Self Induction Questions in English

(B) The self-inductance $L$ of a coil is defined by the relation $\phi = LI$,where $\phi$ is the magnetic flux and $I$ is the current.
Therefore,$L = \frac{\phi}{I}$.
The unit of magnetic flux $\phi$ is Weber $(Wb)$ and the unit of current $I$ is Ampere $(A)$.
Thus,the unit of self-inductance is $\frac{Wb}{A}$,which is defined as Henry $(H)$.

(B) The $SI$ unit of self-inductance $(L)$ is the Henry $(H)$.
From the formula for induced $EMF$,$\varepsilon = -L \frac{di}{dt}$,we have $L = \frac{\varepsilon \cdot dt}{di}$.
The units are $\frac{\text{Volt} \times \text{second}}{\text{Ampere}}$.
Since $\text{Volt} = \frac{\text{Joule}}{\text{Coulomb}}$,substituting this gives $\frac{\text{Joule}}{\text{Coulomb}} \times \frac{\text{second}}{\text{Ampere}}$.
Therefore,the correct expression is $\frac{\text{Joule} \times \text{second}}{\text{Coulomb} \times \text{Ampere}}$.

(A) The dimension of resistance $R$ is given by $V/I$,where $V$ is potential difference and $I$ is current.
The dimension of self-inductance $L$ is given by $V \cdot T/I$,where $T$ is time.
Taking the ratio $R/L$,we get:
$\frac{R}{L} = \frac{V/I}{V \cdot T/I} = \frac{1}{T}$.
Since the dimension of $1/T$ is $T^{-1}$,which is the dimension of frequency,the correct combination is $R/L$.

(D) The formula for the back $e.m.f.$ induced in a coil is given by $e = -L \frac{di}{dt}$.
Here,the change in current $di = i_f - i_i = 0 - 1 = -1 \ A$.
The time interval $dt = 1 \ ms = 10^{-3} \ s$.
The induced $e.m.f.$ is $e = 4 \ V$.
Substituting these values into the formula:
$4 = -L \left( \frac{-1}{10^{-3}} \right)$
$4 = L \times 10^3$
$L = \frac{4}{10^3} = 4 \times 10^{-3} \ H$.

(D) The induced $e.m.f.$ $(e)$ in a coil due to self-inductance is given by the formula: $e = L \cdot \left| \frac{di}{dt} \right|$.
Given:
$e = 5 \,V$
Change in current,$di = 3 \,A - 2 \,A = 1 \,A$
Time interval,$dt = 1 \,ms = 1 \times 10^{-3} \,s$
Substituting the values into the formula:
$5 = L \cdot \frac{1 \,A}{1 \times 10^{-3} \,s}$
$5 = L \cdot 10^3$
$L = \frac{5}{10^3} \,H = 5 \times 10^{-3} \,H = 5 \,mH$.
Therefore,the correct option is $D$.

(A) The induced $e.m.f.$ $(e)$ in a coil due to self-induction is given by the formula: $e = -L \frac{di}{dt}$.
Here,the inductance $L = 5 \, H$.
The rate of change of current $\frac{di}{dt} = -2 \, A/s$ (since the current is decreasing).
Substituting these values into the formula:
$e = -5 \times (-2) = 10 \, V$.
Therefore,the $e.m.f.$ developed across the coil is $10 \, V$.

(C) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 N^2 A / l$.
Since the number of turns per unit length $n = N / l$ is constant,we can write $N = nl$.
Substituting this into the formula,we get $L = \mu_0 (nl)^2 A / l = \mu_0 n^2 l A$.
Here,$l$ represents the length of the coil and $A$ represents the cross-sectional area.
When all linear dimensions are increased by a factor of $2$,the length $l$ becomes $2l$ and the area $A$ (which is proportional to the square of linear dimensions) becomes $2^2 A = 4A$.
Substituting these new values into the expression for $L$,the new inductance $L'$ is $L' = \mu_0 n^2 (2l) (4A) = 8 (\mu_0 n^2 l A) = 8L$.
Therefore,the self-inductance increases by a factor of $8$.

