CBSE Class 12 Physics EMI And AC Worksheet

EMI And AC MCQ Questions with Answers Class 12 Physics

More question-

1. State Faraday’s laws of EMI.

Ans: e = - dɸ/dt

2. Write S.I. unit of magnetic flux. Is it a scalar or a vector?

Ans: Weber, Scalar.

3. On what factor does the self-inductance of a solenoid depends?

Ans: (a) No of turns in the solenoid (b) Length of the solenoid (c) Core inside the solenoid(d) Area of the cross-section.

4. Define self–inductance of a coil. Give its S.I. unit.

Ans: Magnetic Flux link in the coil when unit current flows through it. SI unit is henry.

5. Define self–induction of a coil. Give one example.

Ans: The phenomenon of electric current in a coil due to growth or detail of current in the coil.

Example: Back emf of motor. 

Easy and Scoring Areas

• Faraday’s law of EMI
• Lenz’s law
• Problems based on Lenz’s law
• Eddy currents
• Applications of eddy currents
• Self-induction
• Mutual Induction
• Expression for impedance of LCR series circuit
• Numericals based on impedance and resonance
• AC Generator
• Transformer

1 Mark Questions

Question 1. State Faraday’s laws of EMI.
Answer: According to Faraday's laws of electromagnetic induction, an electromotive force (emf) is induced in a circuit whenever there is a change in the magnetic flux linked with it. The magnitude of this induced emf is equal to the time rate of change of the magnetic flux through the circuit:
\( e = -\frac{d\Phi}{dt} \)
The negative sign indicates the direction of the induced emf, as described by Lenz's law.
In simple words: Whenever the magnetic field passing through a loop of wire changes, it generates electricity in the wire. The faster the magnetic field changes, the more electricity is produced.

Exam Tip: In exams, always state both qualitative and quantitative parts of the law, and write the mathematical formula with the negative sign, explaining what the variables represent.

Question 2. Write S.I. unit of magnetic flux. Is it a scalar or a vector?
Answer: The SI unit used to measure magnetic flux is the weber (Wb). It is classified as a scalar quantity.
In simple words: Magnetic flux is measured in a unit called the weber. It has magnitude but no direction, so it is a scalar.

Exam Tip: Remember that magnetic flux is a scalar quantity, even though it is calculated from two vector quantities: magnetic field and area vector.

Question 3. On what factor does the self-inductance of a solenoid depends?
Answer: The self-inductance of a solenoid is influenced by several parameters:
(a) The total number of turns in the winding of the solenoid.
(b) The physical length of the solenoid.
(c) The magnetic permeability of the core material placed inside the solenoid.
(d) The cross-sectional area of the solenoid.
In simple words: How much a coil resists changes in current depends on how many loops it has, its overall length, its thickness, and the material inside the coil.

Exam Tip: Use the formula \( L = \frac{\mu_0 \mu_r N^2 A}{l} \) to quickly recall and justify all four factors affecting the self-inductance of a solenoid.

Question 4. Define self-inductance of a coil. Give its S.I. unit.
Answer: The self-inductance of a coil is defined as the total magnetic flux linkage associated with the coil when a unit electric current flows through it. Mathematically, \( L = \frac{\Phi}{I} \). The SI unit of self-inductance is the henry (H).
In simple words: Self-inductance measures how much magnetic flux a coil creates when one ampere of current passes through it. It is measured in henries.

Exam Tip: Alternatively, you can define self-inductance as the induced emf set up in the coil when the rate of change of current through it is unity. Both definitions are accepted.

Question 5. Define self-induction of a coil. Give one example.
Answer: Self-induction is the phenomenon where an electromotive force (emf) is induced in a single isolated coil due to a change in the current flowing through that same coil. This induced emf opposes the growth or decay of the current.
Example: The generation of back emf in an electric motor when it is running.
In simple words: When the current in a coil changes, the coil creates its own voltage to oppose that change. This is called self-induction, and it is why electric motors produce a counter-voltage.

Exam Tip: Make sure to clearly distinguish between self-inductance (the property) and self-induction (the phenomenon).

Question 6. How does the self-inductance of an air coil change, when (i) the number of turns in the coil is solenoid increased (ii) an iron rod is introduced in the coil.
Answer: (i) The self-inductance \( L \) of a solenoid is directly proportional to the square of its number of turns (\( L \propto N^2 \)). Consequently, increasing the number of turns will result in a significant increase in its self-inductance.
(ii) Placing an iron rod inside the coil increases the self-inductance because iron has a very high magnetic permeability compared to air, which enhances the magnetic flux density inside the coil.
In simple words: (i) Adding more loops of wire to the coil increases its inductance by a lot. (ii) Inserting an iron rod inside the coil makes it much easier for magnetic fields to form, which increases the inductance.

Exam Tip: Always write down the formula \( L = \frac{\mu_0 \mu_r N^2 A}{l} \) to show that \( L \propto N^2 \) and \( L \propto \mu_r \) to gain full marks for justification.

Question 7. Bulb is connected in a closed circuit containing an air cored solenoid an a battery. How does the brightness of the bulb change, when (i) the number of turns in the coil is increased (ii)an iron rod is introduced in the coil.
Answer: (i) The brightness of the bulb will temporarily decrease during the transient state when the number of turns is increased. This is because a higher number of turns increases the self-induction, which opposes the growth of current and increases the time taken to reach steady-state. Once the steady-state is established (where \( \frac{di}{dt} = 0 \)), the brightness returns to its original level.
(ii) Similarly, introducing an iron rod increases the self-inductance, which only momentarily delays the growth of current. In the steady-state, the brightness remains the same as before.
In simple words: Because the circuit uses a battery (DC), changing the coil or adding an iron rod only affects the light right as the switch is flipped, making it take slightly longer to reach full brightness. Once the current settles, the bulb shines just as brightly as before.

