10th Class Physics Notes 2026 — Chapter 18
Chapter 18: Electromagnetic Induction and Electromagnetic Waves
Class: 10th
Subject: Physics
Board: Punjab
Exercise Solutions
A. Multiple Choice Questions (MCQs)
18.1 What happens to the induced e.m.f. in a coil if the rate of change of the magnetic field increases?
(a) Decreases
(b) Remains the same
(c) Increases ✓
(d) Becomes zero
18.2 In an A.C. generator, slip rings are used to:
(a) Reverse current direction periodically ✓
(b) Convert A.C. to D.C.
(c) Prevent induced current from flowing through the core
(d) Increase magnetic flux
18.3 The sky appears blue during the day due to:
(a) Reflection of light
(b) Scattering of shorter wavelengths ✓
(c) Absorption of blue light
(d) Refraction of light
18.4 Light can be considered as massless particles called:
(a) Electrons
(b) Protons
(c) Photons ✓
(d) Neutrons
18.5 In a transformer, energy is transferred from the primary coil to the secondary coil through:
(a) Direct electrical connection
(b) A magnetic field in the iron core ✓
(c) Electrons
(d) Heat conduction
B. Short Answer Questions
18.1 What is the direction of induced current in Lenz’s law?
Answer:
According to Lenz’s law, the direction of the induced current is always such that the magnetic field produced by it opposes the change in magnetic flux that produces the current.
18.2 Why are slip rings used in an A.C. generator?
Answer:
Slip rings maintain continuous electrical contact between the rotating armature coil of an A.C. generator and the external circuit. They allow the alternating current generated in the rotating coil to flow to the external circuit without twisting or breaking the connecting wires.
18.3 When is induced e.m.f. zero in an A.C. generator?
Answer:
The induced e.m.f. is zero when the plane of the rotating coil is perpendicular to the magnetic field lines, at the \(0^\circ\) and \(180^\circ\) positions in the usual angular convention. At these positions, the rate of change of magnetic flux is momentarily zero.
18.4 Why is the sky red at sunset?
Answer:
At sunset, sunlight travels a much longer path through the Earth’s atmosphere before reaching our eyes. Most of the shorter wavelengths, such as blue and violet, are scattered away in different directions by atmospheric molecules. The longer wavelengths, mainly red and orange, reach the observer more directly, making the sky appear red or orange.
18.5 What is a transformer and on what principle does it work?
Answer:
Transformer: A transformer is an electrical device used to increase or decrease alternating voltage.
Principle: It works on the principle of mutual induction between two coils linked by a common changing magnetic flux.
C. Constructed Response Questions
18.1 If a magnet is moved quickly into a coil, a larger e.m.f. is induced. Why does speed affect the magnitude of the induced current?
Answer:
According to Faraday’s law of electromagnetic induction, the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux.
Where:
\(\varepsilon\) = induced e.m.f.
\(N\) = number of turns in the coil
\(\Delta\Phi\) = change in magnetic flux
\(\Delta t\) = time interval
Moving the magnet faster reduces the time taken for the same change in magnetic flux. This increases the rate of change of flux and produces a larger induced e.m.f. If the resistance remains constant, the induced current also increases.
18.2 Why is an iron core used in transformers instead of wood or plastic? Explain briefly.
Answer:
An iron core is a ferromagnetic material with high magnetic permeability. It provides an efficient path for magnetic flux, helping link the primary and secondary coils while reducing magnetic flux leakage.
Wood and plastic are non-magnetic materials and cannot concentrate or guide magnetic flux as effectively as iron. Therefore, an iron core improves the efficiency of a transformer.
18.3 How does the concept of photons help explain the working of solar sails in space?
Answer:
Light consists of packets of energy called photons. Photons have zero rest mass but carry momentum. When sunlight strikes a reflective solar sail, photons transfer momentum to the sail through radiation pressure. This force pushes the spacecraft forward without requiring onboard fuel.
18.4 Why can a transformer work only on alternating current and not on direct current?
Answer:
A transformer relies on changing magnetic flux to induce voltage in the secondary coil through mutual induction.
