Chapter 21: Electromagnetic Induction

Introduction

Key Points

• Moving electric charges produce magnetic fields (Oersted, Ampere).
• Electric current is induced in a coil by changing magnetic fields (Faraday, Henry).
• Electromagnetic induction: generation of electric current by varying magnetic fields.

Solved Examples

Ex 1

• Given a coil and a moving magnet, when the magnet moves, current is induced.
• Use Faraday's law: induced emf \( \mathcal{E} = -\frac{d\Phi_B}{dt} \).
• Current flows as long as magnetic flux changes.

Practice Set

• Q1: What causes electromagnetic induction?
  • A. Constant magnetic field
  • B. Moving electric charges
  • C. Varying magnetic field
  • D. Static electric field
• Q2: Who demonstrated electric current induced by changing magnetic fields?
  • A. Oersted
  • B. Faraday and Henry
  • C. Ampere
  • D. Coulomb
• Q3: Which of the following is true?
  • A. Stationary magnet near coil induces current
  • B. Moving magnet near coil induces current
  • C. No current induced by changing magnetic field
  • D. Current induced only by battery

Answer Key

• Q1: C - Electromagnetic induction occurs due to varying magnetic fields.
• Q2: B - Faraday and Henry demonstrated induction by changing magnetic fields.
• Q3: B - Moving magnet changes magnetic flux, inducing current.

Memory Aids

• "Changing magnetic flux induces current" (Faraday's law).
• Moving charges → magnetic field (Oersted).
• Induced emf \( \propto \) rate of change of magnetic flux.

Experiments of Faraday and Henry

Experiment 1: Moving Magnet and Coil

• Bar magnet moved towards coil connected to galvanometer.
• Galvanometer deflects indicating induced current.
• Deflection lasts only while magnet moves.

Explanation

• Changing magnetic field through coil induces emf.
• Direction of induced current reverses when magnet direction reverses.

Experiment 2: Current Carrying Coil and Induced Current

• Current in coil C₂ induces current in nearby coil C₁ due to motion.

Explanation

• Changing magnetic field from moving current-carrying coil induces current in adjacent coil.

Experiment 3: Two Coils with Battery and Galvanometer

• Coil C₂ connected to battery and key K; coil C₁ connected to galvanometer G.
• Closing key K causes momentary galvanometer deflection.
• Deflection disappears if key remains closed; opposite deflection when key is opened.

Explanation

• Changing current in C₂ changes magnetic field, inducing current in C₁.
• Steady current produces no change, so no induced current.
• Opening key reverses change, causing opposite induced current.

Solved Examples

Ex 1

• Calculate induced emf when magnetic flux changes from 0.1 Wb to 0 in 0.05 s.
• Use \( \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} = -\frac{0 - 0.1}{0.05} = 2 \text{ V} \).

Practice Set

• Q1: What causes galvanometer deflection in Experiment 1?
  • A. Stationary magnet
  • B. Moving magnet
  • C. Battery
  • D. No magnetic field
• Q2: In Experiment 3, why does galvanometer deflect only momentarily?
  • A. Constant magnetic field
  • B. Changing magnetic field when key is pressed/released
  • C. Battery voltage
  • D. Coil resistance
• Q3: If magnetic flux changes at rate 0.5 Wb/s, induced emf is?
  • A. 0.5 V
  • B. 1 V
  • C. 2 V
  • D. 0 V

Answer Key

• Q1: B - Moving magnet changes magnetic flux inducing current.
• Q2: B - Only changing magnetic field induces emf; steady current causes no deflection.
• Q3: A - Induced emf equals rate of change of flux: 0.5 V.

Memory Aids

• Induced emf \( \propto \) rate of change of magnetic flux.
• Deflection only during change in magnetic field/current.
• Direction of induced current reverses with flux change direction.

Magnetic Flux Definition

Concept and Formula

• Magnetic flux (\( \phi \)) measures the magnetic field passing through a surface.
• Defined as the dot product of magnetic field \( \vec{B} \) and area vector \( \vec{A} \): \( \phi = \vec{B} \cdot \vec{A} \).
• Formula: \( \phi = BA \cos \theta \), where:
• \( B \) = magnetic field strength (Tesla, T)
• \( A \) = area of the surface (m\(^2\))
• \( \theta \) = angle between \( \vec{B} \) and \( \vec{A} \)
• Area vector \( \vec{A} \) is perpendicular to the surface.

Solved Examples

• Ex 1: Calculate magnetic flux if \( B = 0.5\,T \), \( A = 2\,m^2 \), and \( \theta = 60^\circ \).
  • Use \( \phi = BA \cos \theta = 0.5 \times 2 \times \cos 60^\circ = 0.5 \times 2 \times 0.5 = 0.5\,Wb \).

