CHAPTER 6: ELECTROMAGNETIC INDUCTION
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
SECTION A (16 Marks)
Q.1. The current in the primary coil of a pair of coils changes from $\displaystyle {7\text{ A}}$ to $\displaystyle {3\text{ A}}$ in $\displaystyle {0.04\text{ s}}$. The mutual inductance between the two coils is $\displaystyle {0.5\text{ H}}$. The induced EMF in the secondary coil is:
Explanation: $\displaystyle {|\varepsilon| = M {\left( {\frac{{\Delta I}}{{\Delta t}}} \right)} = 0.5 \times {\left( {\frac{{7 - 3}}{{0.04}}} \right)} = 0.5 \times {\left( {\frac{{4}}{{0.04}}} \right)} = 0.5 \times 100 = 50\text{ V}}$.
Q.2. A square shaped coil of side $\displaystyle {10\text{ cm}}$ having 100 turns is placed perpendicular to a magnetic field increasing at $\displaystyle {1\text{ T/s}}$. The induced EMF in the coil is:
Explanation: Area $\displaystyle {A = (0.1\text{ m})^2 = 0.01\text{ m}^2}$. $\displaystyle {|\varepsilon| = N A {\frac{{dB}}{{dt}}} = 100 \times 0.01 \times 1 = 1.0\text{ V}}$.
Q.3. Lenz's law is in strict accordance with the law of conservation of:
Explanation: Mechanical work done in moving a magnet against the repulsive or attractive magnetic force induced in the coil is converted into electrical energy.
Q.4. A coil of area $\displaystyle {0.1\text{ m}^2}$ is placed perpendicular to a uniform magnetic field of $\displaystyle {0.2\text{ T}}$. The magnetic flux through the coil is:
Explanation: $\displaystyle {\Phi = B A \cos 0^\circ = 0.2 \times 0.1 \times 1 = 0.02\text{ Wb}}$.
Q.5. The self-inductance of a long solenoid does NOT depend on:
Explanation: Self-inductance $\displaystyle {L = {\frac{{\mu_0 \mu_r N^2 A}}{{l}}}}$ depends on geometry ($N, A, l$) and permeability of core ($\mu_r$), but not on wire resistivity/material.
Q.6. A coil's self-inductance is a quantitative measure of its:
Explanation: Self-inductance opposes any change (growth or decay) of electric current in a circuit, acting analogous to mass (inertia) in mechanics.
Q.7. A transformer works on the basic principle of:
Explanation: Alternating current in the primary winding produces a changing magnetic flux that links with and induces an EMF in the secondary winding.
Q.8. The direction of induced current in a conductor moving through a magnetic field is given by:
Explanation: Lenz's law gives direction based on flux opposition; Fleming's Right-Hand rule determines direction for motional EMF in conductors.
Q.9. The magnetic energy stored in an inductor carrying steady current $\displaystyle {I}$ is:
Explanation: Work done to establish current against back EMF is stored as magnetic potential energy: $\displaystyle {U = \int_0^I L i \, di = {\frac{{1}}{{2}}} L I^2}$.
Q.10. The SI unit of self-inductance and mutual inductance is:
Explanation: $\displaystyle {1\text{ Henry (H)} = 1\text{ V}\cdot\text{s/A} = 1\text{ Wb/A}}$.
Q.11. Which of the following electrical devices fundamentally utilizes electromagnetic induction?
Q.12. When a metallic rod moves perpendicular to a uniform magnetic field, the induced EMF generated across its ends is called:
Explanation: Lorentz force on free charge carriers inside the moving conductor creates charge separation, inducing EMF $\displaystyle {\varepsilon = B l v}$.
Directions for Q.13 to Q.16:
(a) Both Assertion and Reason are true and Reason is correct explanation.
(b) Both Assertion and Reason are true but Reason is NOT correct explanation.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
Q.13. Assertion (A): The induced current always flows in a direction such that it opposes the change in magnetic flux that produced it.
Reason (R): This direction is determined by Lenz's law, which reflects conservation of energy.
Q.14. Assertion (A): A current is induced in a conducting loop when it is kept stationary inside a strong constant magnetic field.
Reason (R): Induced EMF depends directly on the time rate of change of magnetic flux ($\displaystyle {d\Phi/dt}$).
