RBSE Solutions Class 12 Physics Chapter 9 Electromagnetic Induction

Rbse Class 12 Physics Chapter 9 Multiple Choice Type Questions

Question 1. If a conducting rod moves with a constant velocity v in a magnetic field, emf is induced between both its ends if:
(a) v and B are parallel
(b) v is perpendicular to \( \vec{B} \)
(c) v and B are in opposite direction
(d) All of the options
Answer: (b) v is perpendicular to \( \vec{B} \)
In simple words: For an electromotive force (emf) to be induced, the velocity of the rod must be at a 90-degree angle to the magnetic field. This ensures the maximum change in magnetic flux.

🎯 Exam Tip: Remember that induced emf is maximized when the motion of the conductor, its length, and the magnetic field are all mutually perpendicular.

Question 2. What is the magnetic flux (\( \Phi \)) for a coil if N = 20 turns, A = x (area), and the angular frequency is \( \omega \)?
(a) 20 Bx
(b) 10 Bx\(^2\)
(c) 20 Bx\(^2\) cos\( \omega \)t
(d) 40 Bx\(^2\)
Answer: (c) 20 Bx\(^2\) cos\( \omega \)t
In simple words: The magnetic flux is calculated using the formula \( \Phi = NBA \cos(\omega t) \). Given N = 20 and A = x\(^2\) (implied from the options as area squared for `Bx^2`), the flux becomes \( 20 B x^2 \cos(\omega t) \).

🎯 Exam Tip: Always remember the formula for magnetic flux through a coil, \( \Phi = NBA \cos\theta \), where \( \theta \) is the angle between the magnetic field and the normal to the coil's area. If \( \theta \) varies with time as \( \omega t \), the formula becomes \( NBA \cos(\omega t) \).

Question 3. The unit of the ratio of magnetic flux to resistance is equal to the unit of which quantity:
(a) Charge
(b) Potential difference
(c) Current
(d) Magnetic field
Answer: (a) Charge
In simple words: The ratio of magnetic flux (\( \Phi \)) to resistance (R) has units that are equivalent to the unit of electric charge (q). This is because induced charge \( q = \frac{\Delta\Phi}{R} \).

🎯 Exam Tip: Understanding dimensional analysis and the relationship between physical quantities helps quickly identify equivalent units in physics problems.

Question 4. Induced emf of electromagnetic induction depends upon:
(a) Resistance on conductor
(b) The value of magnetic field
(c) The direction of conductor w.r.t. the magnetic field
(d) Rate of change of flux linked
Answer: (d) Rate of change of flux linked
In simple words: Faraday's law of electromagnetic induction states that the induced electromotive force (emf) is directly proportional to how quickly the magnetic flux changes over time. The faster the change in flux, the greater the induced emf.

🎯 Exam Tip: The core principle of electromagnetic induction, as per Faraday's Law, is the rate of change of magnetic flux, not just the magnetic field strength or conductor properties alone.

Question 6. A coil of Copper wire is placed parallel to a uniform magnetic field, then the induced emf will be:
(a) Infinite
(b) Zero
(c) Equal to magnetic field
(d) Area of coil
Answer: (b) Zero
In simple words: If a coil is placed parallel to a magnetic field, the angle between the normal to its area and the magnetic field is 90 degrees. Since \( \cos(90^\circ) = 0 \), the magnetic flux (\( \Phi = BA \cos\theta \)) through the coil is zero. If the flux is zero and not changing, no electromotive force (emf) is induced.

🎯 Exam Tip: Induced emf requires a *change* in magnetic flux. If a coil is simply static and parallel to the field, there's no flux linking it, so no emf is induced. If the coil is moving such that the flux doesn't change (e.g., parallel motion along the field lines), no emf is induced.

Question 7. Lenz's law gives:
(a) The magnitude of induced current
(b) Magnitude of induced emf
(c) Direction of induced current
(d) Both direction and magnitude of induced current
Answer: (c) Direction of induced current
In simple words: Lenz's law helps us figure out the direction in which an induced current will flow. It states that the induced current will always flow in a direction that opposes the change in magnetic flux that caused it.

🎯 Exam Tip: While Faraday's law gives the magnitude of induced emf, Lenz's law is crucial for determining the *direction* of the induced current, which is always such as to oppose the cause of induction.

Question 8. A coil of Copper wire and another wire are placed in a given figure. If the current in wire is increased from 1 A to 2 A, then the direction of current will be:
Answer: If the current in the wire increases from 1 A to 2 A, the magnetic flux linked with the coil also increases. According to Lenz's law, the induced current will flow in a direction that opposes this increasing flux. Therefore, the direction of the induced current in the coil will be clockwise.
In simple words: When the current grows, the magnetic field around it gets stronger. The coil reacts by making its own magnetic field to push back against this change, which causes a current to flow clockwise.

🎯 Exam Tip: Apply the right-hand thumb rule to find the direction of the magnetic field from the changing current, then use Lenz's law to determine the direction of the induced current that opposes this change.

Question 9. A metallic disc is moved along its axis. If uniform magnetic field is along the rotational axis, then potential difference between the diameter AB is:
(a) Zero
(b) Half of the potential difference between centre and the end
(c) Double the potential difference between centre and the end
(d) None of the options
Answer: (c) Double the potential difference between centre and the end
In simple words: When a metallic disc rotates in a magnetic field, the potential difference between the center and the circumference is \( \frac{1}{2} B \omega r^2 \). Across a full diameter (A to B), the potential difference is the sum of the potential difference from the center to A and the center to B, which essentially doubles the value from the center to the edge, resulting in \( 2 \times \frac{1}{2} B \omega r^2 = B \omega r^2 \).

🎯 Exam Tip: In problems involving rotating conductors in magnetic fields, remember that the induced emf (potential difference) across the diameter is double that from the center to the circumference due to the additive effect.

