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: Magnetic Energy Density — the energy per unit volume stored in a magnetic field is $u_B = \frac{B^2}{2\mu_0}$.

(a) For a solenoid of length $l$, area $A$, and $n$ turns per unit length, the self-inductance is $L = \mu_0 n^2 A l$. The current $I$ produces a uniform field $B = \mu_0 n I$, so $I = B/(\mu_0 n)$. The stored magnetic energy is

Since $A l$ is the volume, this confirms $U_B = u_B \times \text{volume}$. …

The key insight here is that energy in a magnetic field is distributed throughout the space where the field exists, just like energy in an electric field. For a solenoid, the field is nearly uniform inside and zero outside, so the energy density is constant over the volume $A l$. This lets us write total energy as (energy density) $\times$ (volume).

Let’s build this step by step.

1. Start with the inductance of a solenoid. For a long solenoid of length $l$, cross-sectional area $A$, and $N$ turns, the inductance is

This comes from the flux linkage: $L = N \Phi / I$, where $\Phi = B A$ and $B = \mu_0 (N/l) I$.

2. Energy stored in an inductor. The energy stored when a current $I$ flows is

This is the standard result from integrating $P = V I = L I \, dI/dt$ over time.

3. Express $I$ in terms of $B$. Inside the solenoid, $B = \mu_0 \frac{N}{l} I$, so

4. Substitute into $U = \frac12 L I^2$.

Simplify stepwise:

Notice the $N$ and $l$ cancel beautifully — the result depends only on $B$, $A$, and $l$, not on the number of turns. That’s because $B$ already encodes the effect of the current and geometry.

5. Interpretation: magnetic energy density. The volume inside the solenoid is $V = A l$, so the energy per unit volume is

This is the magnetic energy density — a universal result for any magnetic field in vacuum, not just solenoids.

6. Now compare with the electrostatic case. For a parallel-plate capacitor with plate area $A$, separation $d$, and electric field $E$ between them, the capacitance is $C = \varepsilon_0 A / d$, and the stored energy is

Using $V = E d$, we get

The volume between the plates is $A d$, so the electrostatic energy density is

Method: Energy Density Approach

This method uses the magnetic energy density formula derived from the solenoid's self-inductance.

(a) Magnetic Energy in a Solenoid

Step 1: Recall the energy stored in an inductor

The energy stored in any inductor is:

Step 2: Express $L$ and $I$ in terms of $B$

For a long solenoid:

• $L = \mu_0 n^2 A l$ (where $n$ = turns per unit length)
• Magnetic field inside solenoid: $B = \mu_0 n I$ ⇒ $I = \frac{B}{\mu_0 n}$

Step 3: Substitute and simplify

Step 4: Write the final expression

Magnetic energy stored:

$U_B = \frac{B^2}{2\mu_0} \cdot (A l)$

Here, $A l$ is the volume of the solenoid's interior.

(b) Comparison with Electrostatic Energy in a Capacitor

Parallel plate capacitor:

• Electric field: $E = \frac{V}{d}$
• Energy stored: $U_E = \frac{1}{2} C V^2$
• Using $C = \frac{\varepsilon_0 A}{d}$ and $V = Ed$:

Key comparison:

| Feature | Magnetic (Solenoid) | Electrostatic (Capacitor) |

|---------|-------------------|--------------------------| …

(a) Expression for Magnetic Energy Stored in a Solenoid

We start from the energy stored in an inductor:

For a solenoid of length $l$, area $A$, and $N$ turns:

• Inductance:

• Magnetic field inside:

Substitute $L$ and $I$ into $U$:

Simplify step-by-step:

Cancel $N^2$, one $l$, and one $\mu_0$:

Since $A l$ is the volume of the solenoid:

Magnetic energy density (energy per unit volume):

(b) Comparison with Electrostatic Energy in a Capacitor

For a parallel plate capacitor (area $A$, separation $d$, electric field $E$):

• Capacitance:

• Voltage:

• Energy stored:

So electrostatic energy density:

Key Comparison

Both are quadratic in the field and proportional to volume. The constants $\mu_0$ and $\varepsilon_0$ play symmetric roles.

Common Mistakes & How to Avoid Them

✗ Mistake 1: Forgetting the factor of $\frac{1}{2}$

• Why it happens: Students rush and write $U = L I^2$ or $U = B^2 / \mu_0$.
• How to avoid: Always start from $U = \frac{1}{2} L I^2$ — the $\frac{1}{2}$ comes from integrating $P = VI$ from 0 to final current.

✗ Mistake 2: Wrong substitution for $I$ in terms of $B$

• Why it happens: Using $B = \mu_0 n I$ but forgetting $n = N/l$.
• How to avoid: Write $B = \mu_0 \frac{N}{l} I$ explicitly, then solve for $I$ carefully.

✗ Mistake 3: Mixing up $A$ and $l$ in inductance formula

• Why it happens: $L = \mu_0 N^2 A / l$ — students swap $A$ and $l$.
• How to avoid: Remember: $L$ is proportional to area and inversely proportional to length. Draw the solenoid to visualize. …