Exam Tip: Pay attention to whether the source is AC or DC. For a DC source, the effects of self-induction only occur during the growth of current and do not alter the steady-state brightness of the bulb.

Question 8. The electric current is increasing in a straight wire from A to B .What is the direction of induced current in the metallic loop kept above the wire as shown in the fig.
Answer: According to the right-hand grip rule, the increasing current flowing from A to B creates a magnetic field that points out of the plane of the page in the region of the loop. As this magnetic flux increases, Lenz's law states that the induced current in the loop will oppose this change by producing a magnetic field pointing into the page. To achieve this, the induced current must flow in a clockwise direction.
In simple words: The current in the wire creates a magnetic field pointing out of the page through the loop. Since this field is getting stronger, the loop creates its own clockwise current to push a magnetic field back into the page and oppose the growth.

Exam Tip: Always determine the direction of the initial magnetic field using the right-hand rule first, then apply Lenz's law to find the direction of the induced current.

Question 9. A bar magnet falls from a height ‘h’ through a metal ring. Will its acceleration be equal to g ? Give reason for your answer.
Answer: The acceleration of the falling bar magnet will be less than the acceleration due to gravity (\( g \)). As the magnet falls toward the metal ring, the changing magnetic flux induces an electric current in the ring. According to Lenz's law, this induced current creates a magnetic field that opposes the motion of the falling magnet, resulting in an upward retarding force.
In simple words: As the magnet falls through the ring, it creates electricity in the ring. This electricity turns the ring into a temporary magnet that pushes up against the falling magnet, slowing it down so it falls slower than gravity.

Exam Tip: Remember that the opposition occurs both as the magnet approaches the ring (repulsion) and as it leaves the ring (attraction). In both cases, the acceleration remains less than \( g \).

Question 10. Why does a metallic piece become very hot when it is surrounded by a coil carrying high frequency alternating current?
Answer: The high-frequency alternating current in the coil produces a rapidly changing magnetic field. This changing magnetic field induces strong circulating currents, known as eddy currents, inside the metallic piece. Due to the electrical resistance of the metal, these eddy currents generate a substantial amount of heat (\( H = I^2 R t \)), making the metallic piece very hot.
In simple words: The rapidly changing magnetic field from the coil creates swirling electrical currents inside the metal. Because the metal resists this electricity, it generates a lot of heat, just like a toaster wire.

Exam Tip: Use the term 'eddy currents' and mention Joule heating (\( I^2 R t \)) to explain how electromagnetic energy is converted into heat energy.

Question 11. The instantaneous current in an ac circuit is i=2.0 sin314t, what is (i) frequency and (ii) rms value of the current.
Answer: The given equation for the instantaneous current is:
\( i = 2.0 \sin(314t) \)
Comparing this with the standard alternating current equation \( i = I_0 \sin(\omega t) \), we find:
Peak current, \( I_0 = 2.0\text{ A} \)
Angular frequency, \( \omega = 314\text{ rad/s} \)

(i) Since \( \omega = 2\pi f \), the frequency \( f \) is:
\( f = \frac{\omega}{2\pi} = \frac{314}{2 \times 3.14} = 50\text{ Hz} \)

(ii) The root-mean-square (rms) value of the current is:
\( I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\text{ A} \approx 1.414\text{ A} \)
In simple words: By comparing the equation to the standard AC formula, we find that the electricity cycles back and forth 50 times per second, and the effective (rms) current is about 1.41 amperes.

Exam Tip: Always state the standard equation \( i = I_0 \sin(\omega t) \) and compare terms step-by-step to avoid calculation errors with \( \omega = 2\pi f \).

Question. What is phase difference between voltage and current in LCR series circuit if power factor is 0.707?
Answer: The power factor of an AC circuit is given by \( \cos\phi \), where \( \phi \) is the phase difference between the voltage and the current.
Given, power factor \( = 0.707 \):
\( \cos\phi = 0.707 = \frac{1}{\sqrt{2}} \)

\( \implies \phi = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \) (or \( \frac{\pi}{4}\text{ radians} \))
In simple words: The power factor is the cosine of the phase angle. Since the cosine of 45 degrees is 0.707, the voltage and current are out of step by 45 degrees.

Exam Tip: Remember that a power factor of 0.707 means the circuit has equal resistive and reactive components, resulting in a phase angle of exactly \( 45^\circ \).

Question 12. What is phase difference between voltage and current in LCR series circuit at resonance?
Answer: At resonance, the inductive reactance \( X_L \) and capacitive reactance \( X_C \) cancel each other out (\( X_L = X_C \)), making the circuit purely resistive. Consequently, the phase difference \( \phi \) between the voltage and the current is zero degrees (they are in phase).
In simple words: At resonance, the coil and capacitor cancel each other's resistance out, leaving only plain resistance. This means the voltage and current stay perfectly in sync, with zero delay.

Exam Tip: At resonance, the power factor \( \cos\phi \) is maximum and equal to 1, since \( \phi = 0^\circ \).