Question 10. A conducting wire move towards right in a magnetic field B. Then from the given figure, the direction of induced current is given. Then, what is the direction of magnetic field?
(a) Towards left in the plane of paper
(b) Towards right in the plane of paper
(c) Normally perpendicular downwards in the plane of paper
(d) Normally upwards in the plane of paper
Answer: (c) Normally perpendicular downwards in the plane of paper
In simple words: Using Fleming's right-hand rule, if the thumb points to the right (motion of wire) and the middle finger points towards the induced current, then the forefinger will point downwards, showing the magnetic field's direction.

🎯 Exam Tip: Fleming's right-hand rule (motor effect) or left-hand rule (dynamo effect) are essential for determining the relative directions of force, magnetic field, and current. Ensure you use the correct hand and finger assignments.

Question 11. In an electric communication line, the current flows upwards. Considering the Earth's magnetic field to be negligible, the direction of magnetic field on this electric line is:
(a) Towards East
(b) Towards West
(c) Towards North
(d) Towards South
Answer: (a) Towards East
In simple words: According to the right-hand thumb rule, if you point your thumb upwards (in the direction of current), your fingers curl around the wire. This curling indicates the direction of the magnetic field lines, which would be in the east direction.

🎯 Exam Tip: The right-hand thumb rule is critical for determining the direction of the magnetic field around a current-carrying wire. Practice visualizing the field lines for different current directions.

Question 12. The phase difference between induced emf and magnetic flux linked with a coil rotating in a uniform magnetic field is:
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{2} \)
(c) \( \frac{\pi}{3} \)
(d) \( \pi \)
Answer: (b) \( \frac{\pi}{2} \)
In simple words: When a coil rotates in a magnetic field, the magnetic flux changes as a cosine function. The induced emf, which is the negative rate of change of this flux, will follow a sine function. A sine function is \( \frac{\pi}{2} \) (or 90 degrees) out of phase with a cosine function.

🎯 Exam Tip: Remember that induced emf is proportional to the *rate of change* of flux. If flux is a cosine function, its rate of change (derivative) is a sine function, indicating a \( \frac{\pi}{2} \) phase difference.

Question 13. If in a coil with 2 × 10\(^{-3}\) H coefficient of self-inductance, current flows for 0.1 s is increased at a constant rate by 1 A, then the magnitude of induced emf is:
(a) 2 V
(b)
(c) 0.02 V
Answer: (c) 0.02 V
In simple words: The induced emf is found by multiplying the self-inductance of the coil by the rate at which the current changes. Given L = 2 × 10\(^{-3}\) H and \( \frac{dI}{dt} = \frac{1A}{0.1s} = 10 A/s \), the emf \( \epsilon = L \frac{dI}{dt} = (2 \times 10^{-3} H) \times (10 A/s) = 0.02 V \).

🎯 Exam Tip: For self-induction problems, use the formula \( \epsilon = -L \frac{dI}{dt} \). The negative sign indicates that the induced emf opposes the change in current. Calculate the magnitude carefully.

Question 14. What would be the coefficient of self-inductance of a coil of 100 turns, if 5 A current flows through it? The magnetic flux is of 5 × 10\(^3\) Maxwell.
(a) 0.5 × 10\(^{-3}\) H
(b) 2 × 10\(^{-3}\) H
(c) Zero
(d) 10\(^{-3}\) H
Answer: (d) 10\(^{-3}\) H
In simple words: To find the self-inductance (L), we use the relationship \( L = \frac{N\Phi}{I} \). First, convert Maxwell to Weber (1 Maxwell = 10\(^{-8}\) Wb). So, \( \Phi = 5 \times 10^3 \times 10^{-8} \text{ Wb} = 5 \times 10^{-5} \text{ Wb} \). Now, plug in the values: \( L = \frac{100 \times 5 \times 10^{-5}}{5} = 100 \times 10^{-5} = 10^{-3} \text{ H} \).

🎯 Exam Tip: Ensure all units are consistent (e.g., convert Maxwell to Weber) before applying formulas in physics problems. Self-inductance is a measure of a coil's ability to oppose changes in current.

Question 15. Magnetic flux passing normally through a coil, \( \Phi = 10t^2 + 5t + 1 \) changes with time t is in s, \( \Phi \) m Wb-m. What is the induced emf in coil at t = 5 s?
(a) 1 V
(b) -0.105 V
(c) 2 V
(d) 0 V
Answer: (b) -0.105 V
In simple words: The induced electromotive force (emf) is found by taking the negative derivative of the magnetic flux function with respect to time. For \( \Phi = 10t^2 + 5t + 1 \), the derivative \( \frac{d\Phi}{dt} = 20t + 5 \). So, \( \epsilon = - (20t + 5) \). At \( t = 5 \) s, \( \epsilon = -(20 \times 5 + 5) = -(100 + 5) = -105 \text{ mV} \) or \( -0.105 \text{ V} \).

🎯 Exam Tip: Remember to differentiate the flux function with respect to time to find the induced emf. Pay close attention to the negative sign from Lenz's law and units (milliVolts to Volts).

Rbse Class 12 Physics Chapter 9 Very Short Answer Type Questions

Question 1. How much times would the stored energy become when current in an inductor is doubled?
Answer: The energy stored in an inductor is given by the formula \( W = \frac{1}{2} LI^2 \). If the current (I) is doubled to 2I, the new stored energy becomes \( W' = \frac{1}{2} L (2I)^2 = \frac{1}{2} L (4I^2) = 4 \times (\frac{1}{2} LI^2) = 4W \). Thus, the stored energy becomes 4 times greater.
In simple words: Since stored energy depends on the square of the current, doubling the current makes the energy four times larger.

🎯 Exam Tip: Understand the square relationship between stored energy and current in an inductor. This means small changes in current can lead to significant changes in stored energy.

Question 2. Why is spark produced when we break an electric circuit?
Answer: When an electric circuit is broken, the current tries to drop to zero very quickly. This rapid change in current causes a very large induced electromotive force (emf) across the break due to self-induction. This high induced emf can ionize the air in the gap, leading to a visible spark.
In simple words: A quick stop in current creates a big surge of voltage, which causes a spark as it jumps across the gap.