Practice Set

• Q1: Magnetic flux through a surface of area 3 m\(^2\) in a 1 T field at \( 0^\circ \) angle is:
  • A) 0 Wb
  • B) 1.5 Wb
  • C) 3 Wb
  • D) 6 Wb
• Q2: If magnetic field \( B = 0.2\,T \), area \( A = 5\,m^2 \), and flux \( \phi = 0.5\,Wb \), find \( \theta \).
  • A) 60°
  • B) 45°
  • C) 30°
  • D) 90°
• Q3: A surface of area 4 m\(^2\) is placed in a magnetic field of 0.3 T. The magnetic flux is 0.6 Wb. What is the angle between \( \vec{B} \) and \( \vec{A} \)?
  • A) 60°
  • B) 45°
  • C) 30°
  • D) 90°

Answer Key

• Q1: C) 3 Wb — \( \phi = BA \cos 0^\circ = 1 \times 3 \times 1 = 3\,Wb \).
• Q2: A) 60° — \( \cos \theta = \frac{\phi}{BA} = \frac{0.5}{0.2 \times 5} = 0.5 \Rightarrow \theta = 60^\circ \).
• Q3: A) 60° — \( \cos \theta = \frac{\phi}{BA} = \frac{0.6}{0.3 \times 4} = 0.5 \Rightarrow \theta = 60^\circ \).

Memory Aids

• Magnetic flux \( \phi = BA \cos \theta \) — think "field strength × area × alignment".
• \( \theta = 0^\circ \) means max flux; \( \theta = 90^\circ \) means zero flux.
• Area vector is always perpendicular to the surface.

Faraday's Law

Key Points

• Change in magnetic flux through a coil induces emf.
• Induced emf exists only while flux changes.
• Mathematical form: \( \varepsilon = -\frac{d\phi}{dt} \)
• For coil with N turns: \( \varepsilon = -N \frac{d\phi}{dt} \)
• Polarity of emf given by Lenz's law (opposes change).

Solved Examples

• Ex 1: A coil of 50 turns experiences a flux change of 0.1 Wb in 0.02 s. Find induced emf.
  • Use \( \varepsilon = -N \frac{d\phi}{dt} = -50 \times \frac{0.1}{0.02} = -250 \text{ V} \)

Practice Set

• Q1: What causes emf in a coil?
  • A) Constant magnetic flux
  • B) Change in magnetic flux
  • C) Electric current
  • D) Magnetic field direction
• Q2: If flux changes at 0.05 Wb/s in a 100-turn coil, emf induced is?
  • A) 5 V
  • B) 0.5 V
  • C) 50 V
  • D) 500 V
• Q3: A coil with 200 turns has induced emf 10 V when flux changes. Find rate of flux change.
  • A) 0.05 Wb/s
  • B) 0.5 Wb/s
  • C) 2 Wb/s
  • D) 20 Wb/s

Answer Key

• Q1: B - emf induced only by change in magnetic flux.
• Q2: C - \( \varepsilon = N \frac{d\phi}{dt} \Rightarrow 100 \times 0.05 = 5 \text{ V} \) but options show 5 V as A, so correct is A.
• Q3: A - \( \frac{d\phi}{dt} = \frac{\varepsilon}{N} = \frac{10}{200} = 0.05 \text{ Wb/s} \)

Memory Aids

• "Change in flux → emf" (Faraday's law)
• Emf ∝ rate of flux change
• Negative sign means emf opposes flux change (Lenz's law)

Lenz's Law

Key Points

• Direction of induced current opposes the cause of flux change.
• Ensures conservation of energy.
• When magnet's North pole approaches coil, coil acts as North pole (repels magnet).
• When magnet's North pole moves away, coil acts as South pole (attracts magnet).
• Galvanometer shows current direction changes with magnet movement.

Solved Examples

• Ex 1: Magnet's North pole moves toward coil; what is coil's induced pole?
  • Coil acts as North pole to repel magnet.
• Ex 2: Magnet moves away; coil's induced pole?
  • Coil acts as South pole to attract magnet.

Practice Set

• Q1: Lenz's law states induced current flows to?
  • A) Increase flux change
  • B) Oppose flux change
  • C) Align with magnet
  • D) Stop current
• Q2: When magnet moves toward coil, coil behaves like?
  • A) South pole
  • B) North pole
  • C) Neutral
  • D) No pole
• Q3: Galvanometer deflection reverses when?
  • A) Magnet stationary
  • B) Magnet moves away
  • C) Coil disconnected
  • D) Magnet moves toward coil

Answer Key

• Q1: B - Induced current opposes flux change.
• Q2: B - Coil acts as North pole to repel approaching magnet.
• Q3: B - Direction reverses when magnet moves away.

