$ is $L = \frac{N\phi}{I}$.
Substituting the given values:
$L = \frac{100 \times 2.5 \times 10^{-5}}{1} \,H$.
$L = 2.5 \times 10^{-3} \,H$.
Since $1 \,mH = 10^{-3} \,H$, we have $L = 2.5 \,mH$.

(B) The formula for the self-inductance of a solenoid is given by:
$L = \frac{\mu_0 N^2 A}{l}$
where $\mu_0$ is the permeability of free space,$N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the solenoid.
From the formula,it is clear that $L \propto A$ and $L \propto \frac{1}{l}$.
Therefore,for the self-inductance $L$ to increase,the area $A$ must increase and the length $l$ must decrease.
Thus,the correct option is $B$.

(C) The magnetic field inside the solenoid is given by $B = \frac{\mu_0 NI}{l}$.
The total magnetic flux linkage through the solenoid is $\phi_{total} = N(BA)$.
Substituting the value of $B$,we get $\phi_{total} = N \left( \frac{\mu_0 NI}{l} \right) A = \frac{\mu_0 N^2 IA}{l}$.
By definition,self-inductance $L$ is given by $L = \frac{\phi_{total}}{I}$.
Therefore,$L = \frac{\frac{\mu_0 N^2 IA}{l}}{I} = \frac{\mu_0 N^2 A}{l}$.

(A) The flux linked with each turn of the solenoid is given as $\phi = 2.8 \times 10^{-3} \, Wb$.
The total magnetic flux $(\Phi_{net})$ linked with the solenoid having $N = 1500$ turns is calculated as:
$\Phi_{net} = N \times \phi = 1500 \times 2.8 \times 10^{-3} = 4.2 \, Wb$.
The relationship between total flux, self-inductance $(L)$, and current $(I)$ is given by $\Phi_{net} = L \times I$.
Therefore, the self-inductance $L$ is:
$L = \frac{\Phi_{net}}{I} = \frac{4.2}{3.5} = 1.2 \, H$.

(C) The correct option is $C$.
Concept: The magnetic flux $\phi$ associated with a coil is given by $\phi = B A$,and the coefficient of self-inductance $L$ is defined as $L = \frac{N\phi}{I}$,where $I$ is the current flowing through the coil.
The magnetic field $B$ at the center of a circular coil with radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.
For a coil with $N$ turns,the total magnetic field is $B_N = N \left( \frac{\mu_0 I}{2r} \right)$.
The magnetic flux linked with each turn is $\phi_1 = B_N A = N \left( \frac{\mu_0 I}{2r} \right) A$.
The total flux linkage for $N$ turns is $\Phi = N \phi_1 = N^2 \left( \frac{\mu_0 I A}{2r} \right)$.
Using the definition $L = \frac{\Phi}{I}$,we get $L = N^2 \left( \frac{\mu_0 A}{2r} \right)$.
Therefore,the self-inductance $L$ is proportional to the square of the number of turns,i.e.,$L \propto N^2$.

(D) The magnetic field '$B$' inside a long solenoid is given by $B = \mu_0 n i$,where '$n$' is the number of turns per unit length and '$i$' is the current.
The cross-sectional area '$A$' of the solenoid with diameter '$d$' is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
The magnetic flux '$\phi$' through a single turn is $\phi = B \cdot A = (\mu_0 n i) \left( \frac{\pi d^2}{4} \right)$.
For a length '$l$' of the solenoid,the total number of turns is $N = n \cdot l$.
The total flux '$\Phi$' linked with length '$l$' is $\Phi = N \cdot \phi = (n l) \left( \mu_0 n i \frac{\pi d^2}{4} \right) = \mu_0 n^2 i l \frac{\pi d^2}{4}$.
Since $\Phi = L \cdot i$,the total inductance '$L$' for length '$l$' is $L = \frac{\mu_0 n^2 i l \pi d^2}{4 i} = \frac{\mu_0 n^2 l \pi d^2}{4}$.
Therefore,the inductance per unit length is $\frac{L}{l} = \frac{\mu_0 \pi n^2 d^2}{4}$.

(B) The self-inductance $L$ of a solenoid of length $\lambda$ and area of cross-section $A$ with a fixed number of turns $N$ is given by the formula:
$L = \frac{\mu_0 N^2 A}{\lambda}$
From this expression,we can see that $L$ is directly proportional to the area $A$ and inversely proportional to the length $\lambda$.
Therefore,for $L$ to increase,the area $A$ must increase and the length $\lambda$ must decrease.