(C) In mechanics,mass is a measure of inertia,which is the resistance of an object to a change in its state of motion.
In electricity,inductance $(L)$ is the property of a circuit that opposes any change in the current flowing through it. This is analogous to inertia in mechanics,where mass opposes a change in velocity.
Therefore,inductance is the electrical equivalent of mass.

(C) In mechanics,the momentum $p$ is defined as the product of mass $m$ and velocity $v$,i.e.,$p = m \times v$.
In electricity,the magnetic flux $\phi$ is defined as the product of self-inductance $L$ and current $I$,i.e.,$\phi = L \times I$.
By comparing these two expressions,we observe that mass $m$ is analogous to inductance $L$ (which represents electrical inertia),and velocity $v$ is analogous to current $I$.
Therefore,the expression analogous to momentum in electricity is $L \times I$.

(B) The self-inductance $L$ of a solenoid is given by $L = \frac{\mu N^2 A}{l}$,where $\mu = \mu_0 \mu_r$.
Initially,$L_1 = \frac{\mu_0 N^2 A}{l} = 0.18\, mH$.
When a core of relative permeability $\mu_r = 900$ is inserted,the new self-inductance $L_2$ becomes $L_2 = \frac{\mu_0 \mu_r N^2 A}{l} = \mu_r L_1$.
Substituting the values,we get $L_2 = 900 \times 0.18\, mH = 162\, mH$.

(D) The induced $e.m.f.$ in a coil is given by the formula: $e = -L \frac{di}{dt}$.
Here,the change in current $di = 2 \; A - 8 \; A = -6 \; A$.
The time interval $dt = 3 \times 10^{-2} \; s$.
The induced $e.m.f.$ $e = 2 \; V$.
Substituting these values into the formula:
$2 = -L \left( \frac{-6}{3 \times 10^{-2}} \right)$.
$2 = L \left( \frac{6}{3 \times 10^{-2}} \right)$.
$2 = L \times 200$.
$L = \frac{2}{200} = 0.01 \; H$.
Since $1 \; H = 1000 \; mH$,we have $L = 0.01 \times 1000 = 10 \; mH$.

(B) The self-inductance $L$ of a coil is proportional to the square of the number of turns $N$,given by the formula $L \propto N^2$,assuming the radius and length remain constant.
Therefore,the ratio of self-inductances for two similar coils is given by $\frac{L_B}{L_A} = \left( \frac{N_B}{N_A} \right)^2$.
Given: $N_A = 600$,$L_A = 108 \, mH$,and $N_B = 500$.
Substituting the values into the formula:
$L_B = L_A \times \left( \frac{N_B}{N_A} \right)^2$
$L_B = 108 \times \left( \frac{500}{600} \right)^2$
$L_B = 108 \times \left( \frac{5}{6} \right)^2$
$L_B = 108 \times \frac{25}{36}$
$L_B = 3 \times 25 = 75 \, mH$.
Thus,the self-inductance of the second coil is $75 \, mH$.

(A) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
Since $A$ and $l$ remain constant,we have $L \propto N^2$.
Given that the number of turns is doubled,$N_2 = 2N_1$.
Therefore,the new self-inductance $L_2$ is given by $L_2 = L_1 \left( \frac{N_2}{N_1} \right)^2 = L_1 \left( \frac{2N_1}{N_1} \right)^2 = 4L_1$.
Thus,the self-inductance becomes four times the original value.