Memory Aids

• "Oppose the change" = Lenz's law
• Approach → coil repels (same pole)
• Recede → coil attracts (opposite pole)

Lenz's Law

Explanation

• Induced current flows to oppose the change in magnetic flux causing it.
• Mathematically, \( \varepsilon = -\frac{d\phi}{dt} \), negative sign shows opposition.
• Ensures conservation of energy by opposing flux change.
• External work done against induced emf converts to electrical energy.
• Like poles repel; opposite poles attract, affecting induced current direction.

Solved Examples

• Ex 1: Magnet moved towards loop with N-pole.
  • Induced current direction: anticlockwise to create N-pole opposing magnet.
• Ex 2: Magnet moved away from loop with N-pole.
  • Induced current direction: clockwise to create S-pole attracting magnet.

Practice Set

• Q1: What does Lenz's law state about induced current?
  • A) Flows with the change in flux
  • B) Flows to oppose the change in flux
  • C) Flows randomly
  • D) Flows only when flux is constant
• Q2: When a magnet's N-pole moves towards a loop, the induced current is?
  • A) Clockwise
  • B) Anticlockwise
  • C) Zero
  • D) Alternating
• Q3: The negative sign in \( \varepsilon = -\frac{d\phi}{dt} \) indicates?
  • A) Induced emf supports flux change
  • B) Induced emf opposes flux change
  • C) Induced emf is zero
  • D) Induced emf is constant

Answer Key

• Q1: B - Induced current always opposes the change in magnetic flux.
• Q2: B - Current flows anticlockwise to create a repelling N-pole.
• Q3: B - Negative sign shows emf opposes the change in flux.

Memory Aids

• "Lenz's Law: Induced current fights the flux change!"
• Formula: \( \varepsilon = -\frac{d\phi}{dt} \) (Negative means opposition)
• Like poles repel; induced current creates opposing poles.

Direction of Induced EMF

Explanation

• Magnetic flux \( \phi = B A \cos \theta \), depends on magnetic field \(B\), area \(A\), and angle \(\theta\).
• Flux increases if \(B\) increases and \(\theta = 0^\circ\) (field perpendicular to loop).
• Flux decreases if \(B\) decreases and \(\theta = 0^\circ\).
• Flux is zero or negligible if \(\theta = 90^\circ\) (field parallel to loop), regardless of \(B\) change.
• Induced emf depends on rate of change of flux and angle \(\theta\).
• Induced emf magnitude is proportional to relative speed of coil and magnet.
• Induced emf does not depend on coil resistance.

Solved Examples

• Ex 1: Magnetic field increases perpendicular to loop (\(\theta=0^\circ\)).
  • Flux increases, induced emf opposes increase.
• Ex 2: Magnetic field decreases parallel to loop (\(\theta=90^\circ\)).
  • Flux negligible, induced emf approximately zero.

Practice Set

• Q1: When is magnetic flux maximum?
  • A) \(\theta=0^\circ\)
  • B) \(\theta=90^\circ\)
  • C) \(\theta=45^\circ\)
  • D) \(\theta=180^\circ\)
• Q2: Induced emf depends on?
  • A) Coil resistance
  • B) Rate of change of magnetic flux
  • C) Coil material
  • D) Coil length only
• Q3: If relative speed of coil and magnet doubles, induced emf?
  • A) Halves
  • B) Remains same
  • C) Doubles
  • D) Quadruples

Answer Key

• Q1: A - Flux is maximum when field is perpendicular (\(\theta=0^\circ\)).
• Q2: B - Induced emf depends on rate of change of magnetic flux.
• Q3: C - Induced emf is directly proportional to relative speed.

Memory Aids

• Flux max at \(\theta=0^\circ\), zero at \(90^\circ\).
• Induced emf \(\propto\) speed of coil-magnet movement.
• Resistance does not affect induced emf.

Motional Electromotive Force

Concept and Formulas

• A conducting rod moving through a magnetic field cuts magnetic lines, causing free electrons to move and induce emf (motional emf).
• Emf induced in a small segment \( d\vec{l} \): \( d\varepsilon = (\vec{v} \times \vec{B}) \cdot d\vec{l} \)
• Total emf across rod: \( \varepsilon = \int (\vec{v} \times \vec{B}) \cdot d\vec{l} = v B l \sin \theta \)
• Induced electric field in rod: \( \vec{E}_{in} = - (\vec{v} \times \vec{B}) = (\vec{B} \times \vec{v}) \)
• If any two of \( \vec{v} \), \( \vec{B} \), and \( d\vec{l} \) are parallel or antiparallel, emf \( \varepsilon = 0 \).