(B) The formula for induced emf in a coil due to self-inductance is given by $e = L \left| \frac{dI}{dt} \right|$.
Given:
Induced emf $e = 0.5 \,V$
Initial current $I_1 = 10 \,A$
Final current $I_2 = -10 \,A$ (opposite direction)
Change in current $\Delta I = I_2 - I_1 = -10 \,A - 10 \,A = -20 \,A$
Time interval $\Delta t = 0.8 \,s$
Magnitude of rate of change of current $|\frac{\Delta I}{\Delta t}| = \frac{|-20 \,A|}{0.8 \,s} = \frac{20}{0.8} \,A/s = 25 \,A/s$.
Substituting these values into the formula:
$0.5 = L \times 25$
$L = \frac{0.5}{25} \,H = \frac{1}{50} \,H = 0.02 \,H$.
Converting to millihenry: $0.02 \,H = 0.02 \times 1000 \,mH = 20 \,mH$.

(D) The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$,where $\phi$ is the total magnetic flux linked with the coil and $I$ is the current flowing through it.
For a toroid with $N$ turns,the total flux $\phi$ is given by $\phi = N \cdot A \cdot B$,where $A$ is the cross-sectional area of the toroid and $B$ is the magnetic field inside it.
The cross-sectional area $A = \pi r^2$.
The magnetic field inside a toroid is $B = \mu_0 n I$,where $n = \frac{N}{2 \pi R}$ is the number of turns per unit length.
Substituting these into the flux equation:
$\phi = N (\pi r^2) (\mu_0 \frac{N}{2 \pi R} I) = \frac{\mu_0 N^2 r^2 I}{2 R}$.
Finally,the self-induction $L = \frac{\phi}{I} = \frac{\mu_0 N^2 r^2}{2 R}$.

(C) The magnetic field $B$ at the center of a solenoid is given by $B = \mu_0 (N/l) I$,where $N$ is the number of turns,$l$ is the length,and $I$ is the current.
Total magnetic flux $\phi$ linked with the solenoid is $\phi = B A N = \mu_0 (N^2/l) A I$,where $A$ is the cross-sectional area.
By the definition of self-induction,$\phi = L I$.
Comparing the two expressions,the self-inductance $L$ is given by $L = \frac{\mu_0 N^2 A}{l}$.
From this formula,it is clear that $L$ depends on the number of turns $N$,cross-sectional area $A$,and length $l$,but it is independent of the current $I$ flowing through the solenoid.

(B) The magnetic field $B$ at the center of the solenoid is given by $B = \mu_0 (N/l) i$.
The magnetic flux $\Phi$ linked with the solenoid is $\Phi = N B A = N (\mu_0 N i A / l) = (\mu_0 N^2 A / l) i$.
The self-inductance $L$ is defined as $L = \Phi / i = \mu_0 N^2 A / l$.
Substituting the given values: $N = 500$,$l = 0.3 \ m$,$A = 25 \times 10^{-4} \ m^2$,$\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
$L = \frac{4\pi \times 10^{-7} \times (500)^2 \times 25 \times 10^{-4}}{0.3} \approx 2.618 \times 10^{-3} \ H$.
The average back e.m.f. is $e = L (\Delta i / \Delta t)$.
Given $\Delta i = 2.5 \ A$ and $\Delta t = 10^{-3} \ s$.
$e = (2.618 \times 10^{-3}) \times (2.5 / 10^{-3}) = 2.618 \times 2.5 \approx 6.545 \ V$.
Rounding to the nearest value,$e \approx 6.5 \ V$.

(C) The magnetic field at the center of a circular coil carrying current $I$ is given by $B = \frac{\mu_0 NI}{2R}$.
The total magnetic flux $\phi$ linked with the coil is the product of the number of turns $N$, the magnetic field $B$, and the area of the coil $A = \pi R^2$.
$\phi = N \cdot B \cdot A = N \cdot \left( \frac{\mu_0 NI}{2R} \right) \cdot \pi R^2 = \frac{\mu_0 N^2 \pi R I}{2}$.
The coefficient of self-induction $L$ is defined as $L = \frac{\phi}{I}$.
Substituting the value of $\phi$, we get $L = \frac{\mu_0 N^2 \pi R}{2}$.