🎯 Exam Tip: The principle of self-induction explains the spark. The sudden change in current (large \( \frac{dI}{dt} \)) generates a high induced emf (\( \epsilon = -L \frac{dI}{dt} \)), leading to arcing.

Question 3. How can we increase the coefficient of mutual inductance between two coils?
Answer: The coefficient of mutual inductance (M) between two coils can be increased by:
(i) Increasing the number of turns (N\(_1\), N\(_2\)) in both coils.
(ii) Increasing the common cross-sectional area (A) shared by the coils.
(iii) Placing the coils closer to each other to improve coupling.
(iv) Inserting a soft iron core (or other ferromagnetic material) between or within the coils to increase permeability (\( \mu_r \)).
The mutual inductance is proportional to \( \frac{\mu_0 \mu_r N_1 N_2 A}{l} \), where \( l \) is the length.
In simple words: To make two coils interact more strongly (higher mutual inductance), you can add more loops to them, make their shared area bigger, move them closer, or put an iron core between them.

🎯 Exam Tip: Focus on factors that enhance the magnetic flux linkage between the coils: number of turns, common area, proximity, and the magnetic permeability of the core material.

Question 4. What is the value of self inductance, keeping the number of turns in the coil same and doubling the cross-sectional area?
Answer: The coefficient of self-inductance (L) of a coil is given by the relation \( L = \frac{N^2 \mu_0 \mu_r A}{l} \), where N is the number of turns, A is the cross-sectional area, \( l \) is the length, \( \mu_0 \) is the permeability of free space, and \( \mu_r \) is the relative permeability of the core. If the number of turns (N) and length (l) remain the same, but the cross-sectional area (A) is doubled, then the self-inductance (L) will also become 2 times its original value, as L is directly proportional to A.
In simple words: If you keep everything else about a coil the same but make its cross-sectional area twice as big, its self-inductance will also become twice as big.

🎯 Exam Tip: Remember the direct proportionality of self-inductance (L) to the cross-sectional area (A) and the square proportionality to the number of turns (N\(^2\)).

Question 5. The effect of eddy currents in a galvanometer?
Answer: In a moving coil galvanometer, the coil is wound on a non-magnetic metallic frame. When the coil oscillates in the magnetic field, eddy currents are induced in this metallic frame. These eddy currents produce a magnetic field that opposes the motion of the coil, thereby damping its oscillations. This damping brings the coil to rest quickly, making the galvanometer readings stable and precise. Without eddy current damping, the coil would oscillate for a long time, making measurements difficult.
In simple words: Eddy currents help stop the galvanometer's needle from wiggling too much, allowing it to settle quickly for accurate readings.

🎯 Exam Tip: Eddy currents are deliberately used in galvanometers for electromagnetic damping, which reduces oscillation and helps the pointer settle quickly, thus improving the instrument's usability.

Question 7. Self inductance is called the inertia of electricity. Why?
Answer: Self-inductance is often called the "inertia of electricity" because an inductor opposes any change in the current flowing through it. Just as mass resists a change in its state of motion (inertia), an inductor resists any increase or decrease in current by inducing an emf that opposes the change. This property makes it harder to start or stop current flow immediately.
In simple words: An inductor acts like an electrical "bouncer", resisting any sudden changes in current, much like how heavy objects resist changes in their movement.

🎯 Exam Tip: The analogy of "inertia of electricity" is key for understanding self-inductance. Remember that the induced emf always acts to oppose the *change* in current, whether it's growing or decaying.

Question 8. On what factor does the coefficient of self inductance depend?
Answer: The coefficient of self-inductance (L) for a coil depends on several factors:
(i) **Number of turns (N):** L is directly proportional to the square of the number of turns (\(N^2\)). More turns mean more flux linkage.
(ii) **Cross-sectional area (A):** L is directly proportional to the cross-sectional area (A) of the coil. A larger area means more flux passes through.
(iii) **Length of the coil (l):** L is inversely proportional to the length (l) of the coil.
(iv) **Magnetic permeability of the core material (\( \mu \)):** L is directly proportional to the permeability (\( \mu = \mu_0 \mu_r \)) of the material inside the coil. A ferromagnetic core significantly increases L.
The general relation is \( L = \frac{\mu N^2 A}{l} \).
In simple words: How much self-inductance a coil has depends on how many loops it has, how big its area is, how long it is, and what kind of material is placed inside it.

🎯 Exam Tip: When listing factors affecting self-inductance, ensure you include the number of turns (squared relationship), area, length, and core material (permeability).

Question 9. When current flows through a high voltage cable, a bird sitting on the cable flies. Why?
Answer: A bird sitting on a single high-voltage cable does not experience any significant shock because there is no potential difference across its body. However, if the current in the cable suddenly changes (e.g., due to a fault or switching), an induced current can flow through the bird's body for a brief moment due to electromagnetic induction. This sudden induced current or the magnetic field itself can startle the bird, causing it to fly away. Also, a bird might fly away due to the alternating magnetic field created by the AC current, especially if it senses it.
In simple words: A change in electricity in the wire can create a quick burst of energy or a magnetic field that makes the bird feel unsafe and fly away.

🎯 Exam Tip: For safety, remember that a bird on a single high-voltage wire is safe as long as no potential difference exists across its body. It's the *change* in current or an *external path* for current that poses a risk.

Question 10. Write the dimensional formula of \( \frac{L}{R} \), where L is self inductance and R is resistance.
Answer: The ratio \( \frac{L}{R} \) represents the time constant of an LR circuit. Therefore, its dimensional formula is equivalent to that of time.
Dimension of L (self-inductance) = \( [M^1 L^2 T^{-2} A^{-2}] \)
Dimension of R (resistance) = \( [M^1 L^2 T^{-3} A^{-2}] \)
So, \( \frac{L}{R} = \frac{[M^1 L^2 T^{-2} A^{-2}]}{[M^1 L^2 T^{-3} A^{-2}]} = [T^1] \).
Hence, the dimensional formula for \( \frac{L}{R} \) is \( [M^0 L^0 T^1 A^0] \).
In simple words: The ratio of inductance to resistance gives the time constant, which means its unit is seconds. So, its dimensional formula only has 'T' for time.