Explanation: Inside a constant magnetic field without motion, $\displaystyle {d\Phi/dt = 0}$, so zero EMF is induced despite strong magnetic flux.
Q.15. Assertion (A): A transformer cannot step up or step down a direct current (DC) voltage.
Reason (R): Direct current produces a constant magnetic flux in the core, so no EMF is induced in the secondary coil.
Q.16. Assertion (A): Power loss due to eddy currents in transformer cores is minimized by using laminated soft iron sheets.
Reason (R): Laminations insulated by varnish increase the electrical resistance across eddy current loops.
SECTION B (10 Marks)
Q.17. Define magnetic flux ($\displaystyle {\Phi_B}$). Write its mathematical expression and SI unit.
Q.18. A coil of wire is placed in a uniform magnetic field. Under what geometric condition will maximum EMF be induced in the coil when rotated?
Q.17: Magnetic flux is defined as total number of magnetic field lines passing normally through a given surface area.
$\displaystyle {\Phi_B = \vec{B} \cdot \vec{A} = B A \cos\theta}$. SI Unit: Weber ($\text{Wb}$) or $\text{T}\cdot\text{m}^2$.
OR Q.18: Maximum EMF is induced when the rate of change of flux is maximum, which occurs when the plane of the coil is parallel to field lines ($\displaystyle {\theta = 90^\circ}$ between area vector and field $\displaystyle {B}$, making $\displaystyle {\sin\omega t = 1}$).
Q.19. Derive the expression for motional EMF ($\displaystyle {\varepsilon = B l v}$) induced across a conducting rod of length $\displaystyle {l}$ moving with velocity $\displaystyle {v}$ perpendicular to a uniform magnetic field $\displaystyle {B}$.
Accumulation of electrons at one end creates electric field $\displaystyle {E}$ providing electric force $\displaystyle {F_e = e E}$.
At equilibrium: $\displaystyle {e E = e v B \implies E = v B}$.
Potential difference (EMF) across length $l$: $\displaystyle {\varepsilon = E l = B l v}$.
Q.20. Refer to the diagram of an electrical machine shown below:
(a) Identify the machine.
(b) Name the parts labeled P, Q, and R.
(c) State two ways of increasing the magnitude of output voltage generated.
(a) AC Generator (Alternator).
(b) P $\rightarrow$ Slip rings, Q $\rightarrow$ Carbon brushes, R $\rightarrow$ Armature coil.
(c) Ways to increase output EMF ($\displaystyle {\varepsilon_0 = N B A \omega}$):
1. Increase the number of turns $\displaystyle {N}$ in the armature coil.
2. Increase the rotational speed (angular frequency $\displaystyle {\omega}$) of armature.
Q.21. State two physical factors on which the self-inductance of a coil depends.
1. Number of turns $\displaystyle {N}$ ($\displaystyle {L \propto N^2}$).
2. Relative magnetic permeability $\displaystyle {\mu_r}$ of core material inserted inside the coil.
SECTION C (21 Marks)
Q.22. A rectangular coil of area $\displaystyle {0.02\text{ m}^2}$ having 100 turns is rotated at $\displaystyle {60\text{ rpm}}$ in a uniform magnetic field of $\displaystyle {0.1\text{ T}}$. Calculate maximum EMF induced.
$\displaystyle {\varepsilon_{\text{max}} = N B A \omega = 100 \times 0.1 \times 0.02 \times (2\pi) = 0.4\pi\text{ V} \approx 1.26\text{ V}}$.
Q.23. Derive the expression for energy stored in an inductor carrying steady current $\displaystyle {I}$.
Work done in small time $\displaystyle {dt}$: $\displaystyle {dW = P dt = |\varepsilon| i dt = {\left( L {\frac{{di}}{{dt}}} \right)} i dt = L i \, di}$.
Total work done building current from 0 to $I$: $\displaystyle {W = \int_0^I L i \, di = L {\left[ {\frac{{i^2}}{{2}}} \right]}_0^I = {\frac{{1}}{{2}}} L I^2}$.
This work is stored as magnetic field energy: $\displaystyle {U = {\frac{{1}}{{2}}} L I^2}$.
Q.24. What are eddy currents? Mention two practical industrial applications of eddy currents.