(B) Given:
Length of solenoid,$\ell = 31.4 \ cm = 0.314 \ m$
Area of cross-section,$A = 10^{-3} \ m^2$
Total number of turns,$N = 500$
Permeability of free space,$\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
The formula for self-inductance of a solenoid is given by:
$L = \frac{\mu_0 N^2 A}{\ell}$
Substituting the values:
$L = \frac{(4 \times 3.14 \times 10^{-7}) \times (500)^2 \times 10^{-3}}{0.314}$
$L = \frac{4 \times 3.14 \times 10^{-7} \times 250000 \times 10^{-3}}{0.314}$
$L = \frac{4 \times 3.14 \times 10^{-7} \times 2.5 \times 10^5 \times 10^{-3}}{0.314}$
$L = \frac{4 \times 3.14 \times 2.5 \times 10^{-5}}{0.314}$
$L = 4 \times 10 \times 2.5 \times 10^{-5} = 10 \times 10^{-4} = 10^{-3} \ H$

(D) The total magnetic flux linkage in a coil with $n$ turns is given by $N\phi_{total} = L I$,where $N$ is the number of turns,$\phi_{total}$ is the flux per turn,$L$ is the self-inductance,and $I$ is the current.
Given that the flux per turn is $\phi$ and the number of turns is $n$,the total flux linkage is $n\phi$.
Therefore,the relation is $n\phi = LI$.
Rearranging for $\phi$,we get $\phi = \frac{LI}{n}$.

(B) The self-inductance $L$ of a coil is given by the formula $L = \frac{N \phi}{I}$.
For a circular coil or solenoid,the magnetic flux $\phi$ through each turn is proportional to the number of turns $N$ (since $B \propto N$).
Therefore,the total flux linkage $N \phi$ is proportional to $N^2$.
Thus,the self-inductance $L$ is directly proportional to the square of the number of turns,i.e.,$L \propto N^{2}$.

(D) The induced emf $(e)$ in a coil due to self-induction is given by the formula: $e = L \frac{di}{dt}$.
Here, the induced emf $e = 2 \,V$ and the rate of change of current $\frac{di}{dt} = 4 \,A s^{-1}$.
Substituting these values into the formula:
$2 = L \times 4$
$L = \frac{2}{4} \,H$
$L = 0.5 \,H$.
Therefore, the self-inductance of the coil is $0.5 \,H$.

(A) The induced $emf$ $(e)$ in a coil is given by the formula: $e = L \frac{di}{dt}$.
Here,$L_1 = 8 \ mH$ and $L_2 = 2 \ mH$.
The rate of change of current,$\frac{di}{dt}$,is the same for both coils.
Therefore,the ratio of the induced $emf$s is:
$\frac{e_1}{e_2} = \frac{L_1 (di/dt)}{L_2 (di/dt)} = \frac{L_1}{L_2}$.
Substituting the given values:
$\frac{e_1}{e_2} = \frac{8 \ mH}{2 \ mH} = \frac{4}{1}$.
Thus,the ratio is $4: 1$.

(A) The formula for self-inductance $L$ is given by $L = \frac{N \phi}{I}$.
Given:
Number of turns $N = 100$
Current $I = 5 \ A$
Magnetic flux $\phi = 5 \times 10^{-5} \ Wb$
Substituting the values into the formula:
$L = \frac{100 \times 5 \times 10^{-5}}{5}$
$L = 100 \times 10^{-5} \ H$
$L = 10^{-3} \ H = 1 \ mH$.

(C) The formula for self-inductance $L$ is given by $L = \frac{N\phi}{i}$,where $N$ is the number of turns,$\phi$ is the magnetic flux,and $i$ is the current.
Given values are $N = 100$,$\phi = 5 \times 10^{-5} \ Wb$,and $i = 2 \ A$.
Substituting these values into the formula:
$L = \frac{100 \times 5 \times 10^{-5}}{2}$
$L = \frac{500 \times 10^{-5}}{2}$
$L = 250 \times 10^{-5} \ H$
$L = 2.5 \times 10^{-3} \ H$

(B) The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$, where $l$ is the length of the solenoid and $A = \pi r^2$ is the cross-sectional area.
Let $x$ be the total length of the wire required. The wire is wound into $N$ turns, so $x = (2 \pi r) N$, which implies $N = \frac{x}{2 \pi r}$.
Substituting $N$ into the inductance formula:
$L = \frac{\mu_0 (x / 2 \pi r)^2 \times \pi r^2}{l} = \frac{\mu_0 x^2}{4 \pi l}$.
Rearranging for $x$:
$x^2 = \frac{4 \pi L l}{\mu_0} \implies x = \sqrt{\frac{4 \pi L l}{\mu_0}}$.
Given $L = 1 \,mH = 10^{-3} \,H$, $l = 1 \,m$, and $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$:
$x = \sqrt{\frac{4 \pi \times 10^{-3} \times 1}{4 \pi \times 10^{-7}}} = \sqrt{10^4} = 100 \,m$.
Converting to kilometers: $100 \,m = 0.10 \,km$.

