Q.(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field $B$, area $A$ and length $l$ of the solenoid.

Concept and Intuition

For a pure (resistanceless) inductor, the applied voltage sets the rate of change of current, not the current itself. So with a time-varying voltage we must integrate to find $i(t)$ before we can use the energy formula $U=\dfrac12Li^2$, which only depends on the instantaneous current.

Step-by-Step Solution

1. For an inductor, $v(t) = L\dfrac{di}{dt}$, so $\dfrac{di}{dt} = \dfrac{v(t)}{L} = \dfrac{4t}{3}$.
2. Integrate from $t=0$ (current starts at 0, since the voltage is applied at $t=0$): $i(t) = \displaystyle\int_0^t \dfrac{4t'}{3}\,dt' = \dfrac{4}{3}\cdot\dfrac{t^2}{2} = \dfrac{2}{3}t^2$.
3. At $t=3$ s: $i(3) = \dfrac{2}{3}\times9 = 6$ A. …

Concept and Intuition

An inductor stores energy in its magnetic field, analogous to how a capacitor stores energy in its electric field. The formula $U = \tfrac12LI^2$ comes from integrating the back-EMF work done in building up the current from zero to $I$.

Step-by-Step Solution

1. $L = 5\ H$, $I = 4\ mA = 4\times10^{-3}\ A$.
2. $I^2 = (4\times10^{-3})^2 = 16\times10^{-6}\ A^2$. …

Concept and Intuition

Just as an electric field stores energy at density $\dfrac{1}{2}\varepsilon_0 E^2$, a magnetic field stores energy at density $\dfrac{B^2}{2\mu_0}$ (this comes from the energy stored in an inductor's field, analogous to a capacitor's electric field energy). The relation $c=\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}$ links the two constants and lets us re-express $\mu_0$ in terms of $c$ and $\varepsilon_0$.

Step-by-Step Solution

1. Start with the magnetic energy density formula: $u_B=\dfrac{B^2}{2\mu_0}$.
2. From $c^2=\dfrac{1}{\mu_0\varepsilon_0}$, solve for $\mu_0$: $\mu_0=\dfrac{1}{\varepsilon_0 c^2}$. …

Concept and Intuition

An inductor stores energy in its magnetic field while carrying current, analogous to how a capacitor stores energy in its electric field. The stored energy is $U=\dfrac12 LI^2$.

Step-by-Step Solution

1. $L=50\,\text{mH}=0.05$ H, $I=2$ A. …

Concept and Intuition

An inductor's defining property is that the flux linked with it is proportional to the current through it, $\Phi = LI$, with $L$ the self-inductance. Once $L$ is known, the energy stored in the inductor's magnetic field for a given current follows the standard formula $U=\tfrac12LI^2$ — analogous to $\tfrac12CV^2$ for a capacitor's electric field.

Step-by-Step Solution

1. Find the inductance from $\Phi = LI$: $L = \dfrac{\Phi}{I} = \dfrac{32\times10^{-6}}{4\times10^{-3}} = 8\times10^{-3}\ \text{H}$.
2. Energy stored: $U = \dfrac{1}{2}LI^2 = \dfrac{1}{2}(8\times10^{-3})(4\times10^{-3})^2$. …

Concept and Intuition

Once a coil connected to a DC battery reaches steady state, the current no longer changes, so there is no induced back-emf ($\varepsilon = -L\,dI/dt = 0$). The circuit behaves as a pure resistor, and the final steady current is simply $I=V/R$. The energy stored in the magnetic field of an inductor carrying current $I$ is $U = \tfrac12 LI^2$.

Step-by-Step Solution

1. Steady-state current: $I = \dfrac{V}{R} = \dfrac{12}{0.8} = 15\ \mathrm{A}$.
2. Energy stored: $U = \tfrac12 L I^2 = \tfrac12 (0.64\times10^{-3})(15)^2$. …

Concept and Intuition

The energy stored in an inductor is $U=\frac12LI^2$ — quadratic in current, so even a modest current increase can produce a large percentage jump in stored energy.

Step-by-Step Solution

1. $U_1 = \frac12L(2)^2 = 2L$.
2. $U_2 = \frac12L(3)^2 = 4.5L$. …

Concept and Intuition

An inductor doesn't oppose current itself — it opposes a change in current. By Faraday's/Lenz's law, whenever the current through an inductor tries to rise or fall, the inductor generates a back-emf that fights that change. This is conceptually identical to Newton's first law: a mass doesn't oppose velocity itself, it opposes a change in velocity (acceleration). This parallel is exactly why $L$ is treated as the electrical analogue of mass in the standard LC-oscillator ↔ spring-mass analogy.