(B) The formula for the induced $e.m.f.$ $(e)$ in a coil due to self-inductance $(L)$ is given by $e = -L \frac{di}{dt}$.
Taking the magnitude,we have $|e| = L \frac{|di|}{dt}$.
Given:
Change in current,$di = 4 \ A - 2 \ A = 2 \ A$.
Time interval,$dt = 0.05 \ s$.
Induced $e.m.f.$,$|e| = 8 \ V$.
Substituting these values into the formula:
$8 = L \times \frac{2}{0.05}$
$8 = L \times 40$
$L = \frac{8}{40} = 0.2 \ H$.
Therefore,the self-inductance of the coil is $0.2 \ H$.

(B) The formula for the induced $e.m.f.$ in a coil due to self-induction is given by $e = L \cdot \frac{di}{dt}$.
Given:
Induced $e.m.f.$ $(e)$ = $12$ $V$.
Rate of change of current $(\frac{di}{dt})$ = $48$ $A/min$.
First, convert the rate of change of current into $A/s$:
$\frac{di}{dt} = \frac{48 \text{ A}}{60 \text{ s}} = 0.8 \text{ A/s}$.
Now, substitute the values into the formula:
$L = \frac{e}{di/dt} = \frac{12}{0.8} = 15 \text{ H}$.
Therefore, the self-inductance of the coil is $15 \text{ H}$.

(C) When the contact is broken,the current in the circuit decreases rapidly.
According to Lenz's Law,the inductor opposes this change in current by inducing an electromotive force (emf) given by $\varepsilon = -L \frac{di}{dt}$.
Since the current drops to zero very quickly when the switch is opened,the rate of change of current $\frac{di}{dt}$ is very high.
This results in a large induced emf across the inductor,which causes a momentary surge of current through the bulb,making it glow suddenly bright.

(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 cross-sectional area,and $l$ is the length of the solenoid.
From this expression,it is clear that $L \propto A$. Therefore,the self-inductance is directly proportional to the area of the cross-section of the solenoid.

(A) The formula for induced $e.m.f.$ in a coil due to self-induction is given by:
$|e| = L \left| \frac{di}{dt} \right|$
Given:
Self-inductance $L = 5 \, H$
Change in current $di = 2 \, A - 1 \, A = 1 \, A$
Time interval $dt = 5 \, s$
Substituting the values:
$|e| = 5 \times \frac{1}{5} = 1 \, V$
Therefore,the induced $e.m.f.$ is $1 \, V$.

(C) The induced electromotive force $(e)$ in an inductor is given by the formula: $e = L \frac{di}{dt}$.
Rearranging this formula to solve for inductance $(L)$,we get: $L = e \cdot \frac{dt}{di}$.
The unit of electromotive force $(e)$ is Volt $(V)$,the unit of time $(dt)$ is second $(s)$,and the unit of current $(di)$ is ampere $(A)$.
Therefore,the unit of inductance is $\frac{V \cdot s}{A}$,which is Volt-sec/ampere.

(D) The induced $e.m.f.$ $(e)$ in a coil due to self-induction is given by the formula: $e = -L \frac{di}{dt}$.
Given:
Self-inductance $L = 0.4 \, mH = 0.4 \times 10^{-3} \, H$.
Change in current $di = 250 \, mA = 250 \times 10^{-3} \, A$.
Time interval $dt = 0.1 \, s$.
Substituting these values into the formula:
$e = -(0.4 \times 10^{-3} \, H) \times \frac{250 \times 10^{-3} \, A}{0.1 \, s}$.
$e = -(0.4 \times 10^{-3}) \times (2500 \times 10^{-3})$.
$e = -1000 \times 10^{-6} \, V$.
$e = -1 \times 10^{-3} \, V = -1 \, mV$.
Therefore,the induced $e.m.f.$ is $-1 \, mV$.

(A) The formula for induced $e.m.f.$ in a coil due to self-inductance is given by $|e| = L \left| \frac{di}{dt} \right|$.
Given:
Initial current $i_1 = 8 \ A$
Final current $i_2 = 2 \ A$
Change in current $di = |i_2 - i_1| = |2 - 8| = 6 \ A$
Time interval $dt = 3 \times 10^{-3} \ s$
Induced $e.m.f.$ $e = 2 \ V$
Substituting the values into the formula:
$2 = L \times \frac{6}{3 \times 10^{-3}}$
$2 = L \times 2 \times 10^3$
$L = \frac{2}{2 \times 10^3} = 10^{-3} \ H$
Since $1 \ mH = 10^{-3} \ H$,the self-inductance $L = 1 \ mH$.