(B) The inductance of a solenoid is given by $L = \mu_{0} N^{2} \frac{A_{s}}{\ell}$,where $A_{s}$ is the cross-sectional area of the solenoid and $N$ is the number of turns.
Let $x$ be the total length of the wire used for windings and $a$ be the cross-sectional area of the wire.
The resistance $R$ is given by $R = \frac{\rho x}{a}$.
The mass $m$ of the wire is $m = a x D$.
Multiplying these,we get $R m = \left( \frac{\rho x}{a} \right) (a x D) = \rho x^{2} D$.
Thus,$x^{2} = \frac{R m}{\rho D}$.
The total length of the wire $x$ is also equal to $N \times (2 \pi r)$,where $r$ is the radius of the solenoid.
Therefore,$N = \frac{x}{2 \pi r}$.
Substituting $N$ into the inductance formula: $L = \mu_{0} \left( \frac{x}{2 \pi r} \right)^{2} \frac{\pi r^{2}}{\ell} = \mu_{0} \frac{x^{2}}{4 \pi^{2} r^{2}} \frac{\pi r^{2}}{\ell} = \frac{\mu_{0} x^{2}}{4 \pi \ell}$.
Substituting $x^{2} = \frac{R m}{\rho D}$,we get $L = \frac{\mu_{0}}{4 \pi \ell} \left( \frac{R m}{\rho D} \right)$.

(C) The self or mutual inductance of a coil is defined by the flux linkage per unit current,given by the relation:
$L \text{ or } M = \frac{\phi}{I}$
Where $\phi$ is the magnetic flux and $I$ is the current.
The dimensional formula for magnetic flux $\phi$ is $[M L^2 T^{-2} I^{-1}]$.
The dimensional formula for current $I$ is $[I^1]$.
Therefore,the dimension of inductance is:
$[L] = \frac{[\phi]}{[I]} = \frac{[M L^2 T^{-2} I^{-1}]}{[I^1]} = [M L^2 T^{-2} I^{-2}]$

(D) In mechanics,mass $(m)$ represents the inertia of an object,which is the resistance to a change in its state of motion. In an $LC$ circuit,the self-inductance $(L)$ represents the inertia of the electrical system,as it opposes any change in the electric current $(I)$. The energy stored in a mechanical system is $\frac{1}{2}mv^2$,while the energy stored in an inductor is $\frac{1}{2}LI^2$. Thus,self-inductance is the electrical equivalent of mass.

(D) The self-inductance $L$ of a coil is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $\mu_0$ is the permeability of free space,$N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
From this formula,it is evident that the self-inductance $L$ depends only on the geometric parameters of the coil (number of turns,area,and length) and the core material.
It is independent of the current $I$ flowing through the coil.
Therefore,if the current changes from $I$ to $5I$,the self-inductance remains unchanged.
Thus,the new self-inductance is $L$ $H$.

(A) The magnetic field inside an air-cored solenoid is given by $B = \frac{\mu_0 NI}{l}$.
Initial magnetic flux linkage $\phi_1 = N B A = N \left( \frac{\mu_0 NI}{l} \right) A = \frac{\mu_0 N^2 AI}{l}$.
Given values: $N = 500$,$I = 2.5 \ A$,$l = 0.3 \ m$,$A = 25 \times 10^{-4} \ m^2$,$\Delta t = 10^{-3} \ s$,and $\mu_0 = 4\pi \times 10^{-7} \ T \cdot m/A$.
When the current is switched off,the final flux $\phi_2 = 0$.
The magnitude of induced emf is $|\varepsilon| = \frac{|\Delta \phi|}{\Delta t} = \frac{|\phi_2 - \phi_1|}{\Delta t} = \frac{\mu_0 N^2 AI}{l \Delta t}$.
Substituting the values:
$|\varepsilon| = \frac{4\pi \times 10^{-7} \times (500)^2 \times 25 \times 10^{-4} \times 2.5}{0.3 \times 10^{-3}}$.
$|\varepsilon| = \frac{4 \times 3.1416 \times 10^{-7} \times 250000 \times 25 \times 10^{-4} \times 2.5}{3 \times 10^{-4}}$.
$|\varepsilon| = \frac{1.9635 \times 10^{-3}}{3 \times 10^{-4}} \approx 6.54 \ V$.