Applications:
1. Induction Furnaces: High-frequency eddy currents produce intense Joule heating to melt metals.
2. Electromagnetic Braking: Strong eddy currents induced in metal drums of high-speed trains provide smooth braking without mechanical wear.
Q.25. Explain the principle and working of a transformer. Mention two major causes of energy loss in actual transformers.
Principle: Mutual Induction ($\displaystyle {\varepsilon_s / \varepsilon_p = N_s / N_p}$).
Working: AC voltage in primary creates continuous changing flux in soft iron core, inducing proportional AC EMF in secondary coil.
Energy Losses:
1. Copper Loss: Joule heating ($\displaystyle {I^2 R}$) in primary and secondary copper windings.
2. Hysteresis Loss: Energy lost as heat during repeated magnetization cycles of iron core.
Q.26. A circular coil of 50 turns and radius $\displaystyle {10\text{ cm}}$ is placed in a uniform magnetic field of $\displaystyle {1\text{ T}}$. Calculate total magnetic flux through coil when field is normal to coil plane.
Total flux linkage $\displaystyle {\Phi_{\text{total}} = N B A \cos 0^\circ = 50 \times 1 \times 0.0314 \times 1 = 1.57\text{ Wb}}$.
Q.27. Define mutual inductance between two coils. State factors affecting mutual inductance.
Derive the expression for self-inductance of a long air-cored solenoid of length $\displaystyle {l}$, area $\displaystyle {A}$, and total turns $\displaystyle {N}$.
Mutual Inductance $\displaystyle {M}$ is numerical EMF induced in secondary per unit rate of change of current in primary ($\displaystyle {\varepsilon_s = -M dI_p/dt}$). Depends on: (i) turns $N_1, N_2$, (ii) separation/coupling coefficient, (iii) core permeability.
OR Part: Field inside solenoid: $\displaystyle {B = \mu_0 n I = \mu_0 {\left( {\frac{{N}}{{l}}} \right)} I}$.
Flux through single turn: $\displaystyle {\Phi = B A = {\frac{{\mu_0 N A I}}{{l}}}}$.
Total flux linkage: $\displaystyle {N \Phi = {\frac{{\mu_0 N^2 A I}}{{l}}} = L I \implies L = {\frac{{\mu_0 N^2 A}}{{l}}}}$.
Q.28. A solenoid of length $\displaystyle {50\text{ cm}}$ has 1000 turns and carries current $\displaystyle {2\text{ A}}$. Assuming cross-sectional area $\displaystyle {1\text{ cm}^2}$, calculate energy stored in magnetic field inside solenoid.
Self-inductance $\displaystyle {L = {\frac{{\mu_0 N^2 A}}{{l}}} = {\frac{{(4\pi \times 10^{-7}) \times (1000)^2 \times (1 \times 10^{-4})}{{0.5}}} = 2.51 \times 10^{-4}\text{ H}}$.
Stored energy $\displaystyle {U = {\frac{{1}}{{2}}} L I^2 = {\frac{{1}}{{2}}} \times (2.51 \times 10^{-4}) \times (2)^2 \approx 0.050\text{ J}}$.
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: AC Generator Operation & Maximum EMF.
An AC generator converts mechanical energy into electrical energy by rotating a coil in a magnetic field according to Faraday's law of electromagnetic induction.
(i) AC generator operates on the principle of:
(a) Self-induction (b) Electromagnetic induction (c) Mutual coupling
(ii) Current induced in armature coil is maximum when angle $\displaystyle {\theta}$ between area vector and field is:
(a) $\displaystyle {0^\circ}$ (b) $\displaystyle {90^\circ}$ (c) $\displaystyle {180^\circ}$
(iii) An AC generator coil has 1000 turns, area $\displaystyle {10^{-2}\text{ m}^2}$, rotating at $\displaystyle {100\text{ rpm}}$ in field $\displaystyle {4.2 \times 10^{-2}\text{ T}}$. Max EMF is:
(a) $\displaystyle {1.4\text{ V}}$ (b) $\displaystyle {3.4\text{ V}}$ (c) $\displaystyle {4.4\text{ V}}$ (d) $\displaystyle {5.0\text{ V}}$
Q.30. Case Study 2: Induction Cooktops and Eddy Current Heating.