🎯 Exam Tip: Recognizing \( \frac{L}{R} \) as the time constant of an LR circuit immediately tells you its dimension is time, simplifying the dimensional analysis.

Question 11. A rectangular loop is placed in a uniform magnetic field and is moved with uniform velocity. What is the value of induced emf?
Answer: If a rectangular loop moves with uniform velocity in a *uniform* magnetic field, the magnetic flux linked with the loop remains constant over time. According to Faraday's law of electromagnetic induction, induced emf is produced only when there is a change in magnetic flux. Since the flux is not changing, the induced electromotive force (emf) will be zero.
In simple words: If a loop moves steadily in a magnetic field that's everywhere the same, the amount of magnetic field going through the loop doesn't change, so no electricity is generated.

🎯 Exam Tip: The key here is "uniform magnetic field" and "uniform velocity". Induced emf requires a *change* in magnetic flux, not just the presence of a magnetic field or motion.

Question 12. How can we arrange two coils so that maximum value of induced emf is obtained?
Answer: To obtain the maximum value of induced emf between two coils (i.e., maximum mutual inductance), they should be arranged in the following ways:
(i) **Coaxial arrangement:** The two coils should be placed coaxially, meaning they share the same central axis.
(ii) **Close proximity:** The coils should be placed as close to each other as possible, preferably one inside the other.
(iii) **Same orientation:** Their planes should be oriented such that the magnetic flux produced by one coil completely links with the other coil.
(iv) **Ferromagnetic core:** Inserting a soft iron or other ferromagnetic material as a core through both coils significantly increases the magnetic flux linkage and thus the induced emf.
This arrangement ensures that the magnetic flux from the primary coil completely passes through and links with the secondary coil, maximizing mutual induction.
In simple words: To get the most electricity induced from one coil to another, put them right inside each other, make sure their centers line up, and add an iron bar through them.

🎯 Exam Tip: Maximum mutual inductance occurs when the coupling between the coils is maximized. Think about how to get the most magnetic field lines from one coil to pass through the other.

Question 13. Emf is induced when a straight long conducting wire is placed in North-South direction and is allowed to fall freely. Why?
Answer: When a straight long conducting wire falls freely in the North-South direction, it cuts the horizontal component of the Earth's magnetic field. This cutting of magnetic field lines causes a change in magnetic flux, which in turn induces an electromotive force (emf) across the ends of the wire. The induced emf is given by \( \epsilon = Blv \sin\theta \), where B is the Earth's magnetic field, l is the length of the wire, v is its velocity, and \( \theta \) is the angle between the wire and the magnetic field. Since the wire is moving perpendicular to the horizontal component of the Earth's magnetic field, emf is induced.
In simple words: As the wire falls, it cuts through the Earth's invisible magnetic lines. This movement makes a small amount of electricity flow in the wire.

🎯 Exam Tip: Remember that induced emf occurs when a conductor *cuts* magnetic field lines. Even the Earth's magnetic field, though weak, can induce emf in a moving conductor.

Question 14. Emf is induced when a straight long conducting wire is placed in North-South direction and is allowed to fall freely. Why?
Answer: When a straight long conducting wire falls freely in the North-South direction, it cuts the horizontal component of the Earth's magnetic field. This cutting of magnetic field lines causes a change in magnetic flux, which in turn induces an electromotive force (emf) across the ends of the wire. The induced emf is given by \( \epsilon = Blv \sin\theta \), where B is the Earth's magnetic field, l is the length of the wire, v is its velocity, and \( \theta \) is the angle between the wire and the magnetic field. Since the wire is moving perpendicular to the horizontal component of the Earth's magnetic field, emf is induced.
In simple words: As the wire falls, it cuts through the Earth's invisible magnetic lines. This movement makes a small amount of electricity flow in the wire.

🎯 Exam Tip: The crucial condition for induced emf is the conductor cutting magnetic field lines. If the motion is parallel to the field, no lines are cut, and no emf is induced.

Question 15. How are the eddy currents used for damped oscillation in moving coil galvanometer?
Answer: In a moving coil galvanometer, the coil is typically wound on a non-magnetic metallic frame. When the current passes through the coil, it deflects in the magnetic field. As the coil oscillates, eddy currents are induced in this metallic frame due to the changing magnetic flux. According to Lenz's law, these eddy currents flow in a direction that opposes the motion of the coil. This opposing force acts as a damping force, quickly bringing the coil and the pointer to rest without excessive oscillations, which helps in obtaining stable and accurate readings. This is a deliberate use of eddy currents for damping.
In simple words: Eddy currents are created in the galvanometer's metal frame as it moves, and these currents create a force that quickly stops the needle from wobbling, making the reading steady.

🎯 Exam Tip: Eddy current damping is a practical application of electromagnetic induction. It's used in many instruments to quickly stabilize moving parts and achieve accurate measurements.

Rbse Class 12 Physics Chapter 9 Short Answer Type Questions

Question 1. What do you understand by electromagnetic induction? Write the laws of Faraday related with the electromagnetic induction and write the value of induced emf.
Answer: **Electromagnetic Induction:** This is a phenomenon where an electromotive force (emf) and current are induced in a conductor when it is exposed to a changing magnetic field. This means that a changing magnetic field can create an electric current, even without a battery.
Historically, Michael Faraday and Joseph Henry independently performed experiments showing that a time-varying magnetic field can produce an electric current in a closed coil. This discovery led to many applications, such as generators that produce electricity and induction cookers.

**Faraday's Laws of Electromagnetic Induction:**
**First Law:** Whenever the magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. This induced emf lasts only as long as the change in magnetic flux continues.