(B) The formula for the self-inductance $L$ of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$.
From this expression,we can see that $L \propto \frac{A}{l}$.
To increase the self-inductance $L$,the numerator $A$ (area of cross-section) must increase and the denominator $l$ (length of the solenoid) must decrease.
Therefore,the correct option is $B$.

(A) The formula for inductance $L$ is given by $\phi = L \cdot I$,where $\phi$ is magnetic flux and $I$ is current.
Thus,$L = \frac{\phi}{I}$.
The unit of magnetic flux $\phi$ is Weber $(Wb)$ and the unit of current $I$ is Ampere $(A)$.
Therefore,the $SI$ unit of inductance is $\frac{Wb}{A} = Wb \cdot A^{-1}$,which is also known as Henry $(H)$.
Since $V = \frac{d\phi}{dt}$,the unit of flux can also be expressed as $V \cdot s$.
Substituting this,$L = \frac{V \cdot s}{A} = V \cdot s \cdot A^{-1}$.
Comparing these with the given options,$Wb \cdot A^{-1}$ $(D)$,$H$ $(C)$,and $V \cdot s \cdot A^{-1}$ $(B)$ are all valid units of inductance.
Option $A$ $(Wb \cdot s \cdot A^{-1})$ is not a valid unit of inductance.

(A) The self-inductance $L$ is defined by the relation $\phi = LI$,where $\phi$ is magnetic flux and $I$ is current. Thus,$L = \frac{\phi}{I}$.
Units of $L$ are $\frac{\text{weber}}{\text{ampere}}$ (Henry).
Since $\text{weber} = \text{volt} \cdot \text{second}$,the unit is also $\frac{\text{volt} \cdot \text{second}}{\text{ampere}}$.
Since $\text{volt} = \text{ampere} \cdot \text{ohm}$,the unit is $\frac{(\text{ampere} \cdot \text{ohm}) \cdot \text{second}}{\text{ampere}} = \text{ohm} \cdot \text{second}$.
'mho-second' is not a unit of self-inductance,as 'mho' is the unit of conductance $(1/\text{ohm})$.

(B) Given: Current $I = 2 \ A$,rate of change of current $\frac{dI}{dt} = -10^2 \ A \ s^{-1}$ (since it is decreasing),resistance $R = 2 \ \Omega$,$EMF$ $\varepsilon = 12 \ V$,and inductance $L = 5 \ mH = 5 \times 10^{-3} \ H$.
Applying Kirchhoff's voltage law from point $A$ to $B$:
$V_A - IR + \varepsilon - L \frac{dI}{dt} = V_B$
$V_B - V_A = -IR + \varepsilon - L \frac{dI}{dt}$
Substituting the values:
$V_B - V_A = -(2 \times 2) + 12 - (5 \times 10^{-3} \times (-10^2))$
$V_B - V_A = -4 + 12 + 0.5$
$V_B - V_A = 8.5 \ V$

(A) The self-inductance $L$ of a coil is defined by the relation $\phi = L I$,where $\phi$ is the magnetic flux and $I$ is the current flowing through the coil.
Rearranging for $L$,we get $L = \frac{\phi}{I}$.
Given values are:
$\phi = 5 \ \mu Wb = 5 \times 10^{-6} \ Wb$
$I = 1 \ mA = 1 \times 10^{-3} \ A$
Substituting these values into the formula:
$L = \frac{5 \times 10^{-6} \ Wb}{1 \times 10^{-3} \ A} = 5 \times 10^{-3} \ H$.
Therefore,the self-inductance of the coil is $5 \times 10^{-3} \ H$.

