Energy Stored in Inductor Questions in English

(B) The energy $E$ stored in an inductor is given by the formula $E = \frac{1}{2} L i^2$.
Given:
Inductance $L = 50 \, mH = 50 \times 10^{-3} \, H = 0.05 \, H$.
Current $i = 2 \, A$.
Substituting the values into the formula:
$E = \frac{1}{2} \times (0.05 \, H) \times (2 \, A)^2$
$E = \frac{1}{2} \times 0.05 \times 4$
$E = 0.05 \times 2$
$E = 0.1 \, J$.

(D) The energy stored in an inductor is derived from the work done against the induced back electromotive force $(e.m.f.)$.
The induced $e.m.f.$ in an inductor is given by $e = -L \frac{di}{dt}$.
The work done $dW$ to increase the current by $di$ in time $dt$ against this back $e.m.f.$ is $dW = -e \cdot i \cdot dt$.
Substituting the expression for $e$,we get $dW = -(-L \frac{di}{dt}) \cdot i \cdot dt = L \cdot i \cdot di$.
To find the total energy $U$,we integrate from current $0$ to $i$:
$U = \int_0^i L \cdot i \cdot di = L \left[ \frac{i^2}{2} \right]_0^i = \frac{1}{2} L i^2$.

(A) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$,where $L$ is the self-inductance and $i$ is the current flowing through the inductor.
This energy is associated with the magnetic field produced by the current in the inductor.
Therefore,the energy is stored in the form of a magnetic field.

(A) The energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L I^2$.
Given:
Inductance $L = 50 \, mH = 50 \times 10^{-3} \, H$.
Current $I = 4 \, A$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (50 \times 10^{-3}) \times (4)^2$
$U = \frac{1}{2} \times 50 \times 10^{-3} \times 16$
$U = 25 \times 16 \times 10^{-3}$
$U = 400 \times 10^{-3} \, J$
$U = 0.4 \, J$.

(B) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$,where $L$ is the self-inductance and $i$ is the current.
Let the initial current be $i_1$ and the initial energy be $U_1 = \frac{1}{2} L i_1^2$.
When the current is halved,the new current $i_2 = \frac{i_1}{2}$.
The new energy stored is $U_2 = \frac{1}{2} L i_2^2 = \frac{1}{2} L \left( \frac{i_1}{2} \right)^2 = \frac{1}{2} L \left( \frac{i_1^2}{4} \right) = \frac{1}{4} \left( \frac{1}{2} L i_1^2 \right) = \frac{1}{4} U_1$.
Thus,$U_2 = 0.25 U_1$. Therefore,the energy becomes $0.25$ times the previous value.

(C) The energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L I^2$.
Given:
Inductance $L = 100\, mH = 100 \times 10^{-3}\, H = 0.1\, H$.
Current $I = 1\, A$.
Substituting the values into the formula:
$U = \frac{1}{2} \times 0.1 \times (1)^2$
$U = 0.05\, J$.
Therefore,the energy stored in the magnetic field is $0.05\, J$.

(D) The current $i$ flowing through the coil when connected to a $100 \, V$ battery is given by Ohm's law: $i = \frac{V}{R} = \frac{100 \, V}{10 \, \Omega} = 10 \, A$.
The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} L i^2$.
Substituting the given values $L = 5 \, H$ and $i = 10 \, A$:
$U = \frac{1}{2} \times 5 \times (10)^2 = \frac{1}{2} \times 5 \times 100 = 250 \, J$.
Therefore, the energy stored in the coil is $250 \, J$.