Alternating current (A.C.) continuously changes its magnitude and direction, producing a changing magnetic field. This changing field induces an e.m.f. in the secondary coil.
Direct current (D.C.) produces a constant magnetic field after the initial switching period. Since the magnetic flux no longer changes, no continuous e.m.f. is induced in the secondary coil.
Therefore, a transformer operates normally with A.C., not steady D.C.
D. Comprehensive Questions
18.1 Describe how Lenz’s law supports the law of conservation of energy. Give one real-life example to support your explanation.
Answer:
Support for the Law of Conservation of Energy:
Lenz’s law states that the induced current produces a magnetic field that opposes the change in magnetic flux responsible for inducing it. This opposition means that external work is required to produce the change.
If the induced current assisted the change instead of opposing it, energy could be produced without an external energy input, violating the law of conservation of energy. The mechanical work done against the opposing force is converted into electrical energy.
Real-Life Example: Magnetic Braking
In trains, eddy-current brakes use electromagnetic induction to slow moving metal components. When a conductor moves through a magnetic field, eddy currents are induced within it. These currents produce a magnetic effect that opposes the motion, converting kinetic energy into thermal energy without requiring direct friction between the braking surfaces.
18.2 Compare step-up and step-down transformers. Include the difference in number of turns, voltage output, and where each is used.
Answer:
Feature | Step-Up Transformer | Step-Down Transformer |
|---|---|---|
Number of turns | Secondary turns exceed primary turns, \(N_s>N_p\). | Primary turns exceed secondary turns, \(N_p>N_s\). |
Voltage output | Increases voltage, \(V_s>V_p\). | Decreases voltage, \(V_s<V_p\). |
Current output (ideal transformer) | Decreases current, \(I_s<I_p\). | Increases current, \(I_s>I_p\). |
Common applications | Power plants and high-voltage transmission systems. | Substations, mobile phone chargers and domestic power supplies. |
Where:
\(N_p\) = number of primary turns
\(N_s\) = number of secondary turns
\(V_p\) = primary voltage
\(V_s\) = secondary voltage
\(I_p\) = primary current
\(I_s\) = secondary current
18.3 How does the scattering of sunlight cause the sky to appear blue during the day and red at sunset? Explain.
Answer:
Rayleigh Scattering: The Earth’s atmosphere scatters shorter wavelengths of light, such as blue and violet, much more strongly than longer wavelengths, such as red and orange.
Blue Sky During the Day:
When the Sun is high in the sky, sunlight passes through a relatively short path in the atmosphere. Blue light is scattered in different directions by gas molecules. This scattered light reaches our eyes, making the sky appear blue.
Red Sky at Sunset:
At sunset, sunlight travels through a much thicker layer of the atmosphere. Most of the shorter wavelengths are scattered away before reaching our eyes. More of the longer red and orange wavelengths reach the observer directly, making the sky appear red or orange.
18.4 What is an oscilloscope, and how does it help in analyzing electrical signals? Describe its main components.
Answer:
Oscilloscope: An oscilloscope is an electronic test instrument that displays changing electrical voltage signals as waveforms on a screen over time.
Uses in Signal Analysis:
It helps engineers and technicians measure and observe:
Signal amplitude or voltage
Frequency and time period
Phase differences
Waveform shape and distortion
Main Components:
Electron Gun: Emits and focuses a beam of electrons.
Deflection Plates: Deflect the electron beam horizontally and vertically. The horizontal direction represents the time base, while the vertical direction represents the voltage signal.
Display Screen: A fluorescent screen glows where the electron beam strikes it, allowing the waveform to be displayed.
18.5 What is the shape of the path followed by an electron when it moves at right angles to a uniform magnetic field? Explain why.
Answer:
Shape of Path: The electron follows a circular path.
Explanation:
When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic Lorentz force:
Since \(\sin 90^\circ=1\),
Where:
\(F\) = magnetic force
\(q\) = magnitude of the electron’s charge
\(v\) = velocity of the electron
\(B\) = magnetic field strength
The magnetic force is perpendicular to the electron’s velocity. It continuously changes the direction of the electron’s motion and acts as the centripetal force required for circular motion.
