📡 Chapter 18: Electromagnetic Induction & EM Waves – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 18 Electromagnetic Induction & EM Waves including Faraday's law, induced emf, transformers, and power loss calculations. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 18
Electromagnetic Induction covers Faraday's law, Lenz's law, A.C. generators, transformers, cathode rays, CRO, and EM waves.
📐 Numerical Problems & Solutions (PECTAA 2026)
📝 Important Formulas
Ratio formula of transformer: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
Faraday's law of electromagnetic induction: \(\epsilon = -N \frac{\Delta \phi}{\Delta t}\)
18.1 Induced e.m.f. from Change in Flux
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
\(Number\ of\ turns\)\(= N = 50\)
\(Initial\ flux\)\(= \phi_1 = 0.3\ Wb\)
\(Final\ flux\)\(= \phi_2 = 0.8\ Wb\)
\(Time\ interval\)\(= \Delta t = 0.4\ s\)
To Find
\(Induced\ e.m.f. = \epsilon = ?\)
Solution
By using Faraday's law of electromagnetic induction:
\[\begin{aligned}
\epsilon &= -N\frac{\Delta\phi}{\Delta t} \\
\epsilon &= -\frac{N(\phi_2 - \phi_1)}{\Delta t} \\
\epsilon &= -\frac{(50)(0.8 - 0.3)}{0.4} \\
\epsilon &= -\frac{(50)(0.5)}{0.4} \\
\epsilon &= -\frac{25}{0.4} \\
\boldsymbol{\epsilon} &= \boldsymbol{-62.5\ V}
\end{aligned}\]
✅ \(\epsilon = -62.5\ V\)
18.2 Induced e.m.f. and Current in a Circular Loop
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 emf (b) the induced current and its direction (assuming the field points into the page).
Given Data
\(Number\ of\ loops\)\(= N = 1\)
\(Radius\)\(= r = 0.2\ m\)
\(Resistance\)\(= R = 5\ \Omega\)
\(Initial\ magnetic\ field\)\(= B_1 = 0.5\ T\)
\(Final\ magnetic\ field\)\(= B_2 = 0\ T\)
\(Time\ interval\)\(= \Delta t = 0.1\ s\)
To Find
\(Induced\ e.m.f. = \epsilon = ?\)
\(Induced\ current = I = ?\)
Solution
First, calculate the area of the circular loop:
\[\begin{aligned}
A &= \pi r^2 \\
A &= (3.14)(0.2)^2 \\
\boldsymbol{A} &= \boldsymbol{0.1256\ m^2}
\end{aligned}\]
Since the loop is placed perpendicular to the magnetic field, so \(\theta = 0^\circ\)
\[\begin{aligned}
\phi_1 &= B_1 A \cos \theta \\
\phi_1 &= (0.5)(0.1256)\cos 0^\circ \\
\phi_1 &= (0.5)(0.1256)(1) \\
\boldsymbol{\phi_1} &= \boldsymbol{0.0628\ Wb} \\
\phi_2 &= B_2 A \cos \theta \\
\phi_2 &= (0)(0.1256)\cos 0^\circ \\
\boldsymbol{\phi_2} &= \boldsymbol{0\ Wb}
\end{aligned}\]
By using Faraday's law of electromagnetic induction:
\[\begin{aligned}
\epsilon &= -N\frac{\Delta\phi}{\Delta t} \\
\epsilon &= -\frac{N(\phi_2 - \phi_1)}{\Delta t} \\
\epsilon &= -\frac{(1)(0 - 0.0628)}{0.1} \\
\epsilon &= -\frac{-0.0628}{0.1} \\
\boldsymbol{\epsilon} &= \boldsymbol{0.628\ V}
\end{aligned}\]
By using Ohm's law:
\[\begin{aligned}
V &= IR \\
\epsilon &= IR \\
I &= \frac{\epsilon}{R} \\
I &= \frac{0.628}{5} \\
\boldsymbol{I} &= \boldsymbol{0.1256\ A}
\end{aligned}\]
Direction of Induced Current: According to Lenz's Law, the induced current flows clockwise to produce a magnetic field into the page, opposing the decrease in the original magnetic field.
✅ \(\epsilon = 0.628\ V\) | \(I = 0.1256\ A\) (Clockwise)
18.3 Transformer: Secondary Voltage and Type
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
\(Primary\ turns\)\(= N_p = 500\)
\(Secondary\ turns\)\(= N_s = 100\)
\(Primary\ voltage\)\(= V_p = 220\ V\)
To Find
\(Secondary\ voltage = V_s = ?\)
\(Type\ of\ transformer = ?\)
Solution
By using the turns-ratio equation:
\[\begin{aligned}
\frac{V_s}{V_p} &= \frac{N_s}{N_p} \\
\frac{V_s}{220} &= \frac{100}{500} \\
V_s &= \frac{(220)(100)}{500} \\
\boldsymbol{V_s} &= \boldsymbol{44\ V}
\end{aligned}\]
Since \(N_s < N_p\) and \(V_s < V_p\), this is a step-down transformer.