(A) The energy $U$ stored in an inductor with self-inductance $L$ carrying a current $i$ is given by the formula:
$U = \frac{1}{2} L i^2$
Given:
$L = 40 \, mH = 40 \times 10^{-3} \, H$
$i = 2 \, A$
Substituting the values into the formula:
$U = \frac{1}{2} \times (40 \times 10^{-3}) \times (2)^2$
$U = \frac{1}{2} \times 40 \times 10^{-3} \times 4$
$U = 20 \times 10^{-3} \times 4 = 80 \times 10^{-3} \, J$
$U = 0.08 \, J$

(B) The current $i$ flowing through the coil in a steady state is given by $i = \frac{E}{R}$.
Given $E = 100\, V$ and $R = 20\, \Omega$,we have $i = \frac{100}{20} = 5\, A$.
The energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$.
Substituting the values $L = 5\, H$ and $i = 5\, A$:
$U = \frac{1}{2} \times 5 \times (5)^2 = \frac{1}{2} \times 5 \times 25 = \frac{125}{2} = 62.50\, J$.

(C) In a steady-state $D.C.$ circuit, the inductor acts as a short circuit (zero resistance).
First, calculate the steady-state current $i$ flowing through the circuit using Ohm's law:
$i = \frac{V}{R} = \frac{10 \text{ V}}{2 \text{ }\Omega} = 5 \text{ A}$.
The magnetic energy $U$ stored in an inductor is given by the formula:
$U = \frac{1}{2} L i^2$.
Substituting the given values $L = 2 \text{ H}$ and $i = 5 \text{ A}$:
$U = \frac{1}{2} \times 2 \times (5)^2 = 1 \times 25 = 25 \text{ J}$.
Therefore, the magnetic energy stored in the coil is $25 \text{ J}$.

(A) The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} L I^2$.
Given:
Inductance $L = 100 \, mH = 100 \times 10^{-3} \, H = 0.1 \, H$.
Current $I = 10 \, A$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (0.1) \times (10)^2$
$U = \frac{1}{2} \times 0.1 \times 100$
$U = 0.1 \times 50$
$U = 5 \, J$.
Therefore,the energy stored in the inductor is $5 \, J$.

(C) In a steady state, the inductor acts as a short circuit (zero resistance).
The current $i$ flowing through the circuit is given by Ohm's law: $i = \frac{V}{R} = \frac{10 \text{ V}}{2 \text{ }\Omega} = 5 \text{ A}$.
The energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$.
Substituting the values $L = 2 \text{ H}$ and $i = 5 \text{ A}$:
$U = \frac{1}{2} \times 2 \times (5)^2 = 1 \times 25 = 25 \text{ J}$.

(C) The magnetic potential energy $U$ stored in an inductor of inductance $L$ carrying current $I$ is given by the formula:
$U = \frac{1}{2} L I^{2}$
Given:
$U = 25\, mJ = 25 \times 10^{-3}\, J$
$I = 60\, mA = 60 \times 10^{-3}\, A = 6 \times 10^{-2}\, A$
Substituting the values into the formula:
$25 \times 10^{-3} = \frac{1}{2} \times L \times (6 \times 10^{-2})^{2}$
$25 \times 10^{-3} = \frac{1}{2} \times L \times 36 \times 10^{-4}$
$25 \times 10^{-3} = L \times 18 \times 10^{-4}$
$L = \frac{25 \times 10^{-3}}{18 \times 10^{-4}}$
$L = \frac{250}{18} = 13.888...\, H \approx 13.89\, H$.

(C) The potential difference across the coil is given by $V = L \frac{di}{dt}$.
Given $L = 2\,H$ and $\frac{di}{dt} = 4\,A/s$,we have:
$V = (2)(4) = 8\,V$.
The energy stored in the inductor per unit time is the power $P$ delivered to the inductor,which is given by $P = V \cdot i$.
Substituting the values $V = 8\,V$ and $i = 2\,A$:
$P = (8)(2) = 16\,J/s$.

(A) An inductor is a passive electronic component that stores energy in its magnetic field when electric current flows through it. The energy stored is given by $U = \frac{1}{2} L I^2$,where $L$ is the inductance and $I$ is the current. Thus,statement $(a)$ is correct.
$A$ capacitor is a passive electronic component that stores energy in an electric field created by the potential difference between its plates. The energy stored is given by $U = \frac{1}{2} C V^2$,where $C$ is the capacitance and $V$ is the voltage. Thus,statement $(b)$ is correct.
Statements $(c)$ and $(d)$ are incorrect because inductors do not store energy in electric fields (ideally) and capacitors do not store energy in magnetic fields.
Therefore,both statements $(a)$ and $(b)$ are correct.