Question 13. Sketch a graph showing variation of reactance of a capacitor with frequency of the applied voltage.
Answer: The capacitive reactance is inversely proportional to the frequency of the alternating voltage:
\( X_C = \frac{1}{2\pi f C} \implies X_C \propto \frac{1}{f} \)
Therefore, the graph of capacitive reactance against frequency is a rectangular hyperbola, as shown below:
In simple words: As the frequency of the electricity increases, the capacitor's opposition to the current (reactance) drops rapidly. The graph forms a curved line that starts high and slopes downwards.

Exam Tip: Clearly label both axes, with frequency \( f \) on the x-axis and capacitive reactance \( X_C \) on the y-axis, and state the relation \( X_C \propto \frac{1}{f} \).

Question 14. Sketch a graph showing variation of reactance of an inductor with frequency of the applied voltage.
Answer: The inductive reactance is directly proportional to the frequency of the alternating voltage:
\( X_L = 2\pi f L \implies X_L \propto f \)
As a result, the graph of inductive reactance versus frequency is a straight line passing through the origin, as shown below:
In simple words: As the frequency of the electricity increases, the inductor's opposition to the current (reactance) increases at a constant rate. This creates a straight, upward-sloping line.

Exam Tip: Be sure to show the line passing directly through the origin \( (0,0) \) since \( X_L = 0 \) when \( f = 0 \).

Question 15. Why a capacitor blocks dc but allow ac?
Answer: The capacitive reactance is given by the expression \( X_C = \frac{1}{2\pi f C} \). Since direct current (DC) has a frequency of zero (\( f = 0 \)), the capacitive reactance becomes infinite (\( X_C \to \infty \)), meaning the capacitor completely blocks DC. On the other hand, alternating current (AC) has a non-zero frequency, resulting in a finite capacitive reactance that allows AC to pass through.
In simple words: Direct current (DC) doesn't cycle back and forth, so its frequency is zero. This causes the capacitor to offer endless resistance and block it completely. AC cycles constantly, so it easily gets through with much lower resistance.

Exam Tip: In your answer, write down the formula \( X_C = \frac{1}{2\pi f C} \), substitute \( f = 0 \) for DC to show \( X_C = \infty \), and mention that for AC, \( f \neq 0 \) leads to a finite \( X_C \).

Question 16. The frequency of ac is doubled, what happens to (i) inductive reactance (ii) capacitive reactance?
Answer: When the frequency of the AC source is doubled (\( f' = 2f \)):
(i) Inductive Reactance: Since \( X_L = 2\pi f L \), we have \( X_L \propto f \). Therefore, doubling the frequency will cause the inductive reactance to double.
(ii) Capacitive Reactance: Since \( X_C = \frac{1}{2\pi f C} \), we have \( X_C \propto \frac{1}{f} \). Therefore, doubling the frequency will cause the capacitive reactance to be halved.
In simple words: Doubling the speed of the alternating current makes the inductor oppose the current twice as much (reactance doubles), but it cuts the capacitor's opposition in half.

Exam Tip: State the proportionality relations \( X_L \propto f \) and \( X_C \propto \frac{1}{f} \) clearly to justify your answer.

Question 17. Draw the electric field, magnetic field and direction of propagation of an Electromagnetic Wave.
Answer: An electromagnetic wave consists of sinusoidal electric field \( \mathbf{E} \) and magnetic field \( \mathbf{B} \) vectors that oscillate perpendicular to each other and also perpendicular to the direction of wave propagation. In the diagram, the electric field oscillates along the x-axis, the magnetic field oscillates along the y-axis, and the wave propagates along the z-axis.
In simple words: An electromagnetic wave is made of electric and magnetic fields vibrating at right angles to each other, like a pair of waves traveling together along a straight line.

Exam Tip: Make sure to clearly label the mutually perpendicular axes (x, y, z) for the electric field, magnetic field, and wave propagation to score full marks.

Question 18. What is Displacement Current?
Answer: Displacement current is that current which arises due to the changing electric flux with time. Mathematically, it is given by:
\( I_d = \epsilon_0 \frac{d\Phi_e}{dt} \)
where \( \epsilon_0 \) is the permittivity of free space and \( \frac{d\Phi_e}{dt} \) is the rate of change of electric flux.
In simple words: Displacement current is a virtual current that acts like real current when electric fields change over time.

Exam Tip: Do not forget to include the constant \( \epsilon_0 \) in the formula for displacement current, as it represents the electrical permittivity of vacuum.

Question 19. Why a microwave oven heats up a food item containing water molecules most efficiently.
Answer: A microwave oven heats food efficiently because the frequency of the microwaves generated closely matches the natural resonant frequency of rotation of water molecules. This causes the water molecules in the food to absorb the microwave energy via resonance, leading to rapid molecular vibration and heat generation.
In simple words: Microwaves vibrate at the exact speed that water molecules naturally like to spin. This makes the water in the food shake rapidly, generating quick heat through friction.

Exam Tip: Use the keyword 'resonant frequency' and mention that energy transfer is maximum at resonance.

Question 20. Which part of the electromagnetic spectrum has the largest penetrating power?
Answer: The part of the electromagnetic spectrum with the highest penetrating power is gamma rays. This is because they have the shortest wavelengths and the highest frequencies (and thus the highest photon energy).
In simple words: Gamma rays are the most penetrating because they carry the highest energy of any electromagnetic waves.

Exam Tip: Identify 'gamma rays' clearly and explain that high frequency corresponds to high energy, which enables deep penetration.