✅ \(V_s = 44\ V\) | Step-down transformer
18.4 Step-up Transformer: Secondary Voltage
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
\(Turns\ ratio\)\(= N_p:N_s = 2:100\)
\(Primary\ turns\)\(= N_p = 2\)
\(Secondary\ turns\)\(= N_s = 100\)
\(Primary\ voltage\)\(= V_p = 25\ V\)
To Find
\(Secondary\ voltage = V_s = ?\)
Solution
By using ratio formula of transformer:
\[\begin{aligned}
\frac{V_s}{V_p} &= \frac{N_s}{N_p} \\
V_s &= \frac{N_s V_p}{N_p} \\
V_s &= \frac{(100)(25)}{2} \\
\boldsymbol{V_s} &= \boldsymbol{1250\ V}
\end{aligned}\]
✅ \(V_s = 1250\ V\)
18.5 Step-down Transformer: Secondary Voltage
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
\(Turns\ ratio\)\(= N_p:N_s = 100:5\)
\(Primary\ turns\)\(= N_p = 100\)
\(Secondary\ turns\)\(= N_s = 5\)
\(Primary\ voltage\)\(= V_p = 190\ V\)
To Find
\(Secondary\ voltage = V_s = ?\)
Solution
By using ratio formula of transformer:
\[\begin{aligned}
\frac{V_s}{V_p} &= \frac{N_s}{N_p} \\
V_s &= \frac{N_s V_p}{N_p} \\
V_s &= \frac{(5)(190)}{100} \\
\boldsymbol{V_s} &= \boldsymbol{9.5\ V}
\end{aligned}\]
✅ \(V_s = 9.5\ V\)
📘 Examples with Solutions
Ex 18.1 Change in Magnetic Flux from Induced e.m.f.
A coil with 25 turns experiences an induced emf of 2.5 V for 0.20 s. Assuming the e.m.f. is constant during this interval, calculate the total change in magnetic flux (\(\Delta \phi\)) through the coil.
Given Data
\(Number\ of\ turns\)\(= N = 25\)
\(Induced\ e.m.f.\)\(= \epsilon = 2.5\ V\)
\(Time\ interval\)\(= \Delta t = 0.20\ s\)
To Find
\(Change\ in\ magnetic\ flux = \Delta \phi = ?\)
Solution
By using Faraday's law of electromagnetic induction:
\[\begin{aligned}
\epsilon &= -N\frac{\Delta\phi}{\Delta t}
\end{aligned}\]
For calculating the magnitude of the change in flux, we can ignore the negative sign and use:
\[\begin{aligned}
\epsilon &= \frac{N\Delta\phi}{\Delta t} \\
\Delta\phi &= \frac{\epsilon \Delta t}{N} \\
\Delta\phi &= \frac{(2.5)(0.20)}{25} \\
\boldsymbol{\Delta\phi} &= \boldsymbol{0.02\ Wb}
\end{aligned}\]
✅ \(\Delta \phi = 0.02\ Wb\)
Ex 18.2 Power Loss in Transmission Cables
Electricity is transmitted at a voltage of 20,000 V with a current of 5 A through power cables that have a resistance of 10 Ω. Calculate the power loss in the cables?
Given Data
\(Voltage\)\(= V = 20000\ V\)
\(Current\)\(= I = 5\ A\)
\(Resistance\)\(= R = 10\ \Omega\)
To Find
\(Power\ loss = P = ?\)
Solution
By using Joule's law or the power formula:
\[\begin{aligned}
P &= I^2 R \\
P &= (5)^2 (10) \\
P &= (25)(10) \\
\boldsymbol{P} &= \boldsymbol{250\ W}
\end{aligned}\]
✅ \(P = 250\ W\)
📐 Key Formulas – Electromagnetic Induction
Faraday's Law: \(\epsilon = -N \frac{\Delta \phi}{\Delta t}\)
Magnetic Flux: \(\phi = BA \cos \theta\)
Transformer Turns Ratio: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
Ohm's Law: \(V = IR\)
Power Loss: \(P = I^2 R\)
Weber: \(1 Wb = 1 T m^2\)
Unit 10Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13Sound
Unit 14Light
Unit 15Electrostatics
Unit 16Electricity
Unit 17Electromagnetism
Unit 18✨ Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For numerical problems, always write Given Data, To Find, and step-by-step Solution with formulas. Pay attention to units and conversions. These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus