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Electromagnetic Induction - Self and Mutual Induction

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Self-Induction is the phenomenon in which an induced EMF is produced in a coil due to a change in the current flowing through the same coil. This induced EMF is often called 'back EMF' because it opposes the change in current.

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The Self-Inductance (LL) of a coil is defined as the ratio of the magnetic flux linkage to the current flowing through it, Φ=LI\Phi = LI. It is also known as the 'Inertia of Electricity'.

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Mutual Induction is the phenomenon of inducing an EMF in a secondary coil due to a change in current in a primary coil placed near it. The flux linked with the secondary coil is Φs=MIp\Phi_s = M I_p.

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The coefficient of mutual induction (MM) depends on the geometry of the coils, the number of turns, their distance of separation, and the relative orientation (coupling).

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The SI unit for both self and mutual inductance is the Henry (HH), where 1 H=1 Wb/A=1 V⋅s/A1 \text{ H} = 1 \text{ Wb/A} = 1 \text{ V}\cdot\text{s/A}.

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The energy stored in an inductor is in the form of a magnetic field and is given by U=12LI2U = \frac{1}{2} L I^2.

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The coefficient of coupling (kk) between two coils is given by k=ML1L2k = \frac{M}{\sqrt{L_1 L_2}}, where 0≤k≤10 \leq k \leq 1.

📐Formulae

Φ=LI\Phi = L I

e=−Ldidte = -L \frac{di}{dt}

L=μ0N2Al=μ0n2AlL = \frac{\mu_0 N^2 A}{l} = \mu_0 n^2 A l

U=12LI2U = \frac{1}{2} L I^2

Φ2=MI1\Phi_2 = M I_1

e2=−MdI1dte_2 = -M \frac{dI_1}{dt}

M=μ0N1N2AlM = \frac{\mu_0 N_1 N_2 A}{l}

M=kL1L2M = k \sqrt{L_1 L_2}

💡Examples

Problem 1:

A current in a coil changes from 5 A5 \text{ A} to 2 A2 \text{ A} in 0.1 s0.1 \text{ s}. If the average EMF induced in the coil is 150 V150 \text{ V}, calculate the self-inductance of the coil.

Solution:

Given: dI=(2−5) A=−3 AdI = (2 - 5) \text{ A} = -3 \text{ A}, dt=0.1 sdt = 0.1 \text{ s}, and e=150 Ve = 150 \text{ V}. Using the formula e=−LdIdte = -L \frac{dI}{dt}, we get 150=−L−30.1150 = -L \frac{-3}{0.1}. Simplifying this: 150=L×30150 = L \times 30, which gives L=15030=5 HL = \frac{150}{30} = 5 \text{ H}.

Explanation:

The induced EMF is proportional to the rate of change of current. The negative sign in the formula denotes Lenz's law, but for magnitude calculations, we use the absolute change.

Problem 2:

Two solenoids S1S_1 and S2S_2 are wound over each other. S1S_1 has N1=500N_1 = 500 turns and S2S_2 has N2=1000N_2 = 1000 turns. If a current of 2 A2 \text{ A} in S1S_1 produces a magnetic flux of 4×10−4 Wb4 \times 10^{-4} \text{ Wb} through each turn of S2S_2, find the mutual inductance MM.

Solution:

Total flux linked with S2S_2 is Φ2=N2ϕ2=1000×4×10−4=0.4 Wb\Phi_2 = N_2 \phi_2 = 1000 \times 4 \times 10^{-4} = 0.4 \text{ Wb}. The formula for mutual inductance is Φ2=MI1\Phi_2 = M I_1. Substituting the values: 0.4=M×20.4 = M \times 2. Thus, M=0.42=0.2 HM = \frac{0.4}{2} = 0.2 \text{ H}.

Explanation:

Mutual inductance is the ratio of the total magnetic flux linked with the secondary coil to the current flowing in the primary coil.

Self and Mutual Induction - Revision Notes & Key Formulas | CBSE Class 12 Physics