Question 21. Identify the part of the electromagnetic spectrum to which the following wavelength belong:
(a) 10-1 m (b) 10-12 m
Answer: (a) A wavelength of \( 10^{-1}\text{ m} \) (which is \( 10\text{ cm} \)) belongs to the short radio waves (or microwaves) region of the electromagnetic spectrum.
(b) A wavelength of \( 10^{-12}\text{ m} \) lies in the gamma rays region of the electromagnetic spectrum.
In simple words: A wave of 10 centimeters belongs to short radio waves, while a super-tiny wave of \( 10^{-12} \) meters is a gamma ray.

Exam Tip: Memorize the approximate wavelength boundaries of each part of the EM spectrum to easily identify these values in exams.

Question 22. Name the part of electromagnetic spectrum of wavelength 102 m and mention its one application.
Answer: An electromagnetic wave with a wavelength of \( 10^2\text{ m} \) belongs to the radio waves region.
Application: They are widely used for long-distance radio broadcasting and communication.
In simple words: Waves that are 100 meters long are radio waves. They are used to send radio music and signals over long distances.

Exam Tip: State the name of the EM region first, then write a clear, direct application in a separate sentence.

Question 23. The following table gives the wavelength range of some constituents of the electromagnetic spectrum.

select the wavelength range and name the electromagnetic waves that are
(a) widely used in the remote switches of household electronic devices.
(b) produced in nuclear reactions.
Answer: (a) The electromagnetic waves used in remote switches of home appliances are Infrared waves, corresponding to the wavelength range of \( 1\text{ mm} \) to \( 700\text{ nm} \) (S.No. 1).
(b) The electromagnetic waves produced in nuclear reactions are Gamma rays, corresponding to the wavelength range of less than \( 10^{-3}\text{ nm} \) (S.No. 4).
In simple words: (a) Remote controls use infrared waves, which have wavelengths between 1 millimeter and 700 nanometers. (b) Nuclear reactions emit gamma rays, which have extremely short wavelengths of less than \( 10^{-3} \) nanometers.

Exam Tip: When a table is given, always refer to the specific row number or values directly from the table to validate your selection.

Question 24. Name the characteristics of electromagnetic wave that
(a) Increases
(b) decreases
(c) remains constant
Answer: When considering the propagation of an electromagnetic wave under certain media transitions:
(a) The characteristic that increases (or remains independent/depends on source) - the frequency increases in certain specific transitions.
(b) The characteristic that decreases is the wavelength.
(c) The speed of the wave in vacuum remains constant.
In simple words: Under these conditions, the frequency increases, the wavelength decreases, and the speed in a vacuum stays completely constant.

Exam Tip: In standard vacuum propagation, speed remains \( c \). When entering a denser medium, the frequency remains constant while speed and wavelength decrease.

Question 25. From the following identify the electromagnetic waves having the
(i) Maximum (ii) minimum frequency
(a) Radio waves (b) Gamma-rays (c) Visible light
(d) Microwaves (e) Ultraviolet- rays (f) Infrared rays
Answer: (i) The electromagnetic waves with the maximum frequency among the given options are Gamma-rays.
(ii) The electromagnetic waves with the minimum frequency among the given options are Radio waves.
In simple words: Of the choices, gamma rays vibrate the fastest (highest frequency), while radio waves vibrate the slowest (lowest frequency).

Exam Tip: Recall the order of the EM spectrum: Radio, Micro, Infrared, Visible, UV, X-ray, Gamma. Frequency increases as you move from left to right.

Question 26. Identify the following electromagnetic radiations as per the wavelength given below.
(a) 10-3 nm (b) 10-3 m (c) 1 nm
Write one application of each
Answer: (a) \( 10^{-3}\text{ nm} \): This wavelength corresponds to Gamma radiation.
*Application:* Used in radiotherapy for cancer treatment or to initiate nuclear reactions.

(b) \( 10^{-3}\text{ m} \): This wavelength corresponds to Microwaves.
*Application:* Extensively used in RADAR systems for aircraft navigation and speed detection.

(c) \( 1\text{ nm} \): This wavelength corresponds to X-rays.
*Application:* Widely used in medical imaging to detect bone fractures or stones in internal organs.
In simple words: Gamma rays (at \( 10^{-3} \) nm) are used to treat cancer. Microwaves (at \( 10^{-3} \) m) are used in aircraft radar. X-rays (at 1 nm) are used to scan for broken bones.

Exam Tip: When writing applications, use specific technical terms like 'radiotherapy', 'radar', and 'medical diagnostic imaging' to secure maximum marks.

Question 27. Identify the following electromagnetic radiations as per the frequencies given below:
(a) 1020 Hz (b) 109 Hz (c) 1011 Hz
Write one application of each.
Answer: (a) \( 10^{20}\text{ Hz} \): This represents Gamma radiation.
*Application:* It is utilized in the medical treatment of cancer (radiotherapy).

(b) \( 10^9\text{ Hz} \): This represents Radio waves (high frequency range).
*Application:* Used for long-distance radio and television program broadcasting.

(c) \( 10^{11}\text{ Hz} \): This represents Microwaves.
*Application:* Used in microwave ovens for cooking and heating food.
In simple words: At high frequencies like \( 10^{20} \) Hz, gamma rays are used to destroy cancer cells. At \( 10^9 \) Hz, radio waves broadcast signals. At \( 10^{11} \) Hz, microwaves cook food.

Exam Tip: Be ready to identify waves by frequency as well as wavelength. Remember that higher frequency means higher energy.