**Second Law:** The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux linked with the circuit. If the induced emf is \( \epsilon \) and the magnetic flux is \( \Phi_B \), then mathematically:
\( \epsilon = -\frac{d\Phi_B}{dt} \)
If a coil has N turns and the flux changes uniformly through each turn, the total induced emf is:
\( \epsilon = -N \frac{d\Phi_B}{dt} \)
The magnetic flux \( \Phi_B \) is given by \( \Phi_B = BA \cos\theta \), where B is the magnetic field, A is the area, and \( \theta \) is the angle between B and the normal to A.
The flux \( \Phi_B \) in a coil can change in three ways:
(i) By changing the magnetic field (B) in the coil.
(ii) By changing the total area (A) of the coil that is within the magnetic field (e.g., expanding or contracting the coil, or moving it into/out of the field).
(iii) By changing the angle (\( \theta \)) between the magnetic field (B) and the normal to the plane of the coil (e.g., by rotating the coil).
In simple words: Electromagnetic induction is when a changing magnetic field makes electricity flow in a wire. Faraday's laws say that this happens when the magnetic "flow" (flux) through a circuit changes, and how much electricity is made depends on how fast that change happens. The induced electricity is equal to the negative rate of change of magnetic flux.

🎯 Exam Tip: For a comprehensive answer, define electromagnetic induction, state both of Faraday's laws clearly (first qualitatively, second quantitatively), and explain the factors affecting magnetic flux change.

Question 2. A coil in a magnetic field is removed with: (i) fast speed, (ii) slow speed. In which case, is the induced emf and work done more?
Answer: The magnitude of induced electromotive force (emf) is given by Faraday's law: \( \epsilon = -\frac{d\Phi}{dt} \). This means emf is proportional to the rate of change of magnetic flux.
(i) **Fast speed:** If the coil is removed at a fast speed, the change in magnetic flux (\( d\Phi \)) happens over a very short time interval (\( dt \)). A smaller \( dt \) results in a larger \( \frac{d\Phi}{dt} \), and therefore a larger induced emf.
(ii) **Slow speed:** If the coil is removed at a slow speed, the same change in magnetic flux (\( d\Phi \)) happens over a longer time interval (\( dt \)). A larger \( dt \) results in a smaller \( \frac{d\Phi}{dt} \), and therefore a smaller induced emf.

**Conclusion:**
* **Induced emf:** The induced emf is more when the coil is removed at a **fast speed**.
* **Work done:** More work is done when the coil is removed at a **fast speed**. This is because a larger induced emf leads to a larger induced current. The opposing force due to this induced current (Lenz's law) will be greater, requiring more mechanical work to overcome it and remove the coil. This work is then converted into electrical energy and eventually into heat.
In simple words: When you pull a coil out of a magnetic field quickly, more electricity is made, and you have to work harder to pull it. If you pull it out slowly, less electricity is made, and it's easier to pull.

🎯 Exam Tip: The magnitude of induced emf is inversely proportional to the time taken for the flux change. Remember that more induced current (from a larger emf) means more opposing force, requiring more work to move the coil.

Question 3. Explain the Lenz's law related to electromagnetic induction. How does Lenz's law follow the law of conservation of energy?
Answer: **Lenz's Law:**
Lenz's law provides the direction of the induced electromotive force (emf) or induced current in a circuit. It states that the direction of the induced current is always such that it opposes the cause that produces it. This "cause" is the change in magnetic flux. If the magnetic flux through a coil increases, the induced current will create a magnetic field that tries to decrease the flux. Conversely, if the flux decreases, the induced current will create a magnetic field that tries to increase it.
For example, if the North pole of a bar magnet is moved towards a coil, the induced current will flow to create a North pole on the coil face facing the magnet, repelling it. If the South pole is moved towards the coil, the induced current creates a South pole, repelling it. This opposition is why there's a negative sign in Faraday's law: \( \epsilon = -\frac{d\Phi_B}{dt} \).

**Lenz's Law and Conservation of Energy:**
Lenz's law is a direct consequence of the law of conservation of energy. For an induced current to flow, some energy must be converted into electrical energy. This energy doesn't appear out of nowhere; it comes from the mechanical work done against the opposing force.
Consider moving a magnet towards a coil. According to Lenz's law, the induced current creates a repulsive force. To move the magnet closer, you must apply an external force and do mechanical work against this repulsion. This mechanical work is then converted into electrical energy (induced emf and current) in the coil, which is dissipated as heat (Joule heating) or stored in the magnetic field.
If Lenz's law were not true (i.e., if the induced current aided the motion), the magnet would accelerate without external work, generating more and more current, which would violate the law of conservation of energy by creating energy out of nothing. Thus, the opposing nature described by Lenz's law ensures that energy is conserved.
In simple words: Lenz's law says that induced electricity always tries to fight the change that made it. This is important because it means we have to put in work to make electricity, like pushing two magnets together, which perfectly follows the rule that energy cannot be created or destroyed.

🎯 Exam Tip: Clearly state Lenz's law (opposing the cause) and then explain its connection to energy conservation. Emphasize that mechanical work is converted into electrical energy, not created.

Question 4. Why does an opposing force experienced while pulling or pushing a metallic plate in a uniform magnetic field?
Answer: When a metallic plate is pulled or pushed through a uniform magnetic field, the magnetic flux linked with different parts of the plate changes. This change in magnetic flux induces circulating currents within the body of the metallic plate, known as eddy currents. According to Lenz's law, these induced eddy currents will create their own magnetic fields that oppose the motion of the metallic plate. This opposition manifests as an opposing force, making it harder to pull or push the plate through the magnetic field.
In simple words: Moving a metal plate through a magnetic field creates swirling electric currents inside it (eddy currents). These currents then make a magnetic force that pushes back against the movement, making it feel like something is resisting.

🎯 Exam Tip: The key concepts are "change in magnetic flux," "induced eddy currents," and "Lenz's law (opposing force)." Ensure you explain how these lead to the observed opposing force.

