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# Self-Inductance Notes | Study Physics For JEE - JEE

## Document Description: Self-Inductance for JEE 2022 is part of Electromagnetic Induction and Alternating Currents for Physics For JEE preparation. The notes and questions for Self-Inductance have been prepared according to the JEE exam syllabus. Information about Self-Inductance covers topics like and Self-Inductance Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Self-Inductance.

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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 ΔI in a time interval Δt, the average emf induced in the coil is given as

ε =  =

The instantaneous emf is given as

ε = - = -

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 = μ0 nl. The magnetic flux through one turn of solenoid f = μ0 n l A.

The total magnetic flux through the solenoid = N

= N μ0 n l A

= μ0 n2 l A l

Therefore, L = μ0 n2 l A = μ0 n2 V

= μ0 n i pr2 (n l)

L =  = μ0 n2 πr2 l

Inductance per unit volume = μ0 n2

Ex.38 The current in a coil of self-inductance L = 2H is increasing according to the law i = 2 sin t2. 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 = τ

Then 2 = 2 sin τ2 ⇒ τ =

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

W =  =

W =  =

Let θ = 2t2

Differentiating dθ = 4t dt

Therefore, W =

= L (- cosθ) = - L cos 2t2

W =

= 2 L = 2 × 2 = 4 joule

7. Inductor

It is represent by

electrical equivalence of loop

⇒

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

VA - L  = VB

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 vA - vB 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

VA - 1 - 5 - 2 i = VB

(i) Put i = 2,  = 0

VA - 5 - 4 = VB    Therefore, VA - VB = 9 volt

(ii) Put i = 2,  = 1 ;

VA - 1 - 5 - 4 = VB or VA - VB = 10 V0

(iii) Put i = 2,  = - 1

VA + 1 - 5 - 2 × 2 = VB

VA = 8 volt

Ex.40 Find current i, i1 and i2 in the following circuit.

Sol. at t = 0

ii2 and i1 = 0

at t = ∞

⇒ i1 = i2 =

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 =

⇒ 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 =

⇒  ⇒

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