2 Marks and 3 Marks Questions

Question 1. What are eddy currents? Give their one use.
Answer: When the magnetic flux linking a bulk metallic conductor changes over time, circulating closed-loop currents are induced within the body of the conductor. These swirling currents are called eddy currents.
**Use:** They are utilized in the electromagnetic braking systems of high-speed trains.
In simple words: When magnetic fields change inside a solid block of metal, they create swirling loops of electricity called eddy currents. These currents are used to smoothly slow down trains using magnetic brakes.

Exam Tip: In your definition, use key phrases like 'changing magnetic flux' and 'bulk/thick conductor' to ensure full marks.

Question 2. State Lenz’s law. Show that it is in accordance with the law of conservation of energy.
Answer: Lenz's law states that the direction of the induced electromotive force (emf) or current in a circuit is always such that it opposes the change in magnetic flux that produced it.

**Lenz's Law and Conservation of Energy:**
Consider a bar magnet being pushed towards a closed metallic loop with its North (N) pole facing the loop. The changing magnetic flux induces a current in the loop. According to Lenz's law, the face of the loop facing the magnet develops a North pole to oppose the approach of the magnet. To overcome this magnetic repulsion, mechanical work must be done to push the magnet closer. This mechanical energy expended in moving the magnet is converted directly into electrical energy, generating the induced current. If Lenz's law were not true, the magnet would be attracted automatically, creating energy out of nothing, which violates the law of conservation of energy.
In simple words: Lenz's law says that induced current always fights against whatever change created it. When you push a magnet toward a loop, the loop pushes back, meaning you have to do physical work. That physical effort of yours is what gets transformed into the electrical energy.

Exam Tip: Always describe the physical process of pushing a magnet towards a coil to clearly explain how mechanical work converts into electrical energy.

Question 3. Two identical loops, one of copper and the other of aluminum, are rotated with the same angular speed in the same magnetic field. Compare (i) the induced emf and (ii) the current produced in the two coils. Justify your answer.
Answer: (i) **Induced EMF:** The induced electromotive force (emf) is given by \( e = -\frac{d\Phi}{dt} \). Since both loops are identical in geometry and are rotated at the same angular speed in the same magnetic field, the rate of change of magnetic flux is identical for both. Therefore, the induced emf in both loops is equal.
(ii) **Induced Current:** The induced current is given by \( I = \frac{e}{R} \). The electrical resistance of a loop is \( R = \rho \frac{l}{a} \), where \( \rho \) is the resistivity. Since copper has a much lower resistivity than aluminum (\( \rho_{Cu} < \rho_{Al} \)), the copper loop has lower resistance. Consequently, the induced current in the copper loop is greater than that in the aluminum loop.
In simple words: Both loops get the exact same voltage because they are spinning at the same speed in the same magnetic field. However, because copper is a better conductor (has lower resistance) than aluminum, more current flows through the copper loop.

Exam Tip: Clearly state that emf depends only on the rate of change of flux (geometry and speed), while current depends additionally on the material's resistance.

Question 4. Prove that average power consumed over a complete cycle of ac through an ideal inductor is zero.
Answer: For an ideal inductor, the alternating voltage leads the alternating current by a phase angle of \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).
Let the instantaneous current be:
\( I = I_0 \sin(\omega t) \)
Then, the instantaneous voltage is:
\( V = V_0 \sin\left(\omega t + \frac{\pi}{2}\right) = V_0 \cos(\omega t) \)
The average power \( P_{avg} \) consumed over a complete cycle of time period \( T \) is:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} V I \, dt \]
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} [V_0 \cos(\omega t)] [I_0 \sin(\omega t)] \, dt \]
\[ P_{avg} = \frac{V_0 I_0}{2T} \int_{0}^{T} [2 \sin(\omega t) \cos(\omega t)] \, dt \]
\[ P_{avg} = \frac{V_0 I_0}{2T} \int_{0}^{T} \sin(2\omega t) \, dt \]
Since the integration of a sine function over a complete cycle is zero:
\[ \int_{0}^{T} \sin(2\omega t) \, dt = \left[ -\frac{\cos(2\omega t)}{2\omega} \right]_{0}^{T} = 0 \]
Therefore, the average power consumed over a complete cycle is:
\( P_{avg} = 0 \)
In simple words: Because the current and voltage in an ideal inductor are out of step by exactly 90 degrees, the energy stored during one half of the cycle is returned to the source in the next half. The net power used is zero.

Exam Tip: Clearly state the phase relationship between voltage and current in an ideal inductor at the beginning of the derivation to score full marks.

Question 5. Prove that average power consumed over a complete cycle of ac through an ideal capacitor is zero.
Answer: For an ideal capacitor, the alternating current leads the alternating voltage by a phase angle of \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).
Let the instantaneous voltage be:
\( V = V_0 \sin(\omega t) \)
Then, the instantaneous current is:
\( I = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) = I_0 \cos(\omega t) \)
The average power \( P_{avg} \) consumed over a complete cycle of time period \( T \) is:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} V I \, dt \]
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} [V_0 \sin(\omega t)] [I_0 \cos(\omega t)] \, dt \]
\[ P_{avg} = \frac{V_0 I_0}{2T} \int_{0}^{T} [2 \sin(\omega t) \cos(\omega t)] \, dt \]
\[ P_{avg} = \frac{V_0 I_0}{2T} \int_{0}^{T} \sin(2\omega t) \, dt \]
Since the integration of the sine function over a full cycle is zero:
\[ \int_{0}^{T} \sin(2\omega t) \, dt = 0 \]
Therefore, the average power consumed over a complete cycle by an ideal capacitor is:
\( P_{avg} = 0 \)
In simple words: Just like the inductor, the capacitor and voltage are 90 degrees out of step. It stores charge and then gives it all back, so it uses no net power over a full cycle.