(D) When a current flows through a coil,it generates a magnetic field around it.
According to the principles of electromagnetism,the energy associated with an inductor or a current-carrying coil is stored in the magnetic field created by the current.
Unlike a capacitor,which stores energy in an electric field,an inductor stores energy in the form of a magnetic field.
Therefore,the energy in a current-carrying coil is stored in the form of a magnetic field only.

(C) The magnetic field $B$ produced by a solenoid is given by $B = \mu_0 n I$. When the current $I$ is reversed,the direction of the magnetic field $B$ is reversed,but its magnitude remains the same.
The magnetic field energy density $u$ is given by the formula $u = \frac{B^2}{2\mu_0}$.
Since the energy density depends on the square of the magnetic field magnitude $(B^2)$,the sign of $B$ does not affect the energy density.
Therefore,the total magnetic field energy stored in the solenoid,$U = u \times V$ (where $V$ is the volume),remains unchanged.
However,the Reason states that the energy density is proportional to the magnetic field,which is incorrect because it is proportional to the square of the magnetic field $(B^2)$.
Thus,the Assertion is correct,but the Reason is incorrect.

(N/A) The magnetic energy stored in an inductor is given by $U_{B} = \frac{1}{2} L I^{2}$.
For a solenoid,the magnetic field is $B = \mu_{0} n I$,where $n = N/l$ is the number of turns per unit length.
Thus,$I = \frac{B}{\mu_{0} n}$.
The self-inductance of a solenoid is $L = \mu_{0} n^{2} A l$.
Substituting these into the energy formula:
$U_{B} = \frac{1}{2} (\mu_{0} n^{2} A l) \left(\frac{B}{\mu_{0} n}\right)^{2} = \frac{1}{2} (\mu_{0} n^{2} A l) \left(\frac{B^{2}}{\mu_{0}^{2} n^{2}}\right) = \frac{B^{2} A l}{2 \mu_{0}}$.
$(b)$ The magnetic energy density (energy per unit volume) is $u_{B} = \frac{U_{B}}{V} = \frac{U_{B}}{A l} = \frac{B^{2}}{2 \mu_{0}}$.
The electrostatic energy density stored in a parallel plate capacitor is $u_{E} = \frac{1}{2} \varepsilon_{0} E^{2}$.
In both cases,the energy density is proportional to the square of the field strength ($B^{2}$ or $E^{2}$). These expressions are general and valid for any region of space where electric or magnetic fields exist.

(A) An inductor is a passive electrical component that stores energy in its magnetic field when electric current flows through it,characterized by its self-inductance $L$.
When the current $I$ in an inductor changes,a back electromotive force (emf) $\varepsilon$ is induced,which opposes the change in current according to Lenz's Law. The magnitude of this back emf is given by $|\varepsilon| = L \frac{dI}{dt}$.
To establish a current $I$ in the inductor,work must be done against this back emf. The rate of doing work is given by:
$\frac{dW}{dt} = |\varepsilon| I = L I \frac{dI}{dt}$
Integrating this expression to find the total work done $W$ to increase the current from $0$ to $I$:
$W = \int dW = \int_{0}^{I} L I' dI' = L \left[ \frac{I'^2}{2} \right]_{0}^{I} = \frac{1}{2} LI^2$
This work done is stored as magnetic potential energy $U$ in the inductor:
$U = \frac{1}{2} LI^2$
This expression is analogous to the kinetic energy of a particle,$K = \frac{1}{2}mv^2$,where $L$ acts as electrical inertia.

