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.

Setting up the energy expression

For a wire of length $L$, cross-sectional area $A$ and Young's modulus $Y$, stretched by a tension $F$, the extension is

and the stored elastic potential energy is

Both wires are of the same material ($Y$ common) and are pulled by the same tension $F$. Hence

Putting in the given ratios

• Length ratio: $\dfrac{L_A}{L_B} = \dfrac{2}{1}$. …

Initial charge on the $10\,\mu\text{F}$ capacitor:

Initial stored energy:

On connecting to an uncharged $30\,\mu\text{F}$ capacitor, charge is conserved and the two share a common voltage:

Final total energy: …

Inductance measures a coil's opposition to changes in current. When current through an inductor changes, it generates a back EMF that opposes the change. The defining relationship is Faraday's law for inductors:

where $L$ is the inductance, $\mathcal{E}$ is the induced EMF, and $\frac{dI}{dt}$ is the rate of change of current. Rearranging, $L = \frac{\mathcal{E}}{dI/dt}$, which tells us inductance has dimensions of voltage divided by (current per time).

Let me build this systematically from known dimensions.

1. Find the dimensions of EMF (voltage). EMF is work done per unit charge, so $\mathcal{E} = \frac{W}{Q}$. Work has dimensions of energy: $[W] = [\mathrm{M L^2 T^{-2}}]$. Charge is current × time: $[Q] = [\mathrm{A T}]$. Therefore:

2. Find the dimensions of $\frac{dI}{dt}$. This is simply current per unit time:

3. Combine to find dimensions of inductance. …

Concept. Coercivity is the magnetising field intensity $H$ needed to reduce a material's magnetisation to zero. A solenoid produces $H=nI$, where $n$ is the number of turns per unit length and $I$ the current.

Turns per unit length.

$n=\frac{N}{L}=\frac{60}{0.12\ \text{m}}=500\ \text{turns m}^{-1}.$ …

Concept & Intuition

For an ideal solenoid (length >> diameter), the field inside is nearly uniform and directed along the axis. The formula $B = \mu_0 n I$ comes from Ampère’s law: the line integral of $\mathbf{B}$ around a rectangular loop threading the solenoid equals $\mu_0$ times the total current enclosed. The key is that only the turns per unit length matter — the diameter and the number of layers are irrelevant except for counting total turns. Many students mistakenly use the diameter or treat each layer separately, but the field simply adds because all layers contribute the same current in the same direction.

Step-by-step solution

1. Find the total number of turns. The solenoid has 5 layers, each with 400 turns.

2. Determine the length of the solenoid.

   Given length $L = 80$ cm = $0.80$ m.

3. Compute the number of turns per unit length $n$.

4. Apply the formula for the magnetic field inside a long solenoid.

where $\mu_0 = 4\pi \times 10^{-7}~\mathrm{T\cdot m/A}$ and $I = 8~\mathrm{A}$.

5. Calculate the numerical value.

First, $2500 \times 8 = 20000$.

Then $4\pi \times 10^{-7} \times 20000 = 4\pi \times 10^{-7} \times 2\times10^{4} = 8\pi \times 10^{-3}$.

Using $\pi \approx 3.14$,

The magnetic field inside a long, tightly wound solenoid is remarkably uniform. The key idea is that each turn of wire contributes a tiny bit to the total field, and the field strength depends on how many turns are packed into each unit of length.

The formula for the magnetic field $B$ inside an ideal solenoid (in vacuum or air) is:

where:

• $\mu_0$ is the permeability of free space (a constant),
• $n$ is the number of turns per unit length (turns per meter),
• $I$ is the current flowing through the wire.

This relation comes from Ampere’s law. Imagine a rectangular loop that goes along the axis inside the solenoid and closes outside. The only contribution to the line integral of the magnetic field comes from the side inside the solenoid, and the enclosed current is $n I$ times the length of that side. The result is that $B$ is proportional to $n$ directly.

Now, if you double the number of turns per unit length — meaning you pack twice as many turns into each meter — then $n$ becomes $2n$. Since $I$ and $\mu_0$ are unchanged, the new field is:

So the magnetic field doubles. …

The key idea is that a cylindrical capacitor consists of two coaxial cylinders. The inner cylinder carries a charge $+Q$ and the outer a charge $-Q$. In the annular region between them, the field is produced only by the inner cylinder’s charge, because the outer cylinder’s charge lies outside the Gaussian surface we choose. By symmetry, the field is radial and its magnitude depends only on the distance $r$ from the axis.

Let’s work through it step by step.

1. Choose a Gaussian surface.

   Imagine a cylindrical surface of radius $r$ (where $a < r < b$, with $a$ the inner radius and $b$ the outer radius) and length $L$, coaxial with the capacitor. The electric field $\vec{E}$ is radial and constant in magnitude over the curved surface of this Gaussian cylinder.

2. Apply Gauss’s law.

   Gauss’s law states:

The flux through the flat ends is zero because $\vec{E}$ is parallel to them. Only the curved surface contributes:

3. Find the enclosed charge.

   The Gaussian surface encloses only the inner cylinder’s charge $+Q$ (since the outer cylinder’s charge lies outside $r$). So $Q_{\text{enc}} = Q$.

4. Solve for $E$.

   $E \cdot 2\pi r L = \frac{Q}{\varepsilon_0} \quad \Rightarrow \quad E = \frac{Q}{2\pi \varepsilon_0 L} \cdot \frac{1}{r}$ …