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Electromagnetic Induction and Alternating Currents - Self and Mutual Induction

Grade 12ICSEPhysics

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

🔑Concepts

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Self-Induction: The phenomenon in which an opposing induced EMF is produced in a coil as a result of a change in current flowing through the same coil. It is often referred to as the 'Inertia of Electricity'.

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Coefficient of Self-Inductance (LL): Defined as the magnetic flux linkage with the coil when a unit current flows through it, represented by ϕ=LI\phi = LI. The SI unit is Henry (HH).

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Back EMF: The induced EMF in self-induction is given by e=−LdIdte = -L \frac{dI}{dt}, where the negative sign indicates that the EMF opposes the change in current (Lenz's Law).

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Mutual Induction: The phenomenon of inducing an EMF in a secondary coil due to a change in current in a nearby primary coil.

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Coefficient of Mutual Inductance (MM): Defined as the magnetic flux linked with the secondary coil per unit current in the primary coil, represented by ϕs=MIp\phi_s = MI_p. The induced EMF in the secondary is es=−MdIpdte_s = -M \frac{dI_p}{dt}.

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Factors affecting LL and MM: These include the number of turns (NN), the area of cross-section (AA), the length of the coil (ll), and the relative permeability (μr\mu_r) of the core material.

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Coefficient of Coupling (kk): A measure of the magnetic link between two coils, calculated as k=ML1L2k = \frac{M}{\sqrt{L_1 L_2}}, where 0≤k≤10 \le k \le 1.

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Energy Stored in an Inductor: The magnetic potential energy stored in an inductor when a current II flows through it is U=12LI2U = \frac{1}{2}LI^2.

📐Formulae

ϕ=LI\phi = LI

e=−LdIdte = -L \frac{dI}{dt}

L=μ0μrN2AlL = \frac{\mu_0 \mu_r N^2 A}{l}

ϕs=MIp\phi_s = MI_p

es=−MdIpdte_s = -M \frac{dI_p}{dt}

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

U=12LI2U = \frac{1}{2}LI^2

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 induced EMF is 150 V150\text{ V}, calculate the self-inductance of the coil.

Solution:

Given: dI=(2−5)=−3 AdI = (2 - 5) = -3\text{ A}, dt=0.1 sdt = 0.1\text{ s}, e=150 Ve = 150\text{ V}. Using the formula e=−LdIdte = -L \frac{dI}{dt}, we get 150=−L(−30.1)150 = -L \left(\frac{-3}{0.1}\right). This simplifies to 150=L(30)150 = L(30), hence L=15030=5 HL = \frac{150}{30} = 5\text{ H}.

Explanation:

The self-inductance is calculated by relating the induced EMF to the rate of change of current using Faraday's law applied to self-induction.

Problem 2:

Two coils have a mutual inductance of 0.005 H0.005\text{ H}. The current changes in the first coil according to the equation I=I0sin⁡(ωt)I = I_0 \sin(\omega t), where I0=10 AI_0 = 10\text{ A} and ω=100π rad/s\omega = 100\pi\text{ rad/s}. Find the maximum value of EMF induced in the second coil.

Solution:

Induced EMF es=−MdIpdte_s = -M \frac{dI_p}{dt}. Here, dIdt=ddt(I0sin⁡(ωt))=I0ωcos⁡(ωt)\frac{dI}{dt} = \frac{d}{dt}(I_0 \sin(\omega t)) = I_0 \omega \cos(\omega t). The maximum EMF occurs when cos⁡(ωt)=1\cos(\omega t) = 1, so emax=MI0ωe_{max} = M I_0 \omega. Substituting values: emax=0.005×10×100π=5π≈15.7 Ve_{max} = 0.005 \times 10 \times 100\pi = 5\pi \approx 15.7\text{ V}.

Explanation:

The maximum EMF is found by differentiating the current function to find the maximum rate of change and multiplying it by the coefficient of mutual induction.

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