(A) The formula for the self-inductance $(L)$ of a solenoid is given by:
$L = \frac{\mu_0 N^2 A}{l}$
Given:
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Number of turns $N = 10^3$
Area of cross-section $A = 10^{-3} \ m^2$
Length $l = 31.4 \ cm = 31.4 \times 10^{-2} \ m$
Substituting the values:
$L = \frac{4 \times 3.14 \times 10^{-7} \times (10^3)^2 \times 10^{-3}}{31.4 \times 10^{-2}}$
$L = \frac{4 \times 3.14 \times 10^{-7} \times 10^6 \times 10^{-3}}{31.4 \times 10^{-2}}$
$L = \frac{12.56 \times 10^{-4}}{31.4 \times 10^{-2}} = \frac{12.56}{31.4} \times 10^{-2} = 0.4 \times 10^{-2} \ H$
$L = 4 \times 10^{-3} \ H = 4 \ mH$

(B) The magnetic flux $\phi$ linked with a coil is given by the relation $\phi = L \cdot I$,where $L$ is the self-inductance and $I$ is the current flowing through the coil.
Given:
$\phi = 10 \ \mu Wb = 10 \times 10^{-6} \ Wb$
$I = 2 \ mA = 2 \times 10^{-3} \ A$
Rearranging the formula for self-inductance $L$:
$L = \frac{\phi}{I}$
Substituting the values:
$L = \frac{10 \times 10^{-6}}{2 \times 10^{-3}}$
$L = 5 \times 10^{-3} \ H$
$L = 5 \ mH$
Therefore,the self-inductance of the coil is $5 \ mH$.

(A) The formula for self-induced $EMF$ is given by $\varepsilon = -L \frac{dI}{dt}$.
Given:
$\varepsilon = 5 \ V$
$dI = I_2 - I_1 = 2 \ A - 3 \ A = -1 \ A$
$dt = 1 \ ms = 1 \times 10^{-3} \ s$
Substituting the values into the formula:
$5 = -L \left( \frac{-1 \ A}{1 \times 10^{-3} \ s} \right)$
$5 = L \times 10^3$
$L = \frac{5}{1000} \ H = 5 \times 10^{-3} \ H = 5 \ mH$.

(A) The formula for self-inductance $L$ is given by $L = \frac{\varepsilon dt}{dI}$.
Here,$\varepsilon$ is the induced electromotive force $(EMF)$,which has dimensions $[M^1 L^2 T^{-3} A^{-1}]$.
$dt$ is the time interval with dimensions $[T^1]$.
$dI$ is the change in current with dimensions $[A^1]$.
Substituting these dimensions into the formula:
$L = \frac{[M^1 L^2 T^{-3} A^{-1}] \cdot [T^1]}{[A^1]}$
$L = [M^1 L^2 T^{-2} A^{-2}]$.
Therefore,the correct option is $A$.

(B) Given:
Initial current $I_1 = 5.0 \ A$
Final current $I_2 = 0.0 \ A$
Time interval $dt = 0.1 \ s$
Induced emf $\varepsilon = 200 \ V$
The formula for induced emf in an inductor is given by:
$\varepsilon = -L \frac{dI}{dt}$
Rearranging the formula to solve for self-inductance $L$:
$L = -\frac{\varepsilon \cdot dt}{dI}$
Where $dI = I_2 - I_1 = 0.0 - 5.0 = -5.0 \ A$.
Substituting the values:
$L = -\frac{200 \times 0.1}{-5.0}$
$L = \frac{20}{5} = 4 \ H$
Therefore,the self-inductance of the circuit is $4.0 \ H$.

(B) Given:
Initial current $I_1 = 5.0 \ A$
Final current $I_2 = 0.0 \ A$
Change in current $dI = I_2 - I_1 = 0.0 - 5.0 = -5.0 \ A$
Time interval $dt = 0.1 \ s$
Induced emf $\varepsilon = 100 \ V$
The formula for induced emf due to self-inductance is given by:
$\varepsilon = -L \frac{dI}{dt}$
Substituting the values:
$100 = -L \left( \frac{-5.0}{0.1} \right)$
$100 = L \times 50$
$L = \frac{100}{50} = 2 \ H$
Therefore,the self-inductance of the circuit is $2 \ H$.

(A) The self-inductance $L$ of an inductor is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $\mu_0$ is the permeability of free space,$N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the inductor.
This formula shows that the self-inductance $L$ depends only on the geometric properties and the core material of the inductor.
It is independent of the current $I$ flowing through it.
Therefore,if the current is doubled,the self-inductance remains unchanged at $4 \ H$.

