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Physics Part 1

Electromagnetic Induction

Electromagnetic Induction

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Alternating EMF Waveform in Generator
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Applications of Electromagnetic Induction
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Magnetic Flux
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Back EMF and Electrical Inertia
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Faraday and Henry Experiments
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Induced Current Direction in Loops
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Induced EMF from Time-Varying Magnetic Field
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Inductance and Flux Linkage
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Magnetic Energy Density
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Methods of Changing Magnetic Flux
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Mutual Inductance of Coaxial Solenoids
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Self-Inductance of Long Solenoid
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AC Generator Construction and Working
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Mutual Inductance
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Lenz's Law
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Motional Electromotive Force
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AC Generator
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Electromagnetic Induction
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Energy Stored in Inductors
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Faraday's Law of Induction
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Open-Circuit Induced EMF
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Rotating Rod and Wheel Motional EMF
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Self-Inductance
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Learning Objectives

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Learning Objectives

  • Understand the relationship between electricity and magnetism.
  • Explain the phenomenon of electromagnetic induction.
  • Describe Faraday's laws of induction and their implications.
  • Apply Lenz's law to determine the direction of induced current.
  • Calculate induced emf in various scenarios involving changing magnetic fields.
  • Define and calculate mutual and self-inductance.
  • Analyze the operation of AC generators and their principles.

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Revision Notes & Summary

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Chapter Six: Electromagnetic Induction

6.1 Introduction

  • Electricity and magnetism were historically viewed as separate phenomena.
  • Early experiments by Oersted and Ampere established their interrelation.
  • Moving electric charges produce magnetic fields.
  • Faraday and Henry's experiments demonstrated that changing magnetic fields can induce electric currents.
  • This phenomenon is known as electromagnetic induction.

6.2 Key Concepts

Magnetic Flux

  • Defined as:
    ΦB=B⋅A=BAcos⁡(Θ)\Phi_B = B \cdot A = BA \cos(\Theta)
    where Θ\Theta is the angle between the magnetic field BB and the area AA.

Faraday's Laws of Induction

  1. The induced emf in a coil of NN turns is related to the rate of change of magnetic flux:
    ϵ=−NdΦBdt\epsilon = -N \frac{d\Phi_B}{dt}
  2. If the circuit is closed, the current II is given by:
    I=ϵRI = \frac{\epsilon}{R}
    where RR is the resistance of the circuit.

Lenz's Law

  • The polarity of the induced emf opposes the change in magnetic flux that produces it.

Induced EMF Formulas

  • Motional emf for a rod moving in a magnetic field:
    ϵ=Blu\epsilon = B l u
    where ll is the length of the rod and uu is its velocity.
  • Self-inductance:
    ϵ=−LdIdt\epsilon = -L \frac{dI}{dt}
    where LL is the self-inductance of the coil.
  • Mutual inductance:
    ϵ1=−M12dI2dt\epsilon_1 = -M_{12} \frac{dI_2}{dt}
    where M12M_{12} is the mutual inductance.

6.3 Examples

  • Example 6.3: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside. If the current changes from 2.0 A to 4.0 A in 0.1 S, calculate the induced emf.
  • Example 6.4: A rectangular wire loop moving out of a uniform magnetic field of 0.3 T. Calculate the emf developed across a cut in the loop.
  • Example 6.5: A metallic rod rotated in a magnetic field of 0.5 T. Calculate the emf developed between the center and the ring.

6.4 Important Formulas and Definitions

QuantitySymbolUnitsDimensionsEquations
Magnetic FluxΦB\Phi_BWb (weber)[ML²T⁻²A⁻¹]ΦB=B⋅A\Phi_B = B \cdot A
EMFϵ\epsilonV (volt)[ML²T⁻³A⁻¹]
Mutual InductanceMMH (henry)[M L²T⁻²A⁻²]ϵ1=−M12dI2dt\epsilon_1 = -M_{12} \frac{dI_2}{dt}
Self InductanceLLH (henry)[ML²T⁻²A⁻²]ϵ=−LdIdt\epsilon = -L \frac{dI}{dt}

6.5 Points to Ponder

  • The relationship between electricity and magnetism is fundamental.
  • Induced currents oppose changes in magnetic flux, adhering to conservation of energy principles.
  • The self-inductance of a solenoid depends on its geometry and the permeability of the core material.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

The number of turns in the solenoid

B.