Induction stoves use high-frequency alternating magnetic fields to heat metal cooking vessels directly through induced eddy currents.
(a) Name the basic physics principle utilized in induction heating.
(b) Explain how eddy currents produce heat inside the utensil.
(c) Why do induction cooktops require magnetic/ferromagnetic cooking vessels (e.g. cast iron, stainless steel)?
(a) Electromagnetic induction and Joule heating by eddy currents.
(b) Changing magnetic flux induces strong circulating eddy currents in metal bottom; electrical resistance generates heat ($\displaystyle {P = I^2 R}$).
(c) Ferromagnetic materials strongly concentrate magnetic field lines for maximum flux linkage and possess proper electrical resistivity for rapid heating.
SECTION E (15 Marks)
Q.31. (a) Derive expression for instantaneous EMF $\displaystyle {\varepsilon = \varepsilon_0 \sin\omega t}$ induced in a coil of $\displaystyle {N}$ turns and area $\displaystyle {A}$ rotating at angular speed $\displaystyle {\omega}$ in uniform magnetic field $\displaystyle {B}$.
(b) State expression for maximum EMF $\displaystyle {\varepsilon_0}$.
Explain mutual induction. Derive expression for mutual inductance $\displaystyle {M_{12}}$ between two long coaxial solenoids $\displaystyle {S_1}$ and $\displaystyle {S_2}$ of length $\displaystyle {l}$, radii $\displaystyle {r_1 < r_2}$, and turns $\displaystyle {N_1, N_2}$.
(a) Flux $\displaystyle {\Phi = B A \cos\omega t}$. By Faraday's law: $\displaystyle {\varepsilon = -N {\frac{{d\Phi}}{{dt}}} = -N B A {\frac{{d}}{{dt}}}(\cos\omega t) = N B A \omega \sin\omega t}$.
(b) $\displaystyle {\varepsilon_0 = N B A \omega}$.
OR Part: Current $\displaystyle {I_2}$ in outer solenoid $\displaystyle {S_2}$ produces inner field $\displaystyle {B_2 = \mu_0 {\left( {\frac{{N_2}}{{l}}} \right)} I_2}$.
Flux through inner solenoid $\displaystyle {S_1}$: $\displaystyle {N_1 \Phi_1 = N_1 (B_2 A_1) = N_1 {\left[ \mu_0 {\left( {\frac{{N_2}}{{l}}} \right)} I_2 \right]} (\pi r_1^2) = {\frac{{\mu_0 N_1 N_2 \pi r_1^2}}{{l}}} I_2 = M_{12} I_2$.
Hence $\displaystyle {M_{12} = {\frac{{\mu_0 N_1 N_2 \pi r_1^2}}{{l}}} = \mu_0 n_1 n_2 \pi r_1^2 l}$.
Q.32. A horizontal straight wire $\displaystyle {10\text{ m}}$ long extending East-West falls with speed $\displaystyle {5\text{ m/s}}$ perpendicular to horizontal component of earth field $\displaystyle {B_H = 0.3 \times 10^{-4}\text{ Wb/m}^2}$.
(a) Calculate instantaneous induced EMF.
(b) State direction of induced EMF.
A movable conducting arm PQ of length $\displaystyle {l}$ and resistance $\displaystyle {R}$ slides with velocity $\displaystyle {v}$ on parallel frictionless rails in magnetic field $\displaystyle {B}$. Derive:
(a) Induced EMF $\displaystyle {\varepsilon}$, (b) Magnetic retarding force $\displaystyle {F}$, (c) Mechanical power required $\displaystyle {P}$.
(a) $\displaystyle {\varepsilon = B_H l v = (0.3 \times 10^{-4}) \times 10 \times 5 = 1.5 \times 10^{-3}\text{ V} = 1.5\text{ mV}}$.
(b) Fleming's Right-Hand Rule: Direction of induced current/EMF is from West to East.
OR Part:
(a) $\displaystyle {\varepsilon = B l v}$.
(b) Current $\displaystyle {I = \varepsilon/R = B l v / R}$. Retarding force $\displaystyle {F = I l B = {\frac{{B^2 l^2 v}}{{R}}}}$.
(c) Mechanical Power $\displaystyle {P = F v = {\frac{{B^2 l^2 v^2}}{{R}}}}$.