Question 5. Gives reasons: (i) The wires in the coil of resistance box are turned twice. (ii) In Wheatstone bridge, initially a cell key and later galvanometer is pressed.
Answer: (i) The wires in the coil of a resistance box are turned twice (doubled back on themselves) while coiling to minimize self-inductance. When current flows through such a coil, the current in one half of the wire flows in the opposite direction to the current in the adjacent half. This causes the magnetic fields produced by the two halves to largely cancel each other out, thereby significantly reducing the net magnetic flux linkage and, consequently, the self-inductance. This ensures that the resistance box provides only pure resistance, without unwanted inductive effects, especially in AC circuits or during transient current changes.
(ii) In a Wheatstone bridge experiment, the cell key is pressed *before* the galvanometer key. This sequence is important because when the cell key is pressed, current starts flowing in the bridge, and an induced current might momentarily flow in the galvanometer if its key is pressed first. This induced current could interfere with the main current and cause a temporary deflection or incorrect initial reading, leading to errors in measurement. Pressing the cell key first allows the currents in the bridge arms to stabilize. Once stabilized, pressing the galvanometer key will then show a clear deflection only if the bridge is unbalanced, without any transient inductive effects influencing the reading.
In simple words: (i) Resistance box wires are looped to cancel out magnetic fields, so they only act as pure resistance and don't create extra electrical "inertia". (ii) In a Wheatstone bridge, the main power is turned on first, then the sensitive meter, to make sure the electricity has settled down and the measurement is accurate.

🎯 Exam Tip: (i) The purpose of doubling back wires in a resistance box is to reduce self-inductance. (ii) In a Wheatstone bridge, pressing the cell key first minimizes errors caused by transient induced currents.

Question 6. Write Fleming's right hand rule for determining the direction of induced current.
Answer: **Fleming's Right Hand Rule:**
Fleming's right-hand rule is used to determine the direction of the induced electric current when a conductor moves in a magnetic field. To apply this rule, extend the thumb, forefinger, and middle finger of your right hand so that they are mutually perpendicular to each other.
* **Thumb:** Points in the direction of the **motion** of the conductor.
* **Forefinger:** Points in the direction of the **magnetic field**.
* **Middle Finger:** Points in the direction of the **induced current**.
This rule is especially useful for generators (dynamo effect) where mechanical motion causes current induction.
In simple words: Imagine your right hand: your thumb shows which way the wire moves, your first finger shows the magnetic field, and your middle finger then tells you which way the electricity flows.

🎯 Exam Tip: Clearly distinguish Fleming's right-hand rule (for induced current) from Fleming's left-hand rule (for force on a current-carrying conductor). The key is "motion" for the thumb, "field" for the forefinger, and "current" for the middle finger.

Question 7. Define the coefficient of mutual inductance. Give its unit and dimensional formula.
Answer: **Coefficient of Mutual Inductance (M):**
Mutual inductance is a property of two coils that describes how a changing current in one coil induces an electromotive force (emf) in the other coil. The coefficient of mutual inductance (M) between two coils is defined as the ratio of the magnetic flux linked with the secondary coil to the current flowing in the primary coil, assuming all other factors (orientation, size, shape, and medium) remain constant.
If a current \( I_1 \) flows in the primary coil (C\(_1\)), and it produces a magnetic flux \( \Phi_2 \) that links with the secondary coil (C\(_2\)), then:
\( \Phi_2 \propto I_1 \)
\( \implies \Phi_2 = M I_1 \)
Here, M is the constant of proportionality called mutual inductance.
Alternatively, mutual inductance can also be defined by the induced emf. If the current \( I_1 \) in the primary coil changes with time, an emf \( \epsilon_2 \) is induced in the secondary coil:
\( \epsilon_2 = -M \frac{dI_1}{dt} \)
So, M is also the magnitude of induced emf in the secondary coil when the rate of change of current in the primary coil is unity.

**Unit of Mutual Inductance:**
The S.I. unit of mutual inductance (M) is the **Henry (H)**.
1 Henry = 1 Weber per Ampere (Wb/A) or 1 Volt-second per Ampere (V s/A).

**Dimensional Formula of Mutual Inductance:**
Using the formula \( M = \frac{\Phi_2}{I_1} \):
Dimension of magnetic flux \( \Phi \) = \( [M^1 L^2 T^{-2} A^{-1}] \)
Dimension of current \( I \) = \( [A^1] \)
So, Dimension of M = \( \frac{[M^1 L^2 T^{-2} A^{-1}]}{[A^1]} = [M^1 L^2 T^{-2} A^{-2}] \).
In simple words: Mutual inductance (M) tells us how much an electric current changing in one coil can create electricity in another nearby coil. Its unit is the Henry, and its basic physical formula includes mass, length, time, and electric current.

🎯 Exam Tip: Clearly state the definition of mutual inductance (flux linkage per unit current, or induced emf per unit rate of current change), its S.I. unit (Henry), and correctly derive its dimensional formula.

Question 8. A conducting wire is in North-South direction. It is freely dropped towards Earth. Is an emf induced between its ends? why?
Answer: Yes, an electromotive force (emf) is induced between its ends.
When a conducting wire falls freely towards the Earth while oriented in the North-South direction, it cuts across the horizontal component of the Earth's magnetic field. Since the motion (downwards) is perpendicular to both the length of the wire (North-South) and the Earth's horizontal magnetic field, a change in magnetic flux occurs. According to Faraday's law of electromagnetic induction, this change in flux induces an emf across the ends of the wire. The magnitude of this motional emf is given by \( \epsilon = Blv \sin\theta \), where \( \theta = 90^\circ \) in this case, meaning \( \epsilon = Blv \).
In simple words: Yes, electricity is created. As the wire falls through the Earth's invisible magnetic field lines, this movement generates a small amount of electricity across the wire's ends.

🎯 Exam Tip: For motional emf, the key is relative motion between the conductor and the magnetic field such that the conductor *cuts* the magnetic field lines. If the motion is parallel to the field, no emf is induced.