The rate of change of current in the solenoid

C.

The distance between the solenoid and the coil

D.

The resistance of the coil
Correct Answer: D

Solution:

The induced emf depends on the mutual inductance, which is affected by the number of turns and the rate of change of current, as well as the distance between the solenoid and the coil. The resistance of the coil does not affect the magnitude of the induced emf, but it affects the current that flows as a result.

Chapter Concept:

Magnetic Flux

A.

The self-inductance remains the same.

B.

The self-inductance is doubled.

C.

The self-inductance is quadrupled.

D.

The self-inductance is halved.
Correct Answer: C

Solution:

The self-inductance of a solenoid is given by L=μ0n2AlL = \mu_0 n^2 A l. If the number of turns per unit length nn is doubled, the self-inductance becomes L=μ0(2n)2Al=4μ0n2AlL = \mu_0 (2n)^2 A l = 4 \mu_0 n^2 A l, which means the self-inductance is quadrupled.

Chapter Concept:

Self-Inductance of Long Solenoid

A.

ΦB=B⋅A=BAcos⁡θ\Phi_B = B \cdot A = BA \cos \theta

B.

ΦB=B+Acos⁡θ\Phi_B = B + A \cos \theta

C.

ΦB=B×Asin⁡θ\Phi_B = B \times A \sin \theta

D.

ΦB=B−Acos⁡θ\Phi_B = B - A \cos \theta
Correct Answer: A

Solution:

Magnetic flux through a plane area in a uniform magnetic field is given by ΦB=B⋅A=BAcos⁡θ\Phi_B = B \cdot A = BA \cos \theta, where θ\theta is the angle between the magnetic field and the area vector.

Chapter Concept:

Magnetic Flux

A.

0.024 V

B.

0.048 V

C.

0.072 V

D.

0.096 V
Correct Answer: B

Solution:

The emf induced is given by ϵ=Blv\epsilon = B l v, where BB is the magnetic field, ll is the length of the side perpendicular to the velocity, and vv is the velocity. For the longer side, l=8 cml = 8 \text{ cm}, thus ϵ=0.3×0.08×0.01=0.024 V\epsilon = 0.3 \times 0.08 \times 0.01 = 0.024 \text{ V}.

Chapter Concept:

Self-Inductance

A.

The induced emf is zero because the rod is moving perpendicular to the magnetic field.

B.

The induced emf is given by ε=Blv\varepsilon = Blv, and it causes a current to flow in the circuit.

C.

The induced emf depends on the resistance of the rod.

D.

The induced emf is given by ε=Bv2l\varepsilon = Bv^2l, and it opposes the motion of the rod.
Correct Answer: B

Solution:

The induced emf in a rod moving perpendicular to a magnetic field is given by ε=Blv\varepsilon = Blv, where BB is the magnetic field strength, ll is the length of the rod, and vv is the velocity of the rod. This emf causes a current to flow in the circuit.

Chapter Concept:

Motional Electromotive Force

A.

50 V

B.

100 V

C.

200 V

D.

157 V
Correct Answer: D

Solution:

The emf is given by ϵ=12BωR2\epsilon = \frac{1}{2} B \omega R^2, where B=0.5 TB = 0.5 \text{ T}, ω=400 rad/s\omega = 400 \text{ rad/s}, and R=1 mR = 1 \text{ m}. Thus, ϵ=12×0.5×400×12=100 V\epsilon = \frac{1}{2} \times 0.5 \times 400 \times 1^2 = 100 \text{ V}.

Chapter Concept:

Lenz's Law

A.

ε=Blv\varepsilon = Blv

B.

ε=Bv/l\varepsilon = Bv/l

C.

ε=B/lv\varepsilon = B/lv

D.

ε=Bv2l\varepsilon = Bv^2l
Correct Answer: A

Solution:

The motional emf is given by the expression ε=Blv\varepsilon = Blv when a rod of length ll moves with speed vv perpendicular to a magnetic field BB.

Chapter Concept:

Motional Electromotive Force

A.

The current flows in a direction that supports the increase in magnetic flux.

B.

