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**6. Self induction**

Self induction is induction of emf in a coil due to its own current change. Total flux N _{}passing through a coil due to its own current is proportional to the current and is given as N _{} = L i where L is called coefficient of self induction or inductance. The inductance L is purely a geometrical property i.e., we can tell the inductance value even if a coil is not connected in a circuit. Inductance depends on the shape and size of the loop and the number of turns it has.

If current in the coil changes by Î”I in a time interval Î”t, the average emf induced in the coil is given as

Îµ = =

The instantaneous emf is given as

Îµ = - = -

S.I unit of inductance is wb/amp or Henry (H)

L - self inductance is +ve Quantity.

L depends on : (1) Geometry of loop

(2) Medium in which it is kept. L does not depend upon current. L is a scalar Quantity.

**Brain Teaser**

*If a circuit has large self-inductance, what inference can you draw about the circuit.*

**6.1 Self Inductance of solenoid**

Let the volume of the solenoid be V, the number of turns per unit length be n. Let a current I be flowing in the solenoid. Magnetic field in the solenoid is given as B = Î¼_{0} nl. The magnetic flux through one turn of solenoid f = Î¼_{0} n l A.

The total magnetic flux through the solenoid = N _{}

= N Î¼_{0} n l A

= Î¼_{0} n^{2} l A *l*

Therefore, L = Î¼_{0} n^{2} *l* A = Î¼_{0} n^{2} V

= Î¼_{0} n i pr^{2} (n *l*)

L = = Î¼_{0} n^{2} Ï€r^{2} *l*

Inductance per unit volume = Î¼_{0} n^{2}

*Ex.38 The current in a coil of self-inductance L = 2H is increasing according to the law i = 2 sin t ^{2}. Find the amount of energy spent during the period when the current changes from 0 to 2 ampere.*

**Sol. **Let the current be 2 amp at t = Ï„

Then 2 = 2 sin Ï„^{2} â‡’ Ï„ =

When the instantaneous current is i, the self induced emf is L . If the amount of charge that is displaced in time dt is dÎ¸, then the elementary work done

=

W = =

W = =

Let Î¸ = 2t^{2}

Differentiating dÎ¸ = 4t dt

Therefore, W =

= L (- cosÎ¸) = - L cos 2t^{2}

W =

= 2 L = 2 Ã— 2 = 4 joule

**7. Inductor**

It is represent by

electrical equivalence of loop

â‡’

If current i through the inductor is increasing the induced emf will oppose the **increase **in current and hence will be opposite to the current. If current i through the inductor is decreasing the induced emf will oppose the **decrease **in current and hence will be in the direction of the current.

Over all result

V_{A} - L = V_{B}

**Note**

** **If there is a resistance in the inductor (resistance of the coil of inductor) then :

*Ex.39 A B is a part of circuit. Find the potential difference v _{A} - v_{B} if*

(i) current i = 2A and is constant

(ii) current i = 2A and is increasing at the rate of 1 amp/sec.

(iii) current i = 2A and is decreasing at the rate 1 amp/sec.

**Sol. **

** = **

writing KVL from A to B

V_{A} - 1 - 5 - 2 i = V_{B}

(i) Put i = 2, = 0

V_{A} - 5 - 4 = V_{B} Therefore, V_{A} - V_{B} = 9 volt

(ii) Put i = 2, = 1 ;

V_{A} - 1 - 5 - 4 = V_{B} or V_{A} - V_{B} = 10 V_{0}

(iii) Put i = 2, = - 1

V_{A} + 1 - 5 - 2 Ã— 2 = V_{B}

V_{A} = 8 volt

*Ex.40 Find current i, i _{1} and i_{2} in the following circuit.*

**Sol. at t = 0**

*i* = *i*_{2} = and *i*_{1} = 0

**at t = âˆž**

â‡’ i_{1} = i_{2} = =

**7.1 Energy stored in an inductor :**

If current in an inductor at an instant is i and is increasing at the rate di/dt, the induced emf will oppose the current. Its behaviour is shown in the figure.

Power consumed by the inductor = i L

Energy consumed in dt time = i L dt

Therefore, total energy consumed as the current increases from 0 to I =

= â‡’ U =

**Note :**

This energy is stored in the magnetic field with energy density

Total energy U =

*Ex.41 Find out the energy per unit length ratio inside **the solid long wire having current density J.*

**Sol. **Take a ring of radius r and thickness dr as an element inside the wire

=

using B =

â‡’ â‡’

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