(D) The induced emf in an inductor is given by the formula $|e| = L \left| \frac{dI}{dt} \right|$.
Given self-inductance $L = 6 \text{ mH} = 6 \times 10^{-3} \text{ H}$.
From the graph,at $t_1 = 20 \text{ s}$,the current $I_1 = 4 \text{ A}$,and at $t_2 = 40 \text{ s}$,the current $I_2 = 3 \text{ A}$.
The rate of change of current is $\left| \frac{dI}{dt} \right| = \left| \frac{I_2 - I_1}{t_2 - t_1} \right| = \left| \frac{3 - 4}{40 - 20} \right| = \left| \frac{-1}{20} \right| = 0.05 \text{ A/s}$.
Thus,the induced emf is $|e| = (6 \times 10^{-3} \text{ H}) \times (0.05 \text{ A/s}) = 0.3 \times 10^{-3} \text{ V} = 3 \times 10^{-4} \text{ V}$.
However,checking the options provided,$3 \times 10^{-4} \text{ V}$ is equivalent to $30 \times 10^{-5} \text{ V}$. Re-evaluating the calculation: $6 \times 10^{-3} \times (1/20) = 6/20 \times 10^{-3} = 0.3 \times 10^{-3} = 3 \times 10^{-4} \text{ V}$.
Given the options,$30 \times 10^{-5} \text{ V}$ is not listed,but $30 \times 10^{-4} \text{ V}$ is option $D$. Let's re-examine the graph. If the current at $t=40$ was $0$,the change would be $4/20 = 0.2$. $6 \times 10^{-3} \times 0.2 = 1.2 \times 10^{-3}$. If the current at $t=20$ was $4$ and $t=40$ was $3$,the slope is $1/20$. The calculation $3 \times 10^{-4} \text{ V}$ is correct. Since $30 \times 10^{-5} \text{ V}$ is not an option,and $3 \times 10^{-4} \text{ V}$ is the result,we select the closest magnitude if applicable,or note the discrepancy. Given the options,$30 \times 10^{-5} \text{ V}$ is $3 \times 10^{-4} \text{ V}$. Option $D$ is $30 \times 10^{-4} \text{ V} = 3 \times 10^{-3} \text{ V}$. There might be a typo in the question's options. Assuming the intended answer is $3 \times 10^{-4} \text{ V}$.

(B) Given: Inductance $L = 0.2 \,H$,initial current $I_1 = 5 \,A$,final current $I_2 = 2 \,A$,and time interval $\Delta t = 0.5 \,s$.
The change in current is $\Delta I = I_1 - I_2 = 5 \,A - 2 \,A = 3 \,A$.
The magnitude of the average induced emf $|e|$ in the coil is given by the formula:
$|e| = L \left| \frac{\Delta I}{\Delta t} \right| = 0.2 \,H \times \frac{3 \,A}{0.5 \,s} = 0.2 \times 6 \,V = 1.2 \,V$.

(B) Change in current,$\Delta I = 5 \,A - 2 \,A = 3 \,A$.
Time interval,$\Delta t = 0.3 \,s$.
Induced emf,$e = 1.0 \,V$.
The formula for induced emf in a coil due to self-inductance is given by $e = L \frac{dI}{dt}$.
Considering the magnitude,$e = L \frac{\Delta I}{\Delta t}$.
Substituting the values: $1.0 = L \times \frac{3}{0.3}$.
$1.0 = L \times 10$.
$L = \frac{1.0}{10} = 0.1 \,H$.
Since $1 \,H = 1000 \,mH$,we have $L = 0.1 \times 1000 \,mH = 100 \,mH$.

(C) Given: Number of turns, $N = 500$.
Current, $I = 2 \,A$.
Magnetic flux per turn, $\phi = 4 \times 10^{-3} \,Wb$.
The total magnetic flux linkage is given by $N\phi$.
The self-inductance $L$ is defined by the relation $L = \frac{N\phi}{I}$.
Substituting the given values:
$L = \frac{500 \times 4 \times 10^{-3}}{2}$
$L = \frac{2000 \times 10^{-3}}{2}$
$L = \frac{2}{2} = 1 \,H$.
Therefore, the self-inductance of the solenoid is $1.0 \,H$.