Step-by-Step Solution

1. Induced emf in an inductor: $\varepsilon = -L\dfrac{di}{dt}$.
2. The negative sign (Lenz's law) means this emf always opposes the change in current, not the current itself.
3. Compare with Newton's second law $F = m\dfrac{dv}{dt}$ — mass $m$ is the measure of opposition to a change in velocity. …

Concept and Intuition

An inductor stores energy in its magnetic field, and that stored energy is a state function of the instantaneous current: $U=\dfrac{1}{2}LI^2$. It doesn't matter that $I=2\sin(t^2)$ is a complicated, non-uniformly increasing function of time — the source only ever has to supply enough energy to build up whatever magnetic field corresponds to the final current, minus what corresponds to the initial current. This is exactly analogous to gravitational PE depending only on height, not the path taken.

Step-by-Step Solution

1. Energy stored in an inductor carrying current $I$: $U=\dfrac{1}{2}LI^2$.
2. Given: $L = 2\ H$, current changes from $I_1=0$ to $I_2=2\ A$.
3. Energy spent $= U_2 - U_1 = \dfrac{1}{2}L I_2^2 - \dfrac{1}{2}L I_1^2 = \dfrac{1}{2}(2)(2)^2 - 0 = 4\ J$.

Common Mistakes …

Concept and Intuition

When current in an inductor changes at rate $di/dt$, the induced emf is $v=L\,di/dt$ and the instantaneous power delivered is $P=vi=Li\,\dfrac{di}{dt}$. Because $di/dt$ is identical for both coils, equal power at some instant directly constrains the product $Li$, not $L$ or $i$ alone.

Step-by-Step Solution

1. Equate powers: $L_1 i_1 (di/dt) = L_2 i_2 (di/dt) \Rightarrow L_1 i_1 = L_2 i_2$.
2. So $\dfrac{i_2}{i_1} = \dfrac{L_1}{L_2} = \dfrac{8}{2} = 4$.
3. Energy stored in an inductor: $w = \dfrac12 L i^2$. …

Concept and Intuition

Three quantities are linked here — induced emf ($\varepsilon = L\,dI/dt$), power ($P=\varepsilon I$), and stored energy ($U=\tfrac12 LI^2$) — and each ratio must be derived consistently from the given constraints (same $dI/dt$, same power) rather than assumed independently.

Step-by-Step Solution

1. Same $dI/dt = k$ for both coils, so $\varepsilon_1 = L_1 k$, $\varepsilon_2 = L_2 k$, giving $\varepsilon_1:\varepsilon_2 = L_1:L_2 = 5:1$ — matches (A).
2. Equal power: $\varepsilon_1 I_1 = \varepsilon_2 I_2 \Rightarrow L_1 I_1 = L_2 I_2 \Rightarrow I_1:I_2 = L_2:L_1 = 1:5$ — matches (B).
3. Energy stored: $U = \tfrac12 LI^2$, so $\dfrac{U_1}{U_2} = \dfrac{L_1}{L_2}\left(\dfrac{I_1}{I_2}\right)^2 = 5\times\left(\dfrac15\right)^2 = 5\times\dfrac{1}{25} = \dfrac15$.
4. So $U_1:U_2 = 1:5$ — this matches (C), and directly contradicts (D)'s claim of $9:25$. …

Concept and Intuition

Before grinding through node equations, it always pays to check whether a resistor network is secretly

a Wheatstone bridge, because a balanced bridge carries zero current through its bridge

(galvanometer) element — independent of that element's own resistance, and independent of the source

voltage. The balance condition for a bridge with arms $R_1, R_2$ (from the source's one terminal) and

$R_3, R_4$ (continuing to the other terminal) is

Here the four resistors are $2\,\Omega$, $5\,\Omega$, $10\,\Omega$, $25\,\Omega$, with the coil bridging

the two mid-points — exactly the bridge topology.

Step-by-Step Solution

1. Identify the four arms of the bridge: $2\,\Omega$ and $5\,\Omega$ leave the battery's node and go to the two bridge mid-points; $10\,\Omega$ and $25\,\Omega$ continue from those mid-points back to the battery's other node. The coil (5 H, 2 Ω) bridges the two mid-points.
2. Check the balance ratio: $\dfrac{2}{5}=0.4$ and $\dfrac{10}{25}=0.4$. Since these ratios are equal, the bridge is balanced.
3. In a balanced bridge, both mid-points sit at exactly the same potential (regardless of the source …