Exam Tip: Write out the complete derivation even if the question is similar to the inductor. Do not write 'same as above' in board exams.

Question 6. A copper ring is suspended by a thread in a vertical plane. North Pole of a magnet is brought horizontally towards the ring. Will the magnet affect the position of the ring? Explain.
Answer: Yes, the motion of the magnet will affect the position of the copper ring. As the North pole of the magnet is brought closer to the ring, the magnetic flux through the ring changes, inducing an electric current in it. According to Lenz's law, this induced current flows in a direction that opposes the incoming North pole, creating a temporary North pole on the facing side of the ring. This results in a repulsive force that pushes the suspended copper ring away from the magnet.
In simple words: Yes, as you push the magnet closer, it induces electricity in the copper ring. This electricity creates a magnetic field that repels the magnet, pushing the ring away so it swings slightly.

Exam Tip: Identify that the ring experiences a repulsive force due to Lenz's law and will swing away from the approaching magnet.

Question 6. Derive an expression for average power consumed over a complete cycle of ac through an LCR circuit.
Answer: Let the alternating electromotive force (emf) applied to the LCR circuit be:
\( V = V_0 \sin(\omega t) \)
Let the resulting alternating current in the circuit lag behind the voltage by a phase angle \( \phi \):
\( I = I_0 \sin(\omega t - \phi) \)
The instantaneous power \( P \) delivered by the source is:
\( P = V \times I = V_0 I_0 \sin(\omega t) \sin(\omega t - \phi) \)
Using the trigonometric identity \( \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \):
\( P = \frac{V_0 I_0}{2} [\cos(\omega t - (\omega t - \phi)) - \cos(\omega t + \omega t - \phi)] \)
\( P = \frac{V_0 I_0}{2} [\cos\phi - \cos(2\omega t - \phi)] \)
The average power \( P_{avg} \) consumed over a complete cycle (time period \( T \)) is the time-average of this instantaneous power:
\[ P_{avg} = \frac{1}{T} \int_{0}^{T} P \, dt = \frac{V_0 I_0}{2T} \int_{0}^{T} [\cos\phi - \cos(2\omega t - \phi)] \, dt \]
Since \( \cos\phi \) is a constant, and the average value of the time-varying cosine term \( \cos(2\omega t - \phi) \) over a complete cycle is zero:
\[ P_{avg} = \frac{V_0 I_0}{2} \cos\phi \]
We can rewrite this expression as:
\( P_{avg} = \left(\frac{V_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos\phi \)

\( \implies P_{avg} = V_{rms} I_{rms} \cos\phi \)
Here, the term \( \cos\phi \) is known as the power factor of the AC circuit.
In simple words: By multiplying the voltage and current equations and averaging them over a full cycle, we find that the average power is the RMS voltage times the RMS current, multiplied by a factor called the power factor (cosine of the phase angle).

Exam Tip: Always define \( V_{rms} \), \( I_{rms} \), and the power factor \( \cos\phi \) at the end of the derivation to ensure full marks.

Question 7. What is the effect on the mutual inductance between the pair of coil when (i) the distance between the coils is increased? (ii) the number of turns in each coil is decreased? Justify your answer in each case.
Answer: (i) Increasing the distance: The mutual inductance between the coils decreases. This is because as the distance increases, the magnetic flux leakage increases, and fewer magnetic field lines from the primary coil link with the secondary coil (the coupling coefficient \( k \) decreases).
(ii) Decreasing the number of turns: The mutual inductance decreases. The mutual inductance is directly proportional to the product of the number of turns in both coils (\( M \propto N_1 N_2 \)). Therefore, reducing the number of turns in either or both coils decreases \( M \).
In simple words: (i) Pulling the coils further apart weakens their magnetic link, so the mutual inductance drops. (ii) Having fewer loops of wire reduces the strength of the magnetic field they share, which also lowers the mutual inductance.

Exam Tip: State the mathematical formula \( M = \frac{\mu_0 N_1 N_2 A}{l} \) to show that \( M \propto N_1 N_2 \) and explain that flux linkage decreases with distance.

Question 8. Discuss a series resonant circuit. Derive an expression for resonant frequency and show a graphical variation between current and angular frequency of applied ac. Define quality factor and derive an expression for it.
Answer: A series LCR circuit is in resonance when the inductive reactance and capacitive reactance are equal, which minimizes the total impedance and allows maximum current to flow through the circuit.

Derivation of Resonant Frequency:
At resonance, the inductive reactance \( X_L \) is equal to the capacitive reactance \( X_C \):
\( X_L = X_C \)
\( \omega_0 L = \frac{1}{\omega_0 C} \)
\( \implies \omega_0^2 = \frac{1}{LC} \)
\( \implies \omega_0 = \frac{1}{\sqrt{LC}} \)
Since \( \omega_0 = 2\pi f_0 \), the linear resonant frequency \( f_0 \) is:
\( f_0 = \frac{1}{2\pi\sqrt{LC}} \)
The graph below shows how the current \( I \) varies with the angular frequency \( \omega \), peaking at \( \omega_0 \):
Quality Factor (Q-factor):
The Quality Factor (Q-factor) is a measure of the sharpness of resonance. It is defined as the ratio of the resonant angular frequency to the bandwidth of the circuit:
\( Q = \frac{\omega_0}{2\Delta\omega} \)
Alternatively, it is the ratio of the voltage across the inductor (or capacitor) to the voltage across the resistor at resonance:
\( Q = \frac{V_L}{V_R} = \frac{I \omega_0 L}{I R} = \frac{\omega_0 L}{R} \)
Substituting \( \omega_0 = \frac{1}{\sqrt{LC}} \):
\( Q = \frac{1}{R}\sqrt{\frac{L}{C}} \)
In simple words: Resonance happens when a circuit's parts let the maximum amount of current flow at one particular frequency. The Quality factor tells us how sharp and focused this peak is, and it is calculated using the coil, capacitor, and resistor values.

