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Electromagnetic Induction - AC Generator

Grade 12CBSEPhysics

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

🔑Concepts

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An AC generator (alternator) is a device that converts mechanical energy into electrical energy using the principle of Electromagnetic Induction.

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The basic principle is that when a coil is rotated in a uniform magnetic field, the magnetic flux Φ\Phi linked with the coil changes continuously, which induces an electromotive force (EMF) in the coil.

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The magnetic flux linked with a coil of NN turns, each of area AA, rotated at a constant angular velocity ω\omega in a uniform magnetic field BB is given by Φ=NBAcos⁡(θ)\Phi = NBA \cos(\theta), where θ=ωt\theta = \omega t.

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The induced EMF ε\varepsilon is determined by Faraday's Law of Induction: ε=−dΦdt\varepsilon = -\frac{d\Phi}{dt}.

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The EMF varies sinusoidally with time: ε=ε0sin⁡(ωt)\varepsilon = \varepsilon_0 \sin(\omega t), where ε0\varepsilon_0 is the peak value of the EMF.

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The main components of an AC generator include the Field Magnets (to provide a uniform magnetic field), Armature (the rotating coil), Slip Rings (to maintain contact with the external circuit without tangling wires), and Brushes (carbon blocks that transfer current to the external load).

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The direction of the induced current changes periodically every half-cycle, following Fleming's Right-Hand Rule.

📐Formulae

Φ=NBAcos⁡(ωt)\Phi = NBA \cos(\omega t)

ε=−dΦdt=NBAωsin⁡(ωt)\varepsilon = -\frac{d\Phi}{dt} = NBA\omega \sin(\omega t)

ε0=NBAω\varepsilon_0 = NBA\omega

ω=2πf\omega = 2\pi f

I=εR=ε0Rsin⁡(ωt)=I0sin⁡(ωt)I = \frac{\varepsilon}{R} = \frac{\varepsilon_0}{R} \sin(\omega t) = I_0 \sin(\omega t)

💡Examples

Problem 1:

An AC generator consists of a coil of 100100 turns and cross-sectional area 3 m23 \text{ m}^2, rotating at a constant angular speed of 60 rad/s60 \text{ rad/s} in a uniform magnetic field of 0.04 T0.04 \text{ T}. Calculate the maximum EMF produced in the generator.

Solution:

Given: N=100N = 100 turns A=3 m2A = 3 \text{ m}^2 ω=60 rad/s\omega = 60 \text{ rad/s} B=0.04 TB = 0.04 \text{ T}

Using the formula for peak EMF: ε0=NBAω\varepsilon_0 = NBA\omega ε0=100×0.04×3×60\varepsilon_0 = 100 \times 0.04 \times 3 \times 60 ε0=4×3×60\varepsilon_0 = 4 \times 3 \times 60 ε0=720 V\varepsilon_0 = 720 \text{ V}

Explanation:

The maximum or peak EMF (ε0)(\varepsilon_0) occurs when the plane of the coil is parallel to the magnetic field (i.e., sin⁡(ωt)=1\sin(\omega t) = 1). It is calculated as the product of the number of turns, magnetic field strength, area of the coil, and angular frequency.

Problem 2:

A circular coil of radius 10 cm10 \text{ cm} and 2020 turns is rotated about its vertical diameter with an angular speed of 50 rad/s50 \text{ rad/s} in a uniform horizontal magnetic field of magnitude 3.0×10−2 T3.0 \times 10^{-2} \text{ T}. Obtain the maximum EMF induced in the coil.

Solution:

Radius r=10 cm=0.1 mr = 10 \text{ cm} = 0.1 \text{ m}. Area A=πr2=π(0.1)2=0.01π m2A = \pi r^2 = \pi (0.1)^2 = 0.01\pi \text{ m}^2. N=20N = 20, ω=50 rad/s\omega = 50 \text{ rad/s}, B=3.0×10−2 TB = 3.0 \times 10^{-2} \text{ T}.

ε0=NBAω\varepsilon_0 = NBA\omega ε0=20×(3.0×10−2)×(0.01π)×50\varepsilon_0 = 20 \times (3.0 \times 10^{-2}) \times (0.01\pi) \times 50 ε0=20×50×3.0×10−2×0.01π\varepsilon_0 = 20 \times 50 \times 3.0 \times 10^{-2} \times 0.01\pi ε0=1000×0.03×0.01π\varepsilon_0 = 1000 \times 0.03 \times 0.01\pi ε0=30×0.01π=0.3π≈0.942 V\varepsilon_0 = 30 \times 0.01\pi = 0.3\pi \approx 0.942 \text{ V}

Explanation:

The problem requires calculating the area from the radius first, then substituting all values into the standard peak EMF formula for a rotating coil in a magnetic field.