Solved Examples

• Ex 1: A rod of length 0.5 m moves at 4 m/s perpendicular to a magnetic field of 0.3 T. Find emf induced.
  • Use \( \varepsilon = v B l \sin 90^\circ = 4 \times 0.3 \times 0.5 = 0.6 \text{ V} \)

Practice Set

• Q1: A rod of length 1 m moves at 2 m/s perpendicular to a magnetic field of 0.5 T. What is the emf induced?
  • A) 1 V
  • B) 0.5 V
  • C) 2 V
  • D) 0.25 V
• Q2: If velocity \( \vec{v} \) and magnetic field \( \vec{B} \) are parallel, the emf induced is:
  • A) Maximum
  • B) Zero
  • C) Half of maximum
  • D) Depends on length
• Q3: A rod of length 0.4 m moves at 3 m/s at an angle of 30° to a magnetic field of 0.6 T. Calculate emf induced.
  • A) 0.36 V
  • B) 0.72 V
  • C) 0.6 V
  • D) 0.18 V

Answer Key

• Q1: A) 1 V; \( \varepsilon = v B l = 2 \times 0.5 \times 1 = 1 \text{ V} \)
• Q2: B) Zero; emf is zero when \( \vec{v} \) and \( \vec{B} \) are parallel.
• Q3: D) 0.18 V; \( \varepsilon = v B l \sin 30^\circ = 3 \times 0.6 \times 0.4 \times 0.5 = 0.36 \text{ V} \) (Check options carefully, correct is 0.36 V, so option A)

Memory Aids

• Motional emf \( \propto v, B, l, \sin \theta \).
• Zero emf if velocity and magnetic field are parallel.
• Use right-hand rule for direction of induced emf.

Methods of Producing Induced emf

Changing Magnetic Field

• Changing magnetic field intensity \( B \) in a closed circuit changes magnetic flux \( \phi \), inducing emf.
• Formula: \( \varepsilon = - \frac{d\phi}{dt} = - \frac{d}{dt} (B A \cos \theta) = - A \cos \theta \frac{dB}{dt} \)

Changing Area of Circuit

• Changing area \( A \) inside magnetic field changes flux, inducing emf.
• Formula: \( \varepsilon = - B \cos \theta \frac{dA}{dt} \)
• For rectangular loop moving out of field: \( \varepsilon = B l v \cos \theta \)

Changing Orientation of Coil

• Rotating coil of area \( A \), turns \( N \), angular velocity \( \omega \) in magnetic field \( B \) changes flux.
• Flux: \( \phi = N B A \cos (\omega t) \)
• Induced emf: \( \varepsilon = - \frac{d\phi}{dt} = N B A \omega \sin (\omega t) = \varepsilon_0 \sin (\omega t) \)
• \( \varepsilon_0 = N B A \omega \) is peak emf.

Solved Examples

• Ex 1: A coil with 100 turns, area 0.01 m² rotates at 50 rad/s in 0.2 T field. Find peak emf.
  • \( \varepsilon_0 = N B A \omega = 100 \times 0.2 \times 0.01 \times 50 = 10 \text{ V} \)

Practice Set

• Q1: A coil of 200 turns, area 0.02 m² rotates at 60 rad/s in 0.1 T field. Peak emf is:
  • A) 24 V
  • B) 12 V
  • C) 6 V
  • D) 30 V
• Q2: A rectangular loop moves out of magnetic field at 3 m/s. Length in field is 0.5 m, field 0.4 T, angle 0°. Find emf.
  • A) 0.6 V
  • B) 0.3 V
  • C) 1.2 V
  • D) 0 V
• Q3: Magnetic field changes at rate 0.05 T/s in a loop of area 0.1 m² at 30°. Find emf.
  • A) 0.0043 V
  • B) 0.005 V
  • C) 0.0025 V
  • D) 0.01 V

Answer Key

• Q1: A) 24 V; \( 200 \times 0.1 \times 0.02 \times 60 = 24 \text{ V} \)
• Q2: A) 0.6 V; \( \varepsilon = B l v \cos 0^\circ = 0.4 \times 0.5 \times 3 = 0.6 \text{ V} \)
• Q3: A) 0.0043 V; \( \varepsilon = A \cos 30^\circ \times dB/dt = 0.1 \times 0.866 \times 0.05 = 0.00433 \text{ V} \)

Memory Aids

• Emf \( \propto \) rate of change of flux.
• Flux changes by changing \( B \), area, or angle \( \theta \).
• Rotating coil emf is sinusoidal (AC emf).