(D) The induced $e.m.f.$ $(e)$ in a coil due to self-induction is given by the formula: $e = L \frac{di}{dt}$.
Given:
Inductance $L = 60\,\mu H = 60 \times 10^{-6}\,H$.
Change in current $di = 1.5\,A - 1.0\,A = 0.5\,A$.
Time interval $dt = 0.1\,s$.
Substituting these values into the formula:
$e = (60 \times 10^{-6}) \times \frac{0.5}{0.1}$
$e = (60 \times 10^{-6}) \times 5$
$e = 300 \times 10^{-6}\,V = 3 \times 10^{-4}\,V$.
Therefore,the magnitude of the induced $e.m.f.$ is $3 \times 10^{-4}\,V$.

(A) The magnetic flux linked with a coil is given by $\phi = Li$,where $L$ is the self-inductance and $i$ is the current.
Also,$\phi = NBA$,where $N$ is the number of turns,$B$ is the magnetic field,and $A$ is the area of the coil.
For a circular coil,the magnetic field at the center is $B = \frac{\mu_0 Ni}{2r}$.
The area of the coil is $A = \pi r^2$.
Equating the two expressions for flux: $Li = N \left( \frac{\mu_0 Ni}{2r} \right) (\pi r^2)$.
Simplifying,we get $L = \frac{\mu_0 N^2 \pi r}{2}$.
Given: $N = 500$,$r = 0.05\, m$,and $\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A$.
Substituting the values: $L = \frac{4\pi \times 10^{-7} \times (500)^2 \times \pi \times 0.05}{2}$.
$L = 2 \times 10^{-7} \times 250000 \times \pi^2 \times 0.05$.
Using $\pi^2 \approx 10$,$L \approx 2 \times 10^{-7} \times 250000 \times 10 \times 0.05 = 0.025\, H = 25\, mH$.

(C) The formula for the induced $e.m.f.$ in a coil due to self-induction is given by $e = L \frac{di}{dt}$.
Given:
Self-inductance $L = 0.5 \, H$.
Change in current $\Delta i = 10 \, A - 0 \, A = 10 \, A$.
Time interval $\Delta t = 2 \, s$.
The rate of change of current is $\frac{\Delta i}{\Delta t} = \frac{10 \, A}{2 \, s} = 5 \, A/s$.
Substituting these values into the formula:
$e = 0.5 \, H \times 5 \, A/s = 2.5 \, V$.
Therefore,the generated $e.m.f.$ is $2.5 \, V$.

(A) The magnetic field inside a long solenoid is given by $B = \mu_0 n I$,where $n = \frac{N}{L}$ is the number of turns per unit length.
Thus,$B = \frac{\mu_0 N I}{L}$.
The magnetic flux $\phi$ through each turn is $\phi = B \cdot A = \frac{\mu_0 N I A}{L}$.
The total magnetic flux linkage is $\Phi = N \phi = \frac{\mu_0 N^2 I A}{L}$.
By definition,the self-inductance $L_{ind}$ is given by $\Phi = L_{ind} I$.
Comparing the two expressions,we get $L_{ind} = \frac{\mu_0 N^2 A}{L}$.

(D) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the coil.
From this formula,it is clear that $L \propto N^2$,provided that the length $l$ and area $A$ remain constant.
Given that the number of turns is increased to four times,i.e.,$N' = 4N$.
Therefore,the new self-inductance $L'$ will be $L' \propto (N')^2 = (4N)^2 = 16N^2$.
Thus,$L' = 16L$.