(C) The magnitude of the induced emf in a coil is given by the formula: $|e| = L \left| \frac{dI}{dt} \right|$.
Given:
Initial current $I_1 = 4 \,A$.
Final current $I_2 = 0 \,A$.
Change in current $dI = I_2 - I_1 = 0 - 4 = -4 \,A$.
Time interval $dt = 0.1 \,s$.
Induced emf $|e| = 100 \,V$.
Substituting these values into the formula:
$100 = L \times \frac{|-4|}{0.1}$.
$100 = L \times \frac{4}{0.1}$.
$100 = L \times 40$.
$L = \frac{100}{40} = 2.5 \,H$.
Therefore,the self-inductance of the coil is $2.5 \,H$.

(A) The voltage $(V)$ across an inductor is given by the formula $V = L \frac{dI}{dt}$, where $L$ is the inductance and $\frac{dI}{dt}$ is the rate of change of current with respect to time.
From the graph, the current $(I)$ increases linearly with time $(t)$ for a certain interval, meaning the slope $\frac{dI}{dt}$ is constant and positive.
Since $V = L \times (\text{constant})$, the voltage $(V)$ will be a constant positive value during this interval.
When the current $(I)$ is constant, $\frac{dI}{dt} = 0$, so the voltage $(V)$ becomes $0$.
When the current $(I)$ decreases linearly, the slope $\frac{dI}{dt}$ is constant and negative, so the voltage $(V)$ becomes a constant negative value.
Therefore, the graph of voltage $(V)$ versus time $(t)$ will be a series of horizontal steps corresponding to the slopes of the current-time graph.

(C) The induced emf $\varepsilon$ in an inductor is given by the formula $\varepsilon = -L \frac{di}{dt}$.
Taking the magnitude, we have $L = \frac{|\varepsilon|}{\frac{di}{dt}}$.
Given: Change in current $\Delta i = 6 \,A - 2 \,A = 4 \,A$, time interval $\Delta t = 2 \,s$, and induced emf $\varepsilon = 3 \,V$.
The rate of change of current is $\frac{di}{dt} = \frac{\Delta i}{\Delta t} = \frac{4 \,A}{2 \,s} = 2 \,A/s$.
Substituting these values into the formula:
$L = \frac{3 \,V}{2 \,A/s} = 1.5 \,H$.

(A) Given: Current $I = 2 \ A$, Resistance $R = 7 \ \Omega$, $EMF$ $E = 4 \ V$, Inductance $L = 9 \ mH = 9 \times 10^{-3} \ H$, Rate of change of current $\frac{dI}{dt} = -10^3 \ A \ s^{-1}$ (since current is decreasing).
Applying Kirchhoff's voltage law from $A$ to $B$:
$V_A - I R - E - L \frac{dI}{dt} = V_B$
$V_A - V_B = I R + E + L \frac{dI}{dt}$
Substituting the values:
$V_{AB} = (2 \ A)(7 \ \Omega) + 4 \ V + (9 \times 10^{-3} \ H)(-10^3 \ A \ s^{-1})$
$V_{AB} = 14 \ V + 4 \ V - 9 \ V$
$V_{AB} = 18 \ V - 9 \ V = 9 \ V$
Wait, let's re-evaluate the polarity of the inductor. The induced $EMF$ in the inductor is $\varepsilon = -L \frac{dI}{dt}$. Since current is decreasing, $\frac{dI}{dt} = -10^3 \ A \ s^{-1}$, so $\varepsilon = - (9 \times 10^{-3})(-10^3) = +9 \ V$. The potential drop across the inductor is $V_L = L \frac{dI}{dt} = -9 \ V$.
Path $A \to B$: $V_A - I R + E - V_L = V_B$ (considering the battery polarity in the image: $A \to$ resistor $ o$ negative terminal $ o$ positive terminal $ o$ inductor $ o B$).
$V_A - (2)(7) + 4 - (9 \times 10^{-3})(-10^3) = V_B$
$V_A - 14 + 4 + 9 = V_B$
$V_A - 1 = V_B \Rightarrow V_A - V_B = 1 \ V$.

(D) Given: Number of turns $N = 120$,Inductance $L = 40 \text{ mH} = 40 \times 10^{-3} \text{ H}$,Current $I = 30 \text{ mA} = 30 \times 10^{-3} \text{ A}$.
We know that the total flux linkage is given by $N\phi = LI$.
Therefore,the magnetic flux $\phi$ linked with the coil is given by $\phi = \frac{LI}{N}$.
Substituting the values: $\phi = \frac{40 \times 10^{-3} \times 30 \times 10^{-3}}{120}$.
$\phi = \frac{1200 \times 10^{-6}}{120} = 10 \times 10^{-6} \text{ Wb}$.