Exam Tip: In exams, always draw the resonance curve showing the peak current \( I_{max} \) at \( \omega_0 \), and state both definitions of the Quality Factor.

5 Marks Questions

Question 9. Explain with help off a labelled diagram the principal, construction and working of a transformer. Derive a relation between various energy losses in a transformer? Explain the role of transformer in long distance transmission of power?
Answer: **Principle:**
A transformer is a static electromagnetic device that works on the principle of mutual induction. When an alternating current flows through the primary coil, it creates a continuously changing magnetic flux in the iron core, which in turn induces an alternating emf in the secondary coil.

**Construction:**
It consists of a laminated soft iron core to minimize energy losses. Two separate coils of insulated copper wire—the primary coil (having \( N_p \) turns) and the secondary coil (having \( N_s \) turns)—are wound on the core.
**Working and Turns Ratio:**
Let an alternating voltage \( V_p \) be applied to the primary coil. According to Faraday's law of electromagnetic induction, the induced electromotive forces in the primary and secondary windings are:
\( V_p = -N_p \frac{d\Phi}{dt} \)
\( V_s = -N_s \frac{d\Phi}{dt} \)
Dividing these two equations gives the transformer equation:
\( \frac{V_s}{V_p} = \frac{N_s}{N_p} \)
Where \( \frac{N_s}{N_p} \) is called the turns ratio.
For an ideal transformer, input power equals output power:
\( V_p I_p = V_s I_s \implies \frac{V_s}{V_p} = \frac{I_p}{I_s} = \frac{N_s}{N_p} \)

**Energy Losses in a Transformer:**
(i) Copper Loss: Energy is lost as heat (\( I^2 R \)) in the copper windings due to their electrical resistance. This is minimized by using thick wires.
(ii) Iron Loss (Eddy Current Loss): Swirling eddy currents are induced in the iron core by the changing magnetic field, causing heat loss. This is minimized by using a laminated iron core.
(iii) Flux Leakage: Not all magnetic flux created by the primary coil passes through the secondary coil. This is minimized by winding one coil directly over the other.
(iv) Hysteresis Loss: Energy is dissipated as heat during the continuous magnetization and demagnetization cycles of the core. This is minimized by using soft iron, which has low coercivity.

**Role in Long-Distance Power Transmission:**
Power is transmitted over long distances at very high voltages and low currents. A step-up transformer at the power station raises the voltage, which simultaneously lowers the current. Since power loss in transmission cables is \( I^2 R \), reducing the current significantly decreases energy loss. At the destination, step-down transformers lower the voltage back to safe, usable levels.
In simple words: A transformer uses magnetic induction to step voltage up or down. To transmit power over long distances without losing energy, we step the voltage up extremely high (which lowers the current and reduces heat loss in the wires) and then step it back down at our homes.

Exam Tip: In 5-mark questions, draw a neat labeled diagram of the core and windings. Separately list and explain the four main types of energy losses and how they are minimized.

Question 10. Deduce an expression for the self-inductance of a long solenoid of N turns, having air in its core.
Answer: Let us consider a long solenoid of length \( l \), cross-sectional area \( A \), and total number of turns \( N \).
The number of turns per unit length is:
\( n = \frac{N}{l} \)
When a current \( I \) flows through this solenoid, the uniform magnetic field \( B \) produced along its central axis inside is:
\( B = \mu_0 n I = \frac{\mu_0 N I}{l} \)
The magnetic flux \( \Phi \) linked with a single turn of the solenoid is:
\( \Phi = B \times A = \frac{\mu_0 N I A}{l} \)
The total magnetic flux linkage \( \Phi_{total} \) associated with all \( N \) turns of the solenoid is:
\( \Phi_{total} = N \Phi = N \left(\frac{\mu_0 N I A}{l}\right) \)

\( \implies \Phi_{total} = \frac{\mu_0 N^2 I A}{l} \)
By the definition of self-inductance \( L \):
\( \Phi_{total} = L I \)
Comparing the two expressions:
\( L I = \frac{\mu_0 N^2 I A}{l} \)

\( \implies L = \frac{\mu_0 N^2 A}{l} \)
If a magnetic core of relative permeability \( \mu_r \) is inserted inside the solenoid, the self-inductance becomes:
\( L = \frac{\mu_0 \mu_r N^2 A}{l} \)
In simple words: The magnetic field inside a coil depends on the current and the number of loops. By calculating the total magnetic flux linked with all loops and using the definition of self-inductance, we find that the inductance depends on the square of the number of turns, the area, and the length of the coil.

Exam Tip: Always define each term (\( N \), \( l \), \( A \), \( I \)) clearly before starting the mathematical derivation, as examiners check these for step marks.

Question 11. Define mutual inductance and give its SI unit. Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound over the other.
Answer: **Mutual Inductance:** Mutual inductance is the property of a pair of coils by virtue of which an electromotive force (emf) is induced in one coil (secondary) when the current flowing through the neighboring coil (primary) changes over time.
Mathematically, \( \Phi_2 = M I_1 \), where \( M \) is the mutual inductance.
**SI Unit:** The SI unit of mutual inductance is the henry (H).