(A) The induced electromotive force $(EMF)$ in an inductor is given by $\varepsilon = L \frac{dI}{dt}$.
Since the resistance is negligible,the applied voltage $V$ is equal to the induced $EMF$ $\varepsilon$,so $V = L \frac{dI}{dt}$.
Given $V = 3t$ and $L = 2\, H$,we have $3t = 2 \frac{dI}{dt}$.
Rearranging the terms,we get $dI = \frac{3}{2} t \, dt$.
Integrating both sides from $t = 0$ to $t = 4\, s$ and $I = 0$ to $I = I_f$:
$\int_{0}^{I_f} dI = \int_{0}^{4} \frac{3}{2} t \, dt$
$I_f = \frac{3}{2} \left[ \frac{t^2}{2} \right]_{0}^{4} = \frac{3}{4} (16) = 12\, A$.
The energy stored in the coil is given by $U = \frac{1}{2} L I_f^2$.
$U = \frac{1}{2} \times 2 \times (12)^2 = 144\, J$.

(A) The equilibrium current $I$ in the coil is given by Ohm's law: $I = \frac{V}{R} = \frac{10 \, V}{4 \, \Omega} = 2.5 \, A$.
The energy $E$ stored in the magnetic field of an inductor is given by the formula: $E = \frac{1}{2} L I^2$.
Substituting the given values $L = 2 \, H$ and $I = 2.5 \, A$:
$E = \frac{1}{2} \times 2 \times (2.5)^2 = 6.25 \, J$.
To express this in the form $.......... \times 10^{-2} \, J$:
$E = 625 \times 10^{-2} \, J$.
Thus,the value is $625$.

(A) The formula for magnetic energy $U$ stored in an inductor is given by $U = \frac{1}{2} L i^2$.
Given:
Inductance $L = 4\,\mu H = 4 \times 10^{-6}\,H$
Current $i = 2\,A$
Substituting the values into the formula:
$U = \frac{1}{2} \times (4 \times 10^{-6}) \times (2)^2$
$U = \frac{1}{2} \times 4 \times 10^{-6} \times 4$
$U = 8 \times 10^{-6}\,J$
Since $10^{-6}\,J = 1\,\mu J$,the energy is $8\,\mu J$.

(B) The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$.
Given:
Self-inductance $L = 3 \ mH = 3 \times 10^{-3} \ H$.
Current $I = 2 \ A$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (3 \times 10^{-3} \ H) \times (2 \ A)^2$
$U = \frac{1}{2} \times 3 \times 10^{-3} \times 4$
$U = 6 \times 10^{-3} \ J$
Since $1 \ mJ = 10^{-3} \ J$,the energy stored is $6 \ mJ$.

(D) The energy stored in an $LR$ circuit is given by the formula:
$E = \frac{1}{2} LI^2$ ...$(i)$
Given that the current is reduced to half,the new current $I^{\prime}$ is:
$I^{\prime} = \frac{I}{2}$
Substituting this into the energy formula,the new energy $E^{\prime}$ is:
$E^{\prime} = \frac{1}{2} L (I^{\prime})^2$
$E^{\prime} = \frac{1}{2} L \left(\frac{I}{2}\right)^2$
$E^{\prime} = \frac{1}{2} L \left(\frac{I^2}{4}\right)$
$E^{\prime} = \frac{1}{4} \left(\frac{1}{2} LI^2\right)$
Using equation $(i)$,we get:
$E^{\prime} = \frac{1}{4} E$
Therefore,the energy stored becomes $(1/4)^{\text{th}}$ of the original energy.

(D) The energy stored in an inductor when the current reaches its steady state value $I_0$ is given by $U = \frac{1}{2} L I_0^2$.
Here,the steady state current $I_0$ is determined by the Ohm's law for the circuit: $I_0 = \frac{V}{R}$.
Since both circuits have the same e.m.f. $V = 10 \ V$ and the same resistance $R = 40 \ \Omega$,the steady state current $I_0$ is the same for both circuits.
Therefore,the ratio of energy stored in circuit $A$ to circuit $B$ is $\frac{U_A}{U_B} = \frac{\frac{1}{2} L_A I_0^2}{\frac{1}{2} L_B I_0^2} = \frac{L_A}{L_B}$.
Given $L_A = 10 \ H$ and $L_B = 10 \ mH = 10 \times 10^{-3} \ H = 0.01 \ H$.
Thus,the ratio is $\frac{10}{0.01} = 1000$.