(B) The formula for the induced $e.m.f.$ in a coil due to self-induction is given by $e = -L \frac{di}{dt}$.
Here,the magnitude of the induced $e.m.f.$ is $|e| = 8 \,V$.
The change in current is $di = 2 \,A - 0 \,A = 2 \,A$.
The time interval is $dt = 0.05 \,s$.
Substituting these values into the formula: $8 = L \times \frac{2}{0.05}$.
$8 = L \times 40$.
$L = \frac{8}{40} = 0.2 \,H$.
Therefore,the self-inductance of the coil is $0.2 \,H$.

(D) The induced electromotive force $e$ is given by $e = \frac{d\phi}{dt} = L \frac{dI}{dt}$.
From this,the unit of inductance $L$ is $[L] = \frac{[\text{Weber}]}{[\text{Ampere}]}$.
Since $e = \frac{d\phi}{dt}$,we have $[\text{Volt}] = \frac{[\text{Weber}]}{[\text{Second}]}$,which implies $[\text{Weber}] = [\text{Volt} \cdot \text{Second}]$.
Substituting this,we get $[L] = \frac{[\text{Volt} \cdot \text{Second}]}{[\text{Ampere}]}$.
Also,the energy stored in an inductor is $U = \frac{1}{2} L I^2$. Thus,$[\text{Joule}] = [L] \cdot [\text{Ampere}]^2$,which implies $[L] = \frac{[\text{Joule}]}{[\text{Ampere}]^2}$.
Therefore,all the given expressions are equivalent to the henry.

(C) The formula for induced $EMF$ in a coil due to self-inductance is given by $|e| = L \left| \frac{di}{dt} \right|$.
Given:
Change in current,$di = 10 \, A - 0 \, A = 10 \, A$.
Time interval,$dt = 0.5 \, s$.
Induced $EMF$,$|e| = 220 \, V$.
Substituting these values into the formula:
$220 = L \times \frac{10}{0.5}$
$220 = L \times 20$
$L = \frac{220}{20} = 11 \, H$.
Therefore,the self-inductance of the coil is $11 \, H$.

(C) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$, where $N$ is the number of turns, $A$ is the area of cross-section, and $l$ is the length of the solenoid.
Given that the new number of turns $N' = 2N$ and the new length $l' = 2l$, while the area $A$ remains constant.
The new inductance $L'$ is given by:
$L' = \frac{\mu_0 (N')^2 A}{l'} = \frac{\mu_0 (2N)^2 A}{2l} = \frac{\mu_0 (4N^2) A}{2l} = 2 \left( \frac{\mu_0 N^2 A}{l} \right) = 2L$.
Therefore, the inductance is doubled.

(A) The self-inductance $L$ of a coil is given by the relation $L = \frac{N \phi}{I}$.
For a straight conductor,the number of turns $N$ is effectively $0$.
Since the self-inductance is directly proportional to the number of turns ($L \propto N^2$ or $L \propto N$ depending on geometry),for a straight wire where there are no loops or turns,the magnetic flux linkage is zero.
Therefore,the self-inductance of a straight conductor is $0$.

(A) The formula for the induced $e.m.f.$ in a coil due to self-inductance is given by $e = L \left| \frac{di}{dt} \right|$.
Given:
Initial current $i_1 = 4 \, A$
Final current $i_2 = 0 \, A$
Change in current $di = i_1 - i_2 = 4 \, A$
Time interval $dt = 0.1 \, s$
Induced $e.m.f.$ $e = 100 \, V$
Substituting the values into the formula:
$100 = L \times \frac{4}{0.1}$
$100 = L \times 40$
$L = \frac{100}{40} = 2.5 \, H$.
Therefore,the self-inductance of the coil is $2.5 \, H$.