Question 9. A conducting rod of length L is rotated in a magnetic field B with an angular velocity \( \omega \), so that the rotational plane of rod is perpendicular to the magnetic field Determine the induced emf between the ends of the rod.
Answer: Consider a conducting rod OA of length L rotating in a uniform magnetic field B with an angular velocity \( \omega \). The plane of rotation of the rod is perpendicular to the magnetic field, which is directed inwards (represented by crosses).
To determine the induced emf, we consider a small element \( dl \) of the rod at a distance \( l \) from the center O. This element moves with a linear velocity \( v = \omega l \) perpendicular to both its length and the magnetic field.
The induced emf \( d\epsilon \) across this small element \( dl \) is given by:
\( d\epsilon = B v dl \)
Substitute \( v = \omega l \):
\( d\epsilon = B (\omega l) dl \)
To find the total induced emf across the entire rod from O to A, we integrate \( d\epsilon \) from \( l = 0 \) to \( l = L \):
\( \epsilon = \int_{0}^{L} B \omega l dl \)
Since B and \( \omega \) are constant:
\( \epsilon = B \omega \int_{0}^{L} l dl \)
\( \implies \epsilon = B \omega \left[ \frac{l^2}{2} \right]_{0}^{L} \)
\( \implies \epsilon = B \omega \left( \frac{L^2}{2} - 0 \right) \)
\( \implies \epsilon = \frac{1}{2} B \omega L^2 \)
Using Fleming's right-hand rule, if the rod rotates such that its velocity is tangential and the field is inwards, the direction of induced current is from A to O. Therefore, the end O is at a higher potential (positive) and end A is at a lower potential (negative).
In simple words: When a metal rod spins in a magnetic field, electricity is generated across its ends. The amount of electricity depends on half the magnetic field, the speed it spins, and the square of its length. One end becomes positive, the other negative.

🎯 Exam Tip: Remember to integrate the emf generated by each small segment of the rotating rod. The final formula \( \epsilon = \frac{1}{2} B \omega L^2 \) is standard for a rod rotating in a uniform magnetic field with one end fixed.

Question 10. Two coils A and B are perpendicular to each other (in a given figure). There is change in current in one coil. Is current induced in the second coil? Why?
Answer: No, current is not induced in the second coil. This is because when the coils are oriented perpendicularly, a change in current in one coil does not cause a change in the magnetic flux passing through the other coil. For mutual induction to occur, there must be a linkage of magnetic flux between the two coils, which is absent in a perfectly perpendicular arrangement.
In simple words: When the coils are set up at right angles to each other, changing the current in one coil won't create any new magnetic effect in the second coil. So, no electricity will be made in the second coil.

🎯 Exam Tip: Remember that induced current requires a change in magnetic flux through the coil. If the coils are perpendicular and there's no relative motion or flux change, no current is induced due to mutual induction.

Question 11. On what factors do the mutual inductance depend between two coils?
Answer: Mutual inductance depends on several factors: (i) **Relative Permeability (\( \mu_r \)):** The induction between the coils increases significantly if a core made of a material with higher relative permeability (like soft iron) is kept inside them. (ii) **Number of Turns (\( N_1, N_2 \)):** As the number of turns in both the primary coil (\( N_1 \)) and the secondary coil (\( N_2 \)) increases, the mutual inductance also increases proportionally (\( M \propto N_1 N_2 \)). (iii) **Separation:** If the distance between the two coils increases, the magnetic flux linkages between them decrease, which in turn reduces the mutual inductance. (iv) **Coupling:** The mutual inductance \( M \) is related to the self-inductances \( L_1 \) and \( L_2 \) of the two coils by \( M = K \sqrt{L_1 L_2} \), where \( K \) is the coupling coefficient. For 100% coupling (meaning all flux from one coil links with the other), \( K = 1 \), so \( M = \sqrt{L_1 L_2} \). The SI unit of mutual inductance is Weber per Ampere (Wb/A) or Volt-second per Ampere (Vs/A), which is called Henry (H). Its dimensional formula is \( [M^1 L^2 T^{-2} A^{-2}] \).
In simple words: How much two coils affect each other magnetically depends on how many times each coil is wound, their size and shape, how close they are, and what kind of material is placed between them. The unit for this is Henry.

🎯 Exam Tip: When asked for factors, provide a clear list and a brief explanation for each, including the related formula or dependency if possible. Mentioning the unit and dimensional formula adds completeness.

Question 12. The self inductance of a coil is 1 H. What do you understand?
Answer: When the self-inductance of a coil is 1 Henry (H), it means that an induced electromotive force (emf) of 1 volt is produced across the coil when the current flowing through it changes at a rate of 1 ampere per second. It essentially quantifies the coil's ability to resist changes in the current passing through it due to its own magnetic field.
In simple words: If a coil has 1 Henry of self-inductance, it will create 1 volt of electricity across itself if the current going through it changes by 1 ampere every second. It's a measure of how much the coil 'fights' against quick changes in its own current.

🎯 Exam Tip: Clearly define "self-inductance" by relating it to the induced emf and the rate of change of current. Stating the unit (Henry) is crucial for a complete definition.