(C) The magnetic energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$.
Given: Inductance $L = 5 \mu H = 5 \times 10^{-6} \ H$ and current $I = 2 \ A$.
Substituting the values: $U = \frac{1}{2} \times (5 \times 10^{-6}) \times (2)^2$.
$U = \frac{1}{2} \times 5 \times 10^{-6} \times 4$.
$U = 10 \times 10^{-6} \ J = 10 \mu J$.

(C) The energy stored in an inductor is given by the formula: $E = \frac{1}{2} LI^2$.
Given: Energy $E = 64 \times 10^{-3} \ J$ and current $I = 80 \ mA = 80 \times 10^{-3} \ A$.
Rearranging the formula to solve for inductance $L$: $L = \frac{2E}{I^2}$.
Substituting the values: $L = \frac{2 \times 64 \times 10^{-3}}{(80 \times 10^{-3})^2}$.
$L = \frac{128 \times 10^{-3}}{6400 \times 10^{-6}} = \frac{128 \times 10^{-3}}{6.4 \times 10^{-3}} = \frac{128}{6.4} = 20 \ H$.

(B) Given: Current $I = 4 \ A$,Number of turns $N = 400$,Magnetic flux $\phi = 3 \times 10^{-3} \ Wb$.
The magnetic flux linkage is given by $N\phi = LI$,where $L$ is the self-inductance of the coil.
First,calculate the self-inductance $L$:
$L = \frac{N\phi}{I} = \frac{400 \times 3 \times 10^{-3}}{4} = 100 \times 3 \times 10^{-3} = 0.3 \ H$.
The energy stored in an inductor is given by the formula $U = \frac{1}{2} LI^2$.
Substituting the values:
$U = \frac{1}{2} \times 0.3 \times (4)^2$
$U = \frac{1}{2} \times 0.3 \times 16$
$U = 0.3 \times 8 = 2.4 \ J$.

(C) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance and $I$ is the current.
Let the initial current be $I_1 = I$ and the initial energy be $E_1 = \frac{1}{2} L I^2$.
If the current is reduced by $50 \%$,the new current becomes $I_2 = I - 0.5 I = 0.5 I = \frac{I}{2}$.
The new energy stored in the coil is $E_2 = \frac{1}{2} L (I_2)^2 = \frac{1}{2} L (\frac{I}{2})^2 = \frac{1}{2} L (\frac{I^2}{4}) = \frac{1}{4} E_1$.
The change in energy is $\Delta E = E_1 - E_2 = E_1 - \frac{1}{4} E_1 = \frac{3}{4} E_1$.
Since $\frac{3}{4} = 0.75$,the energy decreases by $75 \%$.

(B) The energy stored in an inductor (coil) is given by the formula $E = \frac{1}{2} L I^2$,where $L$ is the self-inductance of the coil.
From this formula,we can see that $E \propto I^2$.
Initially,when the current is $I$,the energy is $E_1 = \frac{1}{2} L I^2$.
When the current is reduced to $I' = I/2$,the new energy $E_2$ is given by $E_2 = \frac{1}{2} L (I/2)^2 = \frac{1}{2} L (I^2/4) = \frac{1}{4} E_1$.
The change in energy is $\Delta E = E_1 - E_2$.
Substituting the value of $E_2$,we get $\Delta E = E_1 - \frac{E_1}{4} = \frac{3E_1}{4}$.

(D) The magnetic potential energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$.
Given:
$U = 25 \ mJ = 25 \times 10^{-3} \ J$
$I = 50 \ mA = 50 \times 10^{-3} \ A$
Rearranging the formula to solve for inductance $L$:
$L = \frac{2U}{I^2}$
Substituting the values:
$L = \frac{2 \times 25 \times 10^{-3}}{(50 \times 10^{-3})^2}$
$L = \frac{50 \times 10^{-3}}{2500 \times 10^{-6}}$
$L = \frac{50 \times 10^{-3}}{2.5 \times 10^{-3}}$
$L = 20 \ H$.