(D) The formula for induced $e.m.f.$ $(e)$ in a coil due to self-inductance $(L)$ is given by $|e| = L \left| \frac{di}{dt} \right|$.
Given:
Induced $e.m.f.$ $(e)$ = $10 \, V$
Change in current $(di)$ = $10 \, A$
Time interval $(dt)$ = $1 \, s$
Substituting these values into the formula:
$10 = L \times \frac{10}{1}$
$10 = 10L$
$L = 1 \, H$
Therefore,the self-inductance of the coil is $1 \, H$.

(A) The relationship between magnetic flux linkage and inductance is given by $N\phi = Li$,where $N$ is the number of turns,$\phi$ is the magnetic flux per turn,$L$ is the inductance,and $i$ is the current.
Given: $N = 400$,$L = 8 \, mH = 8 \times 10^{-3} \, H$,$i = 5 \, mA = 5 \times 10^{-3} \, A$.
Substituting the values into the formula:
$\phi = \frac{Li}{N} = \frac{8 \times 10^{-3} \times 5 \times 10^{-3}}{400} = \frac{40 \times 10^{-6}}{400} = 0.1 \times 10^{-6} = 10^{-7} \, Wb$.
We know that $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$. Therefore,$10^{-7} \, Wb = \frac{\mu_0}{4\pi} \, Wb$.

(B) The magnetic flux $\phi$ through a solenoid is proportional to the current $I$ flowing through it, given by $\phi = LI$, where $L$ is the self-inductance of the solenoid.
According to Faraday's Law, the induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = -L \frac{dI}{dt}$.
Since the current $I$ increases at a constant rate, $\frac{dI}{dt}$ is a positive constant.
Therefore, the induced $EMF$ $\varepsilon = -L \times (\text{constant})$ is a constant value.
According to Lenz's Law, the direction of the induced current is such that it opposes the change in magnetic flux that produced it.
Since the inducing current is increasing, the induced current will flow in the opposite direction to oppose this increase.

(D) The magnetic flux linked with the coil is $\phi = B \cdot A$.
For a coil with $N$ turns,the flux linkage is $N\phi = Li$,where $L$ is the self-inductance.
Differentiating with respect to time $t$,we get: $N \frac{d\phi}{dt} = L \frac{di}{dt}$.
Since $\phi = BA$,we have $NB \frac{dA}{dt} = L \frac{di}{dt}$.
Given: $\frac{dA}{dt} = 5 \, m^2/ms = 5 \times 10^3 \, m^2/s$,$B = 1 \, T$,$di = (2 - 1) \, A = 1 \, A$,$dt = 2 \times 10^{-3} \, s$,and assuming $N = 1$.
Substituting the values: $1 \times 1 \times (5 \times 10^3) = L \times \left( \frac{1}{2 \times 10^{-3}} \right)$.
$5000 = L \times 500$.
$L = \frac{5000}{500} = 10 \, H$.

(B) The formula for the self-inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{l}$.
Given values are:
Length $l = 0.5 \, m$
Area $A = 20 \, cm^2 = 20 \times 10^{-4} \, m^2$
Number of turns $N = 500$
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$.
Substituting these values into the formula:
$L = \frac{4\pi \times 10^{-7} \times (500)^2 \times 20 \times 10^{-4}}{0.5}$
$L = \frac{4 \times 3.14159 \times 10^{-7} \times 250000 \times 20 \times 10^{-4}}{0.5}$
$L = \frac{12.566 \times 10^{-7} \times 500}{0.5} \approx 1.2566 \times 10^{-3} \, H = 1.2566 \, mH$.
Rounding to the nearest provided option, we get $1.25 \, mH$.

(D) The induced $e.m.f.$ $(e)$ in a coil is given by the formula: $e = L \cdot \frac{di}{dt}$.
Here,$e = 12 \, V$.
The rate of change of current is $\frac{di}{dt} = 45 \, A/min$.
First,convert the rate of change of current into $A/s$:
$\frac{di}{dt} = \frac{45 \, A}{60 \, s} = 0.75 \, A/s$.
Now,substitute the values into the formula:
$12 = L \times 0.75$.
$L = \frac{12}{0.75} = \frac{1200}{75} = 16 \, H$.
Therefore,the inductance of the coil is $16 \, H$.