Emf Induced in Revolving Rod

Concept and Formula

• Rod of length \( l \) revolving with angular velocity \( \omega \) in magnetic field \( B \) perpendicular to plane.
• Time period \( T = \frac{2\pi}{\omega} \), frequency \( \nu = \frac{\omega}{2\pi} \).
• Emf induced: \( \varepsilon = - B \frac{dA}{dt} = - \frac{B \pi l^2}{T} = - \frac{B \pi l^2 \omega}{2\pi} = - \frac{B l^2 \omega}{2} = - \pi B l^2 \nu \)

Solved Examples

• Ex 1: Rod length 0.1 m, \( B = 0.5 \) T, \( \nu = 50 \) Hz. Find emf.
  • \( \varepsilon = \pi B l^2 \nu = 3.14 \times 0.5 \times 0.01 \times 50 = 0.785 \text{ V} \)

Practice Set

• Q1: Rod length 0.2 m, \( B = 0.3 \) T, frequency 60 Hz. Emf induced is:
  • A) 2.26 V
  • B) 1.13 V
  • C) 0.68 V
  • D) 3.14 V
• Q2: If angular velocity doubles, emf induced:
  • A) Doubles
  • B) Halves
  • C) Quadruples
  • D) No change
• Q3: Rod length 0.15 m, \( B = 0.4 \) T, \( \omega = 100 \) rad/s. Find emf.
  • A) 0.45 V
  • B) 0.9 V
  • C) 1.35 V
  • D) 0.3 V

Answer Key

• Q1: B) 1.13 V; \( \varepsilon = \pi \times 0.3 \times (0.2)^2 \times 60 = 1.13 \text{ V} \)
• Q2: A) Doubles; emf \( \propto \omega \)
• Q3: B) 0.9 V; \( \varepsilon = \frac{B l^2 \omega}{2} = 0.4 \times 0.0225 \times 100 / 2 = 0.9 \text{ V} \)

Memory Aids

• Emf \( \propto B, l^2, \omega \).
• Frequency \( \nu = \frac{\omega}{2\pi} \).
• Use area swept by rod for emf calculation.

Eddy Currents

Definition and Properties

• Eddy currents are currents induced in the bulk of a conductor when magnetic flux linked with it changes.
• Direction found by Lenz’s law or Fleming’s right hand rule.

Applications

• Induction furnace: heating by eddy currents.
• Speedometer: measures vehicle speed using eddy currents.
• Energy meter: records electricity consumption via eddy currents.

Solved Examples

• Ex 1: Explain why eddy currents cause heating in induction furnace.
  • Changing magnetic flux induces currents in metal, resistance causes heat.

Practice Set

• Q1: Eddy currents are induced when:
  • A) Magnetic flux is constant
  • B) Magnetic flux changes
  • C) No magnetic field
  • D) Conductor is stationary
• Q2: Which device uses eddy currents?
  • A) Transformer
  • B) Induction furnace
  • C) Capacitor
  • D) Resistor
• Q3: Eddy currents can be reduced by:
  • A) Using thicker conductor
  • B) Using laminated core
  • C) Increasing magnetic field
  • D) Increasing frequency

Answer Key

• Q1: B) Magnetic flux changes induce eddy currents.
• Q2: B) Induction furnace uses eddy currents for heating.
• Q3: B) Laminated cores reduce eddy currents by increasing resistance paths.

Memory Aids

• Eddy currents = induced currents in conductor bulk.
• Cause heating and energy loss.
• Reduced by laminations in cores.

Inductance

Concepts and Definitions

• Inductance opposes change in current; analogous to inertia in mechanics.
• Scalar quantity; SI unit: henry (H); dimensions: [ML²T^-2A^-2].
• Self Inductance (L): Voltage induced in the same coil due to change in its current.
  Formula: \( \varepsilon = -L \frac{dI}{dt} \)
• Mutual Inductance (M): Voltage induced in one coil due to change in current in another coil.
  Formulas: \( M_{12} = \frac{N_1 \Phi}{I_2} \), \( M_{21} = \frac{N_2 \Phi}{I_1} \)
• Self inductance of long solenoid: \( L = \frac{\mu_0 \mu_r N^2 A}{l} = \mu_0 \mu_r n^2 V \)
• Mutual inductance of coaxial solenoids: \( M = \frac{\mu_0 N_1 N_2 A}{l} \)

Solved Examples

Ex 1

Calculate self-inductance of a solenoid with \( N=500 \), length \( l=0.5 \) m, cross-sectional area \( A=1 \times 10^{-4} \) m², and \( \mu_r=1 \).

• Use \( L = \frac{\mu_0 \mu_r N^2 A}{l} \)
• \( L = \frac{4\pi \times 10^{-7} \times 1 \times 500^2 \times 1 \times 10^{-4}}{0.5} = 6.28 \times 10^{-3} \) H

Practice Set

Q1

What is the SI unit of inductance?

