Inductance - Self Inductance

The property of a coil by which it opposes the change in its own current and induces an emf in itself — numerically equal to the ratio of magnetic flux (produced due to current in the circuit) linked with the circuit to the current flowing in it, or the ratio of induced emf produced around the circuit to the rate of change of current in it — is called self-inductance.

OR

The self-inductance (L) of a coil is defined as the ratio of the total magnetic flux linkage through the coil to the current flowing in it. Equivalently, it equals the magnitude of the induced EMF per unit rate of change of current.

L = \[\frac{N\Phi_B}{I}\]

ε = -L\[\frac{dI}{dt}\]

Where:

• SI Unit: Henry (H) — named after Joseph Henry
• 1 Henry = 1 Wb A⁻¹ = 1 V · s · A⁻¹ = 1 Ω · s
• Dimensional Formula: [M L² T⁻² A⁻²]

Step 1: Set Up the Solenoid

Consider a long solenoid of:

• Length: l
• Cross-sectional area: A
• Number of turns: N (total), with n = N/l turns per unit length
• Current flowing: I
• Core: Air (or vacuum), permeability μ₀

Step 2: Find the Magnetic Field Inside

Step 3: Find the Flux Linkage

Magnetic flux through one turn:

Total flux linkage through all N turns:

Since N = n l:

Step 4: Apply the Definition

Step 5: With a Magnetic Core (relative permeability μ_r​)

When current builds up from 0 to I in an inductor, work is done against the back-EMF. This energy is stored as magnetic potential energy in the magnetic field:

W = \[\int_0^ILIdI=\frac{1}{2}LI^2\]

Comparison with Capacitor Energy:

Energy stored per unit volume in the magnetic field of a solenoid:

Substituting L = μ₀n²Al and B = μ₀nI:

The Question asks: Find the magnetic energy density inside a solenoid and compare it with the electric energy density.

What's happening step by step:

1. Start with energy stored in the inductor:
   A solenoid with inductance L carrying current I stores energy:

   W = \[\frac {1}{2}\]LI²

2. Substitute the solenoid formula L = μ₀n²Al:

   W = \[\frac{1}{2}\mu_0n^2Al\cdot I^2\]

3. Divide by volume (V = Al) to get energy per unit volume:

   u_B = \[\frac{W}{Al}=\frac{1}{2}\mu_0n^2I^2\]

4. Since B = μ₀nI, substitute to simplify:

   u_B = \[\frac {B^2}{2μ_0}\]