Q.33. (a) Explain the construction and working of an AC generator.
(b) Derive expression for induced EMF.
(c) Mention two advantages of AC over DC for power transmission.
(a) Consists of armature coil, field magnet, slip rings, carbon brushes. Rotation changes flux linkage, generating sinusoidal AC.
(b) $\displaystyle {\varepsilon = N B A \omega \sin\omega t}$.
(c) 1. AC voltage can be easily stepped up/down using transformers, minimizing $\displaystyle {I^2 R}$ power losses during long-distance transmission.
2. AC generators are more robust and require lower maintenance than DC generators.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. A coil of area $\displaystyle {A_0}$ is placed in a magnetic field changing from $\displaystyle {B_0}$ to $\displaystyle {4B_0}$ in time interval $\displaystyle {t}$. Induced EMF is:
Explanation: $\displaystyle {|\varepsilon| = {\frac{{\Delta\Phi}}{{t}}} = {\frac{{(4B_0 - B_0) A_0}}{{t}}} = {\frac{{3 A_0 B_0}}{{t}}}}$.
2. A metallic ring is attached to a wall. When North pole of a bar magnet approaches the ring, induced current viewed from magnet side is:
Explanation: Approaching N-pole induces North polarity on front face to repel magnet, requiring anti-clockwise current.
3. An EMF of $\displaystyle {5\text{ V}}$ is produced in an inductor when current changes from $\displaystyle {3\text{ A}}$ to $\displaystyle {2\text{ A}}$ in $\displaystyle {1\text{ ms}}$. Self-inductance is:
Explanation: $\displaystyle {L = {\frac{{|\varepsilon|}}{{\Delta I / \Delta t}}} = {\frac{{5}}{{(3-2)/10^{-3}}}} = 5 \times 10^{-3}\text{ H} = 5\text{ mH}}$.
4. If current in a coil is halved, the magnetic energy stored in it becomes:
5. The flux linked with a coil varies as $\displaystyle {\Phi = 5 t^2 + 3 t - 2\text{ Wb}}$. Induced EMF at $\displaystyle {t = 2\text{ s}}$ is:
Explanation: $\displaystyle {\varepsilon = -{\frac{{d\Phi}}{{dt}}} = -(10 t + 3)}$. At $\displaystyle {t = 2\text{ s}}$, $\displaystyle {\varepsilon = -(10(2) + 3) = -23\text{ V}}$.
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): Self-inductance is referred to as the inertia of electricity.
Reason (R): Self-inductance produces opposing back EMF resisting any change of current in the circuit.
2. Assertion (A): Acceleration of a bar magnet falling vertically through a long copper solenoid is less than acceleration due to gravity ($\displaystyle {g}$).
Reason (R): Induced current in solenoid produces upward magnetic repulsion opposing downward fall (Lenz's law).
3. Assertion (A): Inductance coils (choke coils) are wound using thick copper wires.
Reason (R): Copper has low electrical resistance, minimizing Joule heat loss $\displaystyle {I^2 R}$.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study 1: Lenz's Law and Conservation of Energy
Lenz's law provides the direction of induced EMF such that induced current opposes the change in flux causing it.
Q1. When South pole of magnet recedes from a coil, the approaching face develops:
(a) North pole (b) South pole (c) Neutral
Q2. If a magnet falls through a cut/broken metal ring, its acceleration is:
(a) Equal to g (b) Less than g (c) Greater than g
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
Self EMF: $\displaystyle {\varepsilon = -L {\frac{{dI}}{{dt}}}}$ | Solenoid Self-Inductance: $\displaystyle {L = {\frac{{\mu_0 \mu_r N^2 A}}{{l}}}}$
Mutual EMF: $\displaystyle {\varepsilon_s = -M {\frac{{dI_p}}{{dt}}}}$ | Coaxial Solenoids: $\displaystyle {M = {\frac{{\mu_0 N_1 N_2 A}}{{l}}}}$
CONCEPTUAL SHORT QUESTIONS
1. A bulb is connected in series with a solenoid across a DC battery. How does bulb brightness change when a soft iron core is inserted into solenoid?
2. Two identical loops, one of copper and another of constantan, are removed from a magnetic field in equal time intervals. In which loop is larger current induced?
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