• A) Ohm
• B) Henry
• C) Tesla
• D) Weber

Q2

Self-induced emf in a coil is given by:

• A) \( \varepsilon = L \frac{dI}{dt} \)
• B) \( \varepsilon = -L \frac{dI}{dt} \)
• C) \( \varepsilon = L I \)
• D) \( \varepsilon = -I \frac{dL}{dt} \)

Q3

Mutual inductance between two coils depends on:

• A) Number of turns and distance between coils
• B) Only number of turns
• C) Only distance between coils
• D) Neither turns nor distance

Answer Key

• Q1: B) Henry - Standard SI unit for inductance.
• Q2: B) \( \varepsilon = -L \frac{dI}{dt} \) - Lenz's law sign included.
• Q3: A) Number of turns and distance between coils - Both affect mutual inductance.

Memory Aids

• Inductance opposes current change like inertia opposes motion change.
• Self-inductance formula: \( \varepsilon = -L \frac{dI}{dt} \).
• Mutual inductance increases with turns, decreases with coil separation.

Self Inductance

Key Points

• Change in current induces emf in the same coil (self induction).
• Induced emf opposes change in current (Lenz's law).
• Formula: \( \varepsilon_s = -L \frac{dI}{dt} \)
• When current is constant, no emf induced.
• Self inductance: \( L = \frac{N \Phi}{I} = \frac{NBA}{I} \)
• Real inductor modeled as resistor \( R \) in series with ideal inductor \( L \).

Solved Examples

Ex 1

Calculate emf induced if current changes at \( 5 \text{ A/s} \) in coil with \( L=2 \text{ H} \).

• \( \varepsilon = -L \frac{dI}{dt} = -2 \times 5 = -10 \text{ V} \)

Practice Set

Q1

What happens to emf if current is constant?

• A) Emf increases
• B) Emf decreases
• C) Emf is zero
• D) Emf is maximum

Q2

Self-inductance depends on:

• A) Number of turns and coil area
• B) Only current
• C) Resistance only
• D) Voltage only

Q3

In a real inductor, resistance is:

• A) Zero
• B) Infinite
• C) Non-zero
• D) Negative

Answer Key

• Q1: C) Emf is zero - No change in current means no induced emf.
• Q2: A) Number of turns and coil area - Affect magnetic flux and inductance.
• Q3: C) Non-zero - Real inductors have resistance causing energy loss.

Memory Aids

• Self-inductance opposes current change in same coil.
• Emf zero if current steady.
• Real inductors = ideal inductor + resistor in series.

Mutual Inductance

Essentials

• Change in current in one coil induces emf in another coil.
• Mutual inductance \( M = \frac{N_2 \Phi_2}{I_1} = \frac{N_2 B_1 A_2}{I_1} \)
• Induced emf in secondary coil: \( e_m = -M \frac{dI_1}{dt} \)
• Mutual inductance depends on number of turns, cross-sectional area, distance, and orientation.
• For coaxial solenoids: \( M = \frac{\mu_0 N_1 N_2 A}{l} \)
• Maximum \( M \) when coils aligned; minimum when perpendicular.

Solved Examples

Ex 1

Find mutual inductance if \( N_2=200 \), \( B_1=0.01 \) T, \( A_2=1 \times 10^{-4} \) m², and \( I_1=2 \) A.

• \( M = \frac{N_2 B_1 A_2}{I_1} = \frac{200 \times 0.01 \times 1 \times 10^{-4}}{2} = 1 \times 10^{-4} \) H

Practice Set

Q1

Mutual inductance is maximum when coils are:

• A) Perpendicular
• B) Parallel and coaxial
• C) Far apart
• D) No current

Q2

Mutual inductance decreases with:

• A) Increasing number of turns
• B) Increasing distance between coils
• C) Increasing cross-sectional area
• D) Increasing current

Q3

Mutual inductance between two solenoids of length \( l \), turns \( N_1, N_2 \), and area \( A \) is:

• A) \( \frac{\mu_0 N_1 N_2 A}{l} \)
• B) \( \frac{\mu_0 N_1 N_2 l}{A} \)
• C) \( \frac{N_1 N_2 A}{\mu_0 l} \)
• D) \( \frac{\mu_0 l}{N_1 N_2 A} \)

Answer Key

• Q1: B) Parallel and coaxial - Maximum flux linkage.
• Q2: B) Increasing distance between coils - Flux linkage reduces.
• Q3: A) \( \frac{\mu_0 N_1 N_2 A}{l} \) - Standard formula for coaxial solenoids.

Memory Aids

• Mutual inductance depends on coil alignment and proximity.
• More turns and area increase \( M \).
• Distance and misalignment decrease \( M \).

Energy Stored in Inductor

Key Formulas

• Power supplied: \( P = \varepsilon_{in} I = L \frac{dI}{dt} I \)
• Energy stored: \( U = \frac{1}{2} L I^2 \)

Solved Examples

Ex 1

Calculate energy stored in inductor with \( L=3 \) H and current \( I=2 \) A.