Equating the magnetic force and centripetal force:
Therefore, the magnetic force keeps the electron moving in a circular path, assuming a uniform magnetic field and no other forces.
E. Numerical Problems
18.1 A coil consisting of 50 tightly wound turns experiences a change in magnetic flux from 0.3 Wb to 0.8 Wb over a time interval of 0.4 s. Determine the average induced electromotive force (e.m.f.) generated in the coil during this time.
Given Data:
Number of turns, \(N=50\)
Initial magnetic flux, \(\Phi_1=0.3\,\text{Wb}\)
Final magnetic flux, \(\Phi_2=0.8\,\text{Wb}\)
Time interval, \(\Delta t=0.4\,\text{s}\)
To Find:
Average induced e.m.f., \(\varepsilon=?\)
Formula:
Calculation:
First, calculate the change in magnetic flux:
Now substitute the values:
Answer:
The average induced e.m.f. is \(-62.5\,\text{V}\). The negative sign indicates the direction specified by Lenz’s law. Its magnitude is \(62.5\,\text{V}\).
18.2 A circular loop of radius 0.2 m and resistance 5 Ω is placed perpendicular to a magnetic field of 0.5 T. If the field drops to zero in 0.1 s, calculate: (a) the induced e.m.f. (b) the induced current and its direction, assuming the field points into the page.
Given Data:
Radius of loop, \(r=0.2\,\text{m}\)
Resistance, \(R=5\,\Omega\)
Initial magnetic field, \(B_1=0.5\,\text{T}\)
Final magnetic field, \(B_2=0\,\text{T}\)
Time interval, \(\Delta t=0.1\,\text{s}\)
To Find:
(a) Induced e.m.f., \(\varepsilon=?\)
(b) Induced current, \(I=?\), and its direction.
Formulae:
Area of the loop:
Change in magnetic flux:
Induced e.m.f. for a single loop:
Induced current:
Calculation:
Step 1: Calculate the area of the loop.
Step 2: Calculate the change in magnetic flux.
Step 3: Calculate the induced e.m.f.
Step 4: Calculate the induced current.
Step 5: Determine the direction of current.
The magnetic field directed into the page is decreasing. According to Lenz’s law, the induced current produces a magnetic field into the page to oppose this decrease. By the right-hand grip rule, a clockwise current produces a magnetic field directed into the page.
Answer:
(a) Induced e.m.f. = \(0.63\,\text{V}\)
(b) Induced current = \(0.126\,\text{A}\), flowing clockwise.
18.3 A transformer has 500 primary turns and 100 secondary turns. If the primary voltage is 220 V, find: (a) secondary voltage (b) is this a step-up or step-down transformer?
Given Data:
Primary turns, \(N_p=500\)
Secondary turns, \(N_s=100\)
Primary voltage, \(V_p=220\,\text{V}\)
To Find:
(a) Secondary voltage, \(V_s=?\)
(b) Type of transformer.
Formula:
For an ideal transformer:
Therefore,
Calculation:
Step 1: Calculate the secondary voltage.
Step 2: Identify the type of transformer.
Since \(N_s<N_p\) and \(V_s<V_p\), the transformer decreases the voltage.
Answer:
(a) Secondary voltage = \(44\,\text{V}\)
(b) It is a step-down transformer.
18.4 A step-up transformer has a turns ratio of 2:100. If an alternating voltage of 25 V is applied to the primary coil, what will be the secondary voltage?
Given Data:
Turns ratio, \(N_p:N_s=2:100\)
Primary voltage, \(V_p=25\,\text{V}\)
To Find:
Secondary voltage, \(V_s=?\)
Formula:
Therefore,
Calculation:
Answer:
The secondary voltage is \(1250\,\text{V}\).
18.5 A step-down transformer has a turns ratio of 100:5. If an A.C. voltage of 190 V is applied to the primary coil, what is the voltage across the secondary coil?
Given Data:
Turns ratio, \(N_p:N_s=100:5\)
Primary voltage, \(V_p=190\,\text{V}\)
To Find:
Secondary voltage, \(V_s=?\)
Formula:
Therefore,
Calculation:
Answer:
The voltage across the secondary coil is \(9.5\,\text{V}\).
Notes by Sir Dilawar
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