(A) The energy $E$ stored in a coil with self-inductance $L$ carrying current $I$ is given by the formula: $E = \frac{1}{2} LI^2$.
Let the initial current be $I_1$ and the initial energy be $E_1 = \frac{1}{2} LI_1^2$.
Let the new current be $I_2 = \frac{I_1}{2}$ and the new energy be $E_2 = \frac{1}{2} LI_2^2$.
Taking the ratio of the two energies:
$\frac{E_2}{E_1} = \frac{\frac{1}{2} LI_2^2}{\frac{1}{2} LI_1^2} = \left( \frac{I_2}{I_1} \right)^2$.
Substituting $I_2 = \frac{I_1}{2}$:
$\frac{E_2}{E_1} = \left( \frac{I_1 / 2}{I_1} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Therefore,$E_2 = \frac{E_1}{4}$.

(B) The energy stored in an inductor is given by the formula $U = \frac{1}{2} L i^2$.
When a coil is connected to a battery of $EMF$ $E$ and resistance $R$,the steady-state current $i$ is given by $i = \frac{E}{R}$.
Substituting the values: $E = 10 \ V$,$L = 4 \ H$,and $R = 5 \ \Omega$.
First,calculate the steady-state current: $i = \frac{10 \ V}{5 \ \Omega} = 2 \ A$.
Now,calculate the stored energy: $U = \frac{1}{2} \times 4 \ H \times (2 \ A)^2$.
$U = 2 \times 4 \ J = 8 \ J$.

(C) An inductor stores energy in the form of a magnetic field created by the current flowing through it.
The energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L I^{2}$,where $L$ is the self-inductance of the inductor and $I$ is the current flowing through it.
Therefore,the correct option is $C$.

(B) The self-inductance $L$ of a coil with $N$ turns is defined by the relation $N\phi = LI$,where $\phi$ is the magnetic flux through each turn.
Thus,the inductance is $L = \frac{N\phi}{I}$.
The magnetic energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2}LI^2$.
Substituting the expression for $L$ into the energy formula:
$U = \frac{1}{2} \left( \frac{N\phi}{I} \right) I^2$.
Simplifying this,we get $U = \frac{1}{2} N\phi I$ or $U = \frac{N\phi I}{2}$.

(B) The energy $E$ stored in an inductor is given by the formula $E = \frac{1}{2} LI^2$.
Given: Inductance $L = 100 mH = 100 \times 10^{-3} H = 0.1 H$ and Current $I = 1 A$.
Substituting the values:
$E = \frac{1}{2} \times 0.1 \times (1)^2$
$E = 0.05 J$.

(A) The energy $U$ stored in an inductor with self-inductance $L$ carrying a current $I$ is given by the formula:
$U = \frac{1}{2} LI^2$
Given:
$L = 200 \ mH = 200 \times 10^{-3} \ H = 0.2 \ H$
$I = 4 \ A$
Substituting the values into the formula:
$U = \frac{1}{2} \times 0.2 \times (4)^2$
$U = 0.1 \times 16$
$U = 1.6 \ J$
Therefore,the energy required is $1.6 \ J$.

(B) The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$.
Given: Self-inductance $L = 0.5 \ mH = 0.5 \times 10^{-3} \ H$ and current $I = 2 \ A$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (0.5 \times 10^{-3}) \times (2)^2$
$U = \frac{1}{2} \times 0.5 \times 10^{-3} \times 4$
$U = 0.5 \times 2 \times 10^{-3}$
$U = 1 \times 10^{-3} \ J = 0.001 \ J$.

(D) The magnetic energy stored in a solenoid is given by $U_B = \frac{1}{2} LI^2$.
For a solenoid,the self-inductance is $L = \mu_0 n^2 Al$ and the magnetic field is $B = \mu_0 nI$,which implies $I = \frac{B}{\mu_0 n}$.
Substituting these into the energy expression:
$U_B = \frac{1}{2} (\mu_0 n^2 Al) \left( \frac{B}{\mu_0 n} \right)^2 = \frac{1}{2} (\mu_0 n^2 Al) \left( \frac{B^2}{\mu_0^2 n^2} \right) = \frac{B^2 Al}{2 \mu_0}$.
The magnetic energy per unit volume $(u_B)$ is defined as the total energy divided by the volume $(V = Al)$:
$u_B = \frac{U_B}{V} = \frac{B^2 Al / (2 \mu_0)}{Al} = \frac{B^2}{2 \mu_0}$.