• \( U = \frac{1}{2} \times 3 \times 2^2 = 6 \) J

Practice Set

Q1

Energy stored in an inductor is proportional to:

• A) Current
• B) Square of current
• C) Inductance only
• D) Resistance

Q2

Energy stored in inductor with \( L=4 \) H and current \( 3 \) A is:

• A) 18 J
• B) 36 J
• C) 54 J
• D) 72 J

Q3

Power supplied to inductor when current changes at \( 5 \text{ A/s} \) and current is \( 2 \) A, \( L=1 \) H:

• A) 5 W
• B) 10 W
• C) 15 W
• D) 20 W

Answer Key

• Q1: B) Square of current - \( U = \frac{1}{2} L I^2 \).
• Q2: C) 54 J - \( \frac{1}{2} \times 4 \times 3^2 = 18 \times 3 = 54 \) J.
• Q3: B) 10 W - \( P = L \frac{dI}{dt} I = 1 \times 5 \times 2 = 10 \) W.

Memory Aids

• Energy stored \( \propto I^2 \).
• Power depends on rate of current change and current.

Combination of Inductances

Formulas

• Series: \( L_s = L_1 + L_2 + \cdots + L_n \)
• Parallel: \( \frac{1}{L_p} = \frac{1}{L_1} + \frac{1}{L_2} + \cdots + \frac{1}{L_n} \)
• For two inductors with mutual inductance \( M \) in series:
  \( V = (L_1 + L_2 + 2M) \frac{di}{dt} \)

Solved Examples

Ex 1

Find total inductance of two inductors \( L_1=2 \) H, \( L_2=3 \) H, \( M=1 \) H in series.

• \( L_s = 2 + 3 + 2 \times 1 = 7 \) H

Practice Set

Q1

Inductors in series add:

• A) Like resistors in parallel
• B) Directly
• C) Inversely
• D) Not at all

Q2

Effective inductance of two inductors in parallel is:

• A) Sum of inductances
• B) Product over sum
• C) Reciprocal of sum of reciprocals
• D) Zero

Q3

For two inductors with mutual inductance \( M \), total emf in series is:

• A) \( (L_1 + L_2) \frac{di}{dt} \)
• B) \( (L_1 + L_2 - 2M) \frac{di}{dt} \)
• C) \( (L_1 + L_2 + 2M) \frac{di}{dt} \)
• D) \( M \frac{di}{dt} \)

Answer Key

• Q1: B) Directly - Inductances add in series.
• Q2: C) Reciprocal of sum of reciprocals - Like resistors in parallel.
• Q3: C) \( (L_1 + L_2 + 2M) \frac{di}{dt} \) - Mutual inductance adds to total emf.

Memory Aids

• Series inductance sums directly.
• Parallel inductance sums inversely.
• Mutual inductance adds twice in series emf.

LR Circuits

Current Growth

• Emf equation: \( V = IR + L \frac{dI}{dt} \)
• Current at time \( t \): \( I = I_0 (1 - e^{-t/\tau}) \), where \( I_0 = \frac{V}{R} \)
• Time constant: \( \tau = \frac{L}{R} \)
• At \( t=0 \), inductor acts as open circuit (infinite resistance).
• At \( t \to \infty \), inductor acts as short circuit (zero resistance).

Current Decay

• Emf equation: \( IR + L \frac{dI}{dt} = 0 \)
• Current at time \( t \): \( I = I_0 e^{-t/\tau} \)
• Time constant: \( \tau = \frac{L}{R} \)
• At \( t=0 \), current is maximum \( I_0 = \frac{V}{R} \)
• At \( t \to \infty \), current approaches zero.

Solved Examples

Ex 1

Calculate time constant for \( L=2 \) H and \( R=4 \) \( \Omega \).

• \( \tau = \frac{L}{R} = \frac{2}{4} = 0.5 \) s

Practice Set

Q1

Time constant \( \tau \) in LR circuit is:

• A) \( L \times R \)
• B) \( \frac{L}{R} \)
• C) \( \frac{R}{L} \)
• D) \( L + R \)

Q2

At \( t=0 \), inductor behaves as:

• A) Short circuit
• B) Open circuit
• C) Resistor
• D) Capacitor

Q3

Current after long time \( t \to \infty \) is:

• A) Zero
• B) \( I_0 \)
• C) Infinite
• D) Negative

Answer Key

• Q1: B) \( \frac{L}{R} \) - Definition of time constant.
• Q2: B) Open circuit - Inductor opposes sudden current change.
• Q3: B) \( I_0 \) - Steady current after transient.

Memory Aids

• Time constant \( \tau = L/R \).
• Inductor blocks sudden current changes.
• Current grows exponentially to \( I_0 \).

AC Generator

Definition and Parts

• Device producing alternating current (AC) by converting mechanical energy into electrical energy.
• Essential parts: Armature (coil), Field magnets (North and South poles), Slip rings, Brushes.

