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Electromagnetic Induction and Electric Motor - Explain electromagnetic induction and working of electric motor and generator

Grade 10CBSE

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

🔑Concepts

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Electromagnetic Induction (EMI): The phenomenon of producing an induced current in a closed circuit by changing the magnetic field linked with it. It was discovered by Michael Faraday.

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Fleming's Right-Hand Rule: Used to determine the direction of induced current. Stretch the thumb, forefinger, and middle finger of the right hand mutually perpendicular. If the thumb points in the direction of MotionMotion and the forefinger in the direction of the MagneticFieldMagnetic Field, then the middle finger points in the direction of the InducedCurrentInduced Current.

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Electric Motor Principle: A motor works on the principle that a current-carrying conductor placed in a magnetic field experiences a mechanical force (Lorentz force). The direction of this force is given by Fleming's Left-Hand Rule.

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Fleming's Left-Hand Rule: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular. If the forefinger represents the MagneticFieldMagnetic Field and the middle finger represents the CurrentCurrent, the thumb points in the direction of the ForceForce or MotionMotion.

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Role of Split Rings (Commutator): In an electric motor, split rings reverse the direction of current flowing through the coil every half rotation, ensuring that the coil continues to rotate in the same direction.

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Electric Generator: A device that converts mechanical energy into electrical energy using the principle of Electromagnetic Induction. It consists of a rotating coil in a magnetic field, where slip rings (for AC) or split rings (for DC) collect the induced current.

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Factors affecting Induced Current: The magnitude of induced current can be increased by: 1. Increasing the number of turns in the coil, 2. Increasing the strength of the magnetic field, 3. Increasing the speed of rotation of the coil.

📐Formulae

F=BIlsin⁡θF = BIl \sin \theta

Φ=B⋅A⋅cos⁡θ\Phi = B \cdot A \cdot \cos \theta

Power of Motor(P)=V×I\text{Power of Motor} (P) = V \times I

Efficiency(η)=Output PowerInput Power×100\text{Efficiency} (\eta) = \frac{\text{Output Power}}{\text{Input Power}} \times 100

💡Examples

Problem 1:

A straight conductor of length 0.4 m0.4\text{ m} is moved with a speed of 5 m/s5\text{ m/s} perpendicular to a magnetic field of induction 0.5 T0.5\text{ T}. Calculate the induced emf in the conductor.

Solution:

Given: l=0.4 ml = 0.4\text{ m}, v=5 m/sv = 5\text{ m/s}, B=0.5 TB = 0.5\text{ T}. Using the formula for motional emf: e=Bvle = Bvl e=0.5×5×0.4e = 0.5 \times 5 \times 0.4 e=1.0 Ve = 1.0\text{ V}

Explanation:

When a conductor moves through a magnetic field, it cuts the magnetic field lines, leading to a change in flux and inducing an electromotive force (emf) across its ends.

Problem 2:

A current-carrying wire of length 2 m2\text{ m} is placed in a magnetic field of 0.1 T0.1\text{ T} such that it is perpendicular to the field lines. If the current flowing is 5 A5\text{ A}, find the force experienced by the wire.

Solution:

Given: l=2 ml = 2\text{ m}, B=0.1 TB = 0.1\text{ T}, I=5 AI = 5\text{ A}, θ=90∘\theta = 90^\circ. Using F=BIlsin⁡θF = BIl \sin \theta F=0.1×5×2×sin⁡(90∘)F = 0.1 \times 5 \times 2 \times \sin(90^\circ) Since sin⁡(90∘)=1\sin(90^\circ) = 1: F=1.0 NF = 1.0\text{ N}

Explanation:

The force on a current-carrying conductor is maximum when the wire is perpendicular to the magnetic field and zero when it is parallel to it.

Problem 3:

An electric motor takes 5 A5\text{ A} from a 220 V220\text{ V} line. If the efficiency is 100%100\%, what is the power of the motor? If the input energy was 1500 J1500\text{ J} and 300 J300\text{ J} was lost as heat, find the useful output work using vertical subtraction.

Solution:

Power P=V×I=220×5=1100 WP = V \times I = 220 \times 5 = 1100\text{ W}. To find useful work: 1500−3001200\begin{array}{r} 1500 \\ - 300 \\ \hline 1200 \end{array} The useful output work is 1200 J1200\text{ J}.

Explanation:

Power is the rate of doing work. In a real-world motor, input energy is always greater than output energy due to heat losses (I2RtI^2Rt).