The current flows in a direction that opposes the increase in magnetic flux.

C.

The current flows in a random direction.

D.

There is no induced current.
Correct Answer: B

Solution:

Lenz's Law states that the direction of the induced current is such that it opposes the change in magnetic flux that produced it. Therefore, when the North pole of a bar magnet is pushed towards a coil, the induced current flows in a direction that opposes the increase in magnetic flux.

Chapter Concept:

Lenz's Law

A.

L=μ0n2AlL = \mu_0 n^2 A l

B.

L=μrμ0n2AlL = \mu_r \mu_0 n^2 A l

C.

L=μ0nAlL = \mu_0 n A l

D.

L=μrμ0nAlL = \mu_r \mu_0 n A l
Correct Answer: B

Solution:

The self-inductance of a long solenoid with a magnetic core is given by L=μrμ0n2AlL = \mu_r \mu_0 n^2 A l, where μr\mu_r is the relative permeability, μ0\mu_0 is the permeability of free space, nn is the number of turns per unit length, AA is the cross-sectional area, and ll is the length of the solenoid.

Chapter Concept:

Self-Inductance of Long Solenoid

A.

There is a continuous deflection in the galvanometer.

B.

The galvanometer shows a momentary deflection.

C.

The galvanometer shows no deflection.

D.

The galvanometer shows a deflection only when the key is released.
Correct Answer: B

Solution:

The galvanometer shows a momentary deflection when the tapping key is pressed because the magnetic flux through the neighboring coil changes, inducing an emf.

Chapter Concept:

Faraday and Henry Experiments

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True or False

Correct Answer: True

Solution:

Electromagnetic induction occurs when there is a change in magnetic flux through a coil, leading to the induction of an emf, as demonstrated by Faraday and Henry.

Chapter Concept :

Electromagnetic Induction

Correct Answer: True

Solution:

An AC generator operates on the principle of electromagnetic induction, where mechanical energy is converted into electrical energy by rotating a coil in a magnetic field.

Chapter Concept :

AC Generator

Correct Answer: False

Solution:

Lenz's Law states that the direction of induced current is such that it opposes the change in magnetic flux that produced it, ensuring conservation of energy.

Chapter Concept :

Lenz's Law

Correct Answer: True

Solution:

The magnetic flux through a coil can be changed by altering its shape, which in turn changes the area or orientation of the coil in a magnetic field, leading to an induced emf according to Faraday's law.

Chapter Concept :

Methods of Changing Magnetic Flux

Correct Answer: True

Solution:

The mutual inductance M12M_{12} of coil 1 with respect to coil 2 is equal to the mutual inductance M21M_{21} of coil 2 with respect to coil 1. This is a fundamental property of mutual inductance and is valid irrespective of the geometry of the coils.

Chapter Concept :

Mutual Inductance

Correct Answer: False

Solution:

Lenz's Law states that the direction of induced current is such that it opposes the change in magnetic flux that produced it, ensuring conservation of energy.

Chapter Concept :

Lenz's Law

Correct Answer: True

Solution:

The motional emf ε=Blv\varepsilon = Blv is generated when a conducting rod of length ll moves with speed vv perpendicular to a magnetic field BB. This can be explained by the Lorentz force acting on the charges in the rod, which causes a separation of charges and induces an emf.

Chapter Concept :

Motional Electromotive Force

Correct Answer: True

Solution:

Magnetic flux is indeed a scalar quantity and its SI unit is the weber (Wb). This is confirmed by the definition of magnetic flux through a plane area in a uniform magnetic field, which is given by ΦB=B⋅A=BAcos⁡θ\Phi_B = B \cdot A = BA \cos \theta, where θ\theta is the angle between the magnetic field and the area vector.

Chapter Concept :

Magnetic Flux

Correct Answer: True

Solution:

Self-inductance is indeed the property by which a changing current in a coil induces a back emf in the same coil, as described by the equation ε=−LdIdt\varepsilon = -L \frac{dI}{dt}, where LL is the self-inductance.

Chapter Concept :

Self-Inductance

Correct Answer: False

Solution:

The number of spokes is immaterial because the emfs across the spokes are in parallel, and thus do not affect the total induced emf.

Chapter Concept :

Rotating Rod and Wheel Motional EMF

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