(A) The formula for the average induced $e.m.f.$ in a coil due to self-inductance is given by $|e| = L \left| \frac{di}{dt} \right|$.
Here,the change in current $di = 10\,A - (-10\,A) = 20\,A$.
The time interval $dt = 0.5\,s$.
The induced $e.m.f.$ $e = 1\,V$.
Substituting these values into the formula:
$1 = L \times \frac{20}{0.5}$
$1 = L \times 40$
$L = \frac{1}{40}\,H = 0.025\,H$.
Converting to millihenry $(mH)$:
$L = 0.025 \times 1000\,mH = 25\,mH$.

(C) The self-inductance $L$ of a solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the cross-sectional area,and $l$ is the length of the solenoid.
From this formula,it is clear that the self-inductance $L$ is directly proportional to the square of the number of turns,i.e.,$L \propto N^2$.
If the number of turns $N$ is doubled $(N' = 2N)$,the new self-inductance $L'$ will be:
$L' \propto (2N)^2 = 4N^2 = 4L$.
Therefore,the self-inductance becomes $4L$ henry.

(A) The induced $e.m.f.$ $(e)$ in a coil due to self-inductance is given by the formula: $|e| = L \frac{di}{dt}$.
Here,the change in current $di = 18 \,A - 2 \,A = 16 \,A$.
The time interval $dt = 0.05 \,s$.
The induced $e.m.f.$ $|e| = 20 \,V$.
Substituting these values into the formula:
$20 = L \times \frac{16}{0.05}$
$20 = L \times 320$
$L = \frac{20}{320} \,H = \frac{1}{16} \,H = 0.0625 \,H$.
To convert the value into $mH$,multiply by $1000$:
$L = 0.0625 \times 1000 \,mH = 62.5 \,mH$.

(B) The induced $e.m.f.$ $(e)$ in an inductor is given by the formula:
$|e| = L \frac{di}{dt}$
Given:
Inductance $L = 10 \, \mu H = 10 \times 10^{-6} \, H$
Change in current $di = 1 \, A - 0 \, A = 1 \, A$
Time interval $dt = 10 \, s$
Substituting the values into the formula:
$|e| = (10 \times 10^{-6} \, H) \times \frac{1 \, A}{10 \, s}$
$|e| = 10^{-6} \, V$
$|e| = 1 \, \mu V$
Therefore,the correct option is $B$.

(C) The self-inductance $L$ is defined by the relation $\phi = LI$,where $\phi$ is magnetic flux and $I$ is current. Thus,$L = \phi / I$. The unit of flux $\phi$ is Weber $(Wb)$ and current $I$ is Ampere $(A)$,so $L$ has units of $Wb/A$ (Henry).
Since $V = L(dI/dt)$,we have $L = V \cdot t / I$. Since $V/I = R$ (Ohm),$L$ has units of $Ohm \cdot s$.
Energy stored in an inductor is $U = \frac{1}{2}LI^2$,so $L = 2U/I^2$. Since $U$ is in Joules $(J)$,$L$ has units of $J/A^2$ or $J \cdot A^{-2}$.
Comparing these with the options,$J \cdot A$ is not a unit of self-inductance. Therefore,option $C$ is correct.

(D) The total magnetic flux linkage $(\Phi_T)$ in a coil is given by the product of the number of turns $(N)$ and the magnetic flux through a single turn $(\phi)$: $\Phi_T = N\phi$.
Given: $N = 100$, $\phi = 10^{-5} \, Wb$, and current $i = 5 \, mA = 5 \times 10^{-3} \, A$.
The formula for self-inductance $(L)$ is $\Phi_T = Li$.
Substituting the values: $N\phi = Li$.
$100 \times 10^{-5} = L \times (5 \times 10^{-3})$.
$10^{-3} = L \times 5 \times 10^{-3}$.
$L = \frac{10^{-3}}{5 \times 10^{-3}} = \frac{1}{5} = 0.2 \, H$.
Converting to $mH$: $0.2 \, H = 0.2 \times 1000 \, mH = 200 \, mH$.
Since $200 \, mH$ is not among the options, the correct choice is $D$.