**Derivation of Mutual Inductance of Two Coaxial Solenoids:**
Consider two long coaxial solenoids \( S_1 \) (inner) and \( S_2 \) (outer) of the same length \( l \).
Let:
\( N_1, N_2 = \) total number of turns in \( S_1 \) and \( S_2 \) respectively.
\( r_1, r_2 = \) radii of \( S_1 \) and \( S_2 \) (where \( r_1 < r_2 \)).
\( A = \pi r_1^2 = \) cross-sectional area of the inner solenoid \( S_1 \).

Let a time-varying current \( I_2 \) flow through the outer solenoid \( S_2 \). The magnetic field \( B_2 \) produced inside \( S_2 \) is:
\( B_2 = \mu_0 n_2 I_2 = \mu_0 \left(\frac{N_2}{l}\right) I_2 \)
Since the inner solenoid \( S_1 \) is completely inside the magnetic field of \( S_2 \), the magnetic flux \( \Phi_1 \) linked with each turn of \( S_1 \) is:
\( \Phi_1 = B_2 \times A = \left(\frac{\mu_0 N_2 I_2}{l}\right) A \)
The total flux linkage with the entire inner solenoid \( S_1 \) having \( N_1 \) turns is:
\( \Phi_{total1} = N_1 \Phi_1 = \frac{\mu_0 N_1 N_2 A I_2}{l} \)
By the definition of mutual inductance \( M_{12} \):
\( \Phi_{total1} = M_{12} I_2 \)
Equating the two:
\( M_{12} I_2 = \frac{\mu_0 N_1 N_2 A I_2}{l} \)

\( \implies M_{12} = \frac{\mu_0 N_1 N_2 A}{l} \)

By symmetry (reciprocity theorem), the mutual inductance \( M_{21} \) of solenoid 2 with respect to solenoid 1 is equal:
\( M_{21} = M_{12} = M = \frac{\mu_0 N_1 N_2 A}{l} \)
In simple words: Mutual inductance is the ability of one coil to induce a voltage in a nearby coil. For two nested coils, the shared magnetic field creates a link between them. The resulting formula shows that the mutual inductance depends on the product of their turns, the inner coil's area, and the total length.

Exam Tip: Remember that the area \( A \) used in the formula is always the cross-sectional area of the *inner* solenoid (\( A = \pi r_1^2 \)), because the magnetic field of the inner solenoid does not exist in the space between the inner and outer solenoid walls.

Question 12. Describe briefly the basic elements of an ac generator. Write expression for the emf produced. Also show graphically the emf produced by ac generator. An a.c. generator consist a coil of 50 turns and area 2.5 m2 rotating at an angular speed of 60rad/s in a uniform magnetic field B = 0.30T between two fixed pole pieces . The resistance of the circuit including that coil the coil is 500W. (i) Find the maximum current drawn from the generator. (ii) What will be the orientation of the coil with respect to the magnetic field to have (a) maximum (b) zero magnetic flux?
Answer: **Basic Elements of an AC Generator:**
(i) **Field Magnet:** A strong permanent magnet or electromagnet that produces a uniform magnetic field between its North and South poles.
(ii) **Armature (Coil):** A rectangular coil consisting of a large number of turns of insulated copper wire wound over a soft iron core, which rotates in the magnetic field.
(iii) **Slip Rings:** Two hollow metallic rings connected to the ends of the armature coil, which rotate along with the coil.
(iv) **Brushes:** Two stationary carbon brushes that remain in light contact with the rotating slip rings to draw the current into the external circuit.
(v) **Load:** The external circuit connected to the brushes to utilize the generated electricity.

**Expression for EMF:**
The instantaneous induced electromotive force (emf) is given by:
\( e = e_0 \sin(\omega t) \)
where \( e_0 = N B A \omega \) is the peak value of the induced emf.

**EMF Graph:**
**Numerical Calculation:**
Given:
Number of turns, \( N = 50 \)
Area of coil, \( A = 2.5\text{ m}^2 \)
Angular speed, \( \omega = 60\text{ rad/s} \)
Magnetic field, \( B = 0.30\text{ T} \)
Resistance, \( R = 500\text{ }\Omega \)

(i) The maximum (peak) current \( I_{max} \) is given by:
\( I_{max} = \frac{e_0}{R} = \frac{N B A \omega}{R} \)
\( I_{max} = \frac{50 \times 0.30 \times 2.5 \times 60}{500} = \frac{2250}{500} = 4.5\text{ A} \)

(ii) **Orientation of Coil for Flux:**
The magnetic flux is \( \Phi = B A \cos\theta \), where \( \theta \) is the angle between the area vector (normal to the coil plane) and the magnetic field.
(a) For maximum magnetic flux, \( \theta = 0^\circ \) (i.e., the plane of the coil is perpendicular to the magnetic field).
(b) For zero magnetic flux, \( \theta = 90^\circ \) (i.e., the plane of the coil is parallel to the magnetic field).
In simple words: An AC generator turns motion into electricity using magnets, a rotating coil, slip rings, and carbon brushes. The maximum current it produces in this setup is 4.5 amperes. To get the most magnetic flux, the coil must face the magnet flat-on, while turning it sideways results in zero flux.

Exam Tip: Remember that maximum flux through the coil occurs when the induced emf is zero (and vice versa), because emf depends on the rate of change of flux (\( \sin\theta \)), whereas flux depends on the absolute value (\( \cos\theta \)).