As this magnetic field passes through the solenoid, the windings of the solenoid are linked by the field lines.

**Self-inductance of a**** ****long solenoid**

Consider a long solenoid
of length *l* and cross-sectional area *A*. Let *n* be the
number of turns per unit length (or turn density) of the solenoid. When an
electric current *i* is passed through the solenoid, a magnetic field is
produced by it which is almost uniform and is directed along the axis of the
solenold as shown in Figure 4.21. The magnetic field at any point inside the
solenoid is given by (Refer section 3.9.3)

*B *=* *Âµ* ni*

As this magnetic field
passes through the solenoid, the windings of the solenoid are linked by the
field lines. The magnetic flux passing through each turn is

The total magnetic flux
linked or flux linkage of the solenoid with *N* turns (the total number of
turns *N* is given by *N *=* n l*) is

Comparing equations
(4.15) and (4.17), we have

From the above equation,
it is clear that inductance depends on the geometry of the solenoid (turn
density *n*, cross-sectional area *A*, length* l*) and the
medium present inside* *the solenoid. If the solenoid is filled with a
dielectric medium of relative permeability Âµ* _{r}* , then

Whenever a current is
established in the circuit, the inductance opposes the growth of the current.
In order to establish a current in the circuit, work is done against this
opposition by some external agency. This work done is stored as magnetic
potential energy.

Let us assume that
electrical resistance of the inductor is negligible and inductor effect alone
is considered. The induced emf Îµ at any instant *t* is

Let *dW* be work
done in moving a charge *dq *in a time* dt *against the opposition,
then

Total work done in establishing the
current i is

This work done is stored
as magnetic potential energy.

The energy density is
the energy stored per unit volume of the space and is given by

**EXAMPLE 4.10**

A solenoid of 500 turns
is wound on an iron core of relative permeability 800. The length and radius of
the solenoid are 40 cm and 3 cm respectively. Calculate the average emf induced
in the solenoid if the current in it changes from 0 to 3 A in 0.4 second.

**Solution**

N = 500 turns; Âµ_{r}
= 800 ;

*l* = 40 cm = 0.4 m; r = 3 cm = 0.03 m;

*di* = 3 â€“ 0 = 3 A; dt = 0.4 s

Self inductance,

**EXAMPLE 4.11**

The self-inductance of
an air-core solenoid is 4.8 mH. If its core is replaced by iron core, then its
self-inductance becomes 1.8 H. Find out the relative permeability of iron.

**Solution**

Tags : Definition, Explanation, Formulas, Solved Example Problems | Electromagnetic Induction , 12th Physics : Electromagnetic Induction and Alternating Current

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12th Physics : Electromagnetic Induction and Alternating Current : Self-inductance of a long solenoid | Definition, Explanation, Formulas, Solved Example Problems | Electromagnetic Induction

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