(A) The self-inductance $L$ of a circular coil is given by the formula $L = \frac{\mu_0 N^2 A}{l}$,where $N$ is the number of turns,$A$ is the area of cross-section,and $l$ is the length of the coil.
From this relation,we see that $L \propto N^2$.
Given that the number of turns $N$ is doubled $(N' = 2N)$,the new inductance $L'$ will be $L' \propto (2N)^2 = 4N^2$.
Therefore,$L' = 4L$.
The resistance of the coil does not affect the self-inductance of the coil,as inductance depends only on the geometry and the number of turns of the coil.
Thus,the inductance becomes $4$ times the initial value.

(C) The induced $e.m.f.$ $(e)$ in a coil is given by the formula: $e = -L \frac{di}{dt}$.
Given,inductance $L = 5\, H$.
The rate of change of current $\frac{di}{dt} = -2\, A/s$ (since the current is decreasing).
Substituting these values into the formula:
$e = -5 \times (-2) = +10\, V$.
Therefore,the induced $e.m.f.$ is $10\, V$.

(C) The formula for self-induced $e.m.f.$ $(e)$ in a coil is given by $e = L \frac{di}{dt}$,where $L$ is the self-inductance and $\frac{di}{dt}$ is the rate of change of current.
Given:
Self-inductance $L = 0.1 \, H$
Rate of change of current $\frac{di}{dt} = 200 \, A/s$
Substituting the values into the formula:
$e = 0.1 \times 200 = 20 \, V$
Therefore,the self-induced $e.m.f.$ is $20 \, V$.

(C) The self-inductance $L$ of a solenoid is given by the formula: $L = \frac{\mu_0 N^2 A}{l}$.
Given:
Number of turns $N = 1000 = 10^3$.
Length $l = 1 \, m$.
Area $A = 10 \, cm^2 = 10 \times 10^{-4} \, m^2 = 10^{-3} \, m^2$.
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$.
Substituting the values:
$L = \frac{4\pi \times 10^{-7} \times (10^3)^2 \times 10^{-3}}{1}$
$L = 4\pi \times 10^{-7} \times 10^6 \times 10^{-3}$
$L = 4\pi \times 10^{-4} \, H$
$L = 4 \times 3.14159 \times 10^{-4} \, H = 12.566 \times 10^{-4} \, H = 1.2566 \times 10^{-3} \, H$
$L = 1.2566 \, mH \approx 1.256 \, mH$.

(A) The formula for induced $e.m.f.$ in a coil due to self-induction is given by $e = L \left| \frac{di}{dt} \right|$.
Given:
Initial current $i_1 = +2 \, A$
Final current $i_2 = -2 \, A$
Change in current $di = i_2 - i_1 = -2 - 2 = -4 \, A$
Time interval $dt = 0.05 \, s$
Induced $e.m.f.$ $e = 8 \, V$
Substituting the values into the formula:
$8 = L \times \frac{|-4|}{0.05}$
$8 = L \times \frac{4}{0.05}$
$8 = L \times 80$
$L = \frac{8}{80} = 0.1 \, H$.

(A) The voltage across an inductor is given by $V = L \frac{di}{dt}$.
When there is a change in current,the inductor opposes this change by generating an induced electromotive force (emf) according to Lenz's Law.
If the current were to rise immediately,the rate of change of current $\frac{di}{dt}$ would be infinite,which would require an infinite induced emf.
Since the circuit cannot support an infinite emf,the current rises gradually,following an exponential growth curve until it reaches its steady-state value.