(A) The magnetic energy $E$ stored in a solenoid with inductance $L$ carrying a current $I$ is given by the formula:
$E = \frac{1}{2} L I^2$
Given:
Inductance $L = 2 \ H$
Current $I = 1 \ A$
Substituting the values into the formula:
$E = \frac{1}{2} \times 2 \times (1)^2$
$E = 1 \times 1 = 1 \ J$
Therefore,the magnetic energy stored in the solenoid is $1 \ J$.

(B) The magnetic potential energy $U$ stored in an inductor with self-inductance $L$ carrying a current $I$ is given by the formula:
$U = \frac{1}{2} L I^{2}$
This energy is stored in the magnetic field generated by the current flowing through the inductor.

(A) Given:
Current $I = 4 \ mA = 4 \times 10^{-3} \ A$
Magnetic flux $\phi = 32 \times 10^{-6} \ Wb$ (or $T \ m^2$)
First,calculate the self-inductance $L$ using the formula $\phi = L \cdot I$:
$L = \frac{\phi}{I} = \frac{32 \times 10^{-6}}{4 \times 10^{-3}} = 8 \times 10^{-3} \ H$
The energy stored in an inductor is given by $U = \frac{1}{2} L I^2$:
$U = \frac{1}{2} \times (8 \times 10^{-3}) \times (4 \times 10^{-3})^2$
$U = 4 \times 10^{-3} \times 16 \times 10^{-6}$
$U = 64 \times 10^{-9} \ J$
Therefore,the correct option is $A$.

(A) The magnetic energy $U$ stored in an inductor is given by the formula:
$U = \frac{1}{2} LI^2$
This implies that $U \propto I^2$.
Let the initial current be $I_1 = 2 \ A$ and the final current be $I_2 = 3 \ A$.
The ratio of the final energy $U_2$ to the initial energy $U_1$ is:
$\frac{U_2}{U_1} = \left( \frac{I_2}{I_1} \right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4} = 2.25$
Thus,$U_2 = 2.25 \ U_1$.
The percentage increase in the stored energy is given by:
$\Delta U \% = \left( \frac{U_2 - U_1}{U_1} \times 100 \right) \%$
$\Delta U \% = \left( \frac{2.25 \ U_1 - U_1}{U_1} \times 100 \right) \% = (1.25 \times 100) \% = 125 \%$

(A) The energy stored in a coil of inductance $L$ carrying a steady current $i$ is given by the formula:
$E = \frac{1}{2} L i^2$
This energy is associated with the magnetic field produced by the current flowing through the inductor.
Therefore,the energy stored in the coil is in the form of magnetic energy.

(A) Given,the inductance $L = 50 mH = 50 \times 10^{-3} H = 5 \times 10^{-2} H$.
The current $I = 4 A$.
The energy $E$ stored in an inductor is given by the formula $E = \frac{1}{2} L I^2$.
Substituting the values,we get $E = \frac{1}{2} \times (5 \times 10^{-2}) \times (4)^2$.
$E = \frac{1}{2} \times 5 \times 10^{-2} \times 16$.
$E = 5 \times 10^{-2} \times 8 = 40 \times 10^{-2} J = 0.4 J$.

(A) The magnetic energy density $u_B$ in a magnetic field $B$ is given by the formula:
$u_B = \frac{B^2}{2 \mu_0}$
Since the energy density is defined as energy per unit volume $(u_B = \frac{U}{V})$,the total energy $U$ stored in a volume $V$ is:
$U = u_B \times V$
The volume $V$ of a solenoid with length $L$ and cross-sectional area $A$ is $V = A \times L$.
Substituting the values,we get:
$U = \left( \frac{B^2}{2 \mu_0} \right) \times (A L) = \frac{1}{2 \mu_0} B^2 AL$