Principle

• Based on electromagnetic induction: changing magnetic flux through coil induces emf.
• Direction of induced current given by Fleming's right-hand rule.

Working

• Coil rotates in magnetic field; magnetic flux linked changes sinusoidally.
• Induced emf varies as \( \varepsilon = \varepsilon_0 \sin \omega t \), where \( \varepsilon_0 = BAN\omega \).
• Current \( I = \frac{\varepsilon}{R} = I_0 \sin \omega t \), with \( I_0 = \frac{\varepsilon_0}{R} \).

Solved Examples

• Ex 1: Calculate maximum emf if \(N=100\), \(B=0.5\,T\), \(A=0.01\,m^2\), \(\omega=100\,rad/s\).
  • \( \varepsilon_0 = BAN\omega = 0.5 \times 0.01 \times 100 \times 100 = 50\,V \)

Practice Set

• Q1: What is the role of slip rings in an AC generator?
  • A) Convert AC to DC
  • B) Maintain electrical connection while coil rotates
  • C) Increase magnetic field
  • D) Reduce resistance
• Q2: The emf induced in a coil rotating in a magnetic field is given by \( \varepsilon = \varepsilon_0 \sin \omega t \). What does \( \omega \) represent?
  • A) Number of turns
  • B) Angular velocity of coil
  • C) Magnetic field strength
  • D) Resistance of coil
• Q3: A coil with 50 turns and area \(0.02\,m^2\) rotates at \(60\,Hz\) in a magnetic field of \(0.3\,T\). Calculate the maximum emf.
  • A) 11.3 V
  • B) 22.6 V
  • C) 33.9 V
  • D) 45.2 V

Answer Key

• Q1: B - Slip rings maintain continuous electrical contact during rotation.
• Q2: B - \( \omega \) is angular velocity of the coil.
• Q3: B - \( \varepsilon_0 = BAN\omega = 0.3 \times 0.02 \times 50 \times 2\pi \times 60 = 22.6\,V \).

Memory Aids

• AC generator emf: \( \varepsilon = BAN\omega \sin \omega t \).
• Slip rings keep connection during rotation.
• Fleming's right-hand rule for current direction.

Induced Electric Field

Concept

• Changing magnetic flux induces emf and current in coil (Faraday's law).
• Induced electric field (\( \vec{E} \)) causes electron flow; magnetic field alone cannot move stationary charges.
• Induced electric field is non-conservative; line integral around closed path \( \oint \vec{E} \cdot d\vec{l} = \varepsilon = -\frac{d\phi}{dt} \).
• Force on charge: \( \vec{F} = q \vec{E} \).
• Electric field magnitude same at all points on symmetrical loop; tangential direction.
• For symmetrical loop: \( E l = |\frac{d\phi}{dt}| = A |\frac{dB}{dt}| \), where \( l \) is loop length, \( A \) is area.
• Direction of induced electric field matches induced current direction.


Solved Examples

• Ex 1: Calculate induced electric field in a circular loop of circumference \(0.5\,m\) with magnetic field changing at \(0.2\,T/s\).
  • \( E = \frac{A |dB/dt|}{l} \), area \( A = \frac{l^2}{4\pi} = \frac{(0.5)^2}{4\pi} = 0.0199\,m^2 \)
  • \( E = \frac{0.0199 \times 0.2}{0.5} = 0.00796\,V/m \)

Practice Set

• Q1: Which field causes current flow in a stationary conductor with changing magnetic flux?
  • A) Magnetic field
  • B) Electrostatic field
  • C) Induced electric field
  • D) Gravitational field
• Q2: The line integral of induced electric field around a closed loop is:
  • A) Zero
  • B) Equal to emf
  • C) Equal to magnetic flux
  • D) Infinite
• Q3: A circular loop of length \(1\,m\) is in a magnetic field changing at \(0.1\,T/s\). Calculate induced electric field.
  • A) 0.01 V/m
  • B) 0.02 V/m
  • C) 0.03 V/m
  • D) 0.04 V/m

Answer Key

• Q1: C - Induced electric field causes current in stationary conductor.
• Q2: B - Line integral of \( \vec{E} \) equals emf induced.
• Q3: A - \( E = \frac{A |dB/dt|}{l} \), for circle \( A = \frac{l^2}{4\pi} = 0.0796\,m^2 \), so \( E = \frac{0.0796 \times 0.1}{1} = 0.00796 \approx 0.01\,V/m \).

Memory Aids

• Induced electric field: non-conservative, \( \oint \vec{E} \cdot d\vec{l} = -\frac{d\phi}{dt} \).
• Force on charge: \( \vec{F} = q \vec{E} \).
• Electric field tangential and uniform on symmetrical loop.