Question 13. Prove that : When magnetic flux linked with a coil changes from \( \phi_1 \) to \( \phi_2 \), then induced charge, \( q = \frac{N}{R}(\phi_1 - \phi_2) \), where N is the number of turns in the coil and R is the resistance of coil.
Answer: From Faraday's Law of electromagnetic induction, the induced electromotive force (\( \varepsilon \)) in a coil with \( N \) turns is given by: \( \varepsilon = -N \frac{d\phi_B}{dt} \) We also know that the induced current \( I \) flowing in the coil is related to the induced emf and resistance \( R \) by Ohm's Law: \( I = \frac{\varepsilon}{R} \) Substituting the expression for \( \varepsilon \): \( I = \frac{-N}{R} \frac{d\phi_B}{dt} \) The induced charge \( dq \) that flows in a small time interval \( dt \) is given by: \( dq = I dt \) Substitute the expression for \( I \): \( dq = \left(\frac{-N}{R} \frac{d\phi_B}{dt}\right) dt \)
\( \implies dq = \frac{-N}{R} d\phi_B \) To find the total induced charge \( q \) when the magnetic flux changes from an initial value \( \phi_1 \) to a final value \( \phi_2 \), we integrate both sides: \( \int dq = \int_{\phi_1}^{\phi_2} \frac{-N}{R} d\phi_B \)
\( \implies q = \frac{-N}{R} [\phi_B]_{\phi_1}^{\phi_2} \)
\( \implies q = \frac{-N}{R} (\phi_2 - \phi_1) \)
\( \implies q = \frac{N}{R} (\phi_1 - \phi_2) \) This equation demonstrates that the total induced charge flowing through a circuit depends on the total change in magnetic flux and the resistance of the coil, but it does not depend on the rate at which the magnetic flux changes.
In simple words: When the magnetic field lines (flux) passing through a coil change, it pushes a certain amount of electric charge through the coil. This total charge depends on how many loops the coil has, how much the magnetic field changes, and the coil's electrical resistance. It doesn't matter how fast the change happens.

🎯 Exam Tip: For proofs, start with the fundamental laws (Faraday's Law in this case), show each step clearly, and explain the significance of the final derived formula, especially the independence of charge from the rate of flux change.

Question 14. Prove that law of conservation of energy is obeyed when a rectangular coil is placed perpendicular in non-uniform magnetic field.
Answer: When a rectangular coil moves with constant velocity perpendicular to a non-uniform magnetic field, an emf is induced, and a current \( I \) flows. Due to the magnetic field, forces act on the current-carrying sides of the coil. If the magnetic field is non-uniform, the forces on the two sides (e.g., \( F_{ab} \) and \( F_{cd} \)) will be different. According to Lenz's law, this results in a net opposing force \( F \) that acts against the motion of the loop. To maintain the coil's constant velocity, external mechanical work \( W \) must be done against this opposing force. This mechanical work input is transformed into electrical energy within the coil. As current \( I \) flows through the coil's resistance \( R \), this electrical energy is then dissipated as heat \( H \) (Joule heating). The rate of mechanical work done by the external agent is equal to the rate at which heat is produced in the coil. By deriving expressions for work done and heat produced, it can be shown that \( W = H \). This equality demonstrates that the energy supplied mechanically to move the coil is entirely converted into heat energy. Thus, the total energy of the system remains conserved, proving that Lenz's law is in accordance with the law of conservation of energy.
In simple words: Moving a coil in an uneven magnetic field requires us to do work against a magnetic push. This work energy doesn't disappear; it gets converted into electricity (current) within the coil, which then turns into heat because of the coil's resistance. Since the energy put in equals the energy that comes out as heat, the total energy is always conserved, just as the law of conservation of energy states.

🎯 Exam Tip: To prove energy conservation for Lenz's law, explain how mechanical work against the opposing force is transformed into electrical energy (induced current) and then into heat, showing no energy is lost.

Question 15. Magnetic flux passing normally through a coil, \( \phi = 10t^2 + 5t + 1 \) changes with time t is in s, \( \phi \) in mWb-m. What is the induced emf in coil at \( t = 5 \) s?
(a) 1 V
(b) -0.105 V
(c) 2 V
(d) 0 V
Answer: (b) -0.105 V
Solution:
Given magnetic flux \( \phi = (10t^2 + 5t + 1) \text{ mWb-m} \)
The induced electromotive force \( \varepsilon \) is given by Faraday's law:
\( \varepsilon = -\frac{d\phi}{dt} \)
First, find the rate of change of flux by differentiating \( \phi \) with respect to \( t \):
\( \frac{d\phi}{dt} = \frac{d}{dt} (10t^2 + 5t + 1) \)
\( \implies \frac{d\phi}{dt} = 20t + 5 \) mWb/s
Now, substitute the given time \( t = 5 \) s into the equation for \( \frac{d\phi}{dt} \):
\( \frac{d\phi}{dt} = 20(5) + 5 = 100 + 5 = 105 \text{ mWb/s} \)
Therefore, the induced emf is:
\( \varepsilon = - (105) \text{ mV} \)
\( \implies \varepsilon = -0.105 \text{ V} \)
In simple words: First, we find how quickly the magnetic field's effect (flux) is changing over time by doing a mathematical step called differentiation. Then, we put the given time value into this result. The final answer, which is the induced voltage (emf), will be negative because it always opposes the change. Since the initial unit was in milliWeber, the emf is in millivolts.

🎯 Exam Tip: Remember to differentiate the flux equation with respect to time to find the induced emf, and don't forget the negative sign from Faraday's law. Then substitute the given time value. Pay close attention to units, as they can determine the scale of the final answer.

RBSE Class 12 Physics Chapter 9 Long Answer Type Questions

Question 1. Determine induced emf due to a moving conducting rod moving with a uniform velocity in a uniform magnetic field. How is the direction of this induced emf determined?
Answer: Consider a straight conducting rod of length \( l \) (say, AB) moving with a uniform velocity \( \vec{v} \) perpendicular to a uniform magnetic field \( \vec{B} \). Assume the magnetic field points out of the page (represented by dots), and the rod moves to the right. The free electrons within the rod also move with velocity \( \vec{v} \) through the magnetic field. Consequently, a magnetic force \( \vec{F}_m \) acts on these electrons. This force is given by \( \vec{F}_m = q (\vec{v} \times \vec{B}) \). Since electrons have a negative charge, the force on them will be opposite to the direction of \( (\vec{v} \times \vec{B}) \). If \( \vec{v} \) is to the right and \( \vec{B} \) is out of the page, \( (\vec{v} \times \vec{B}) \) is downwards. So, the magnetic force \( \vec{F}_m \) on electrons is directed upwards, pushing electrons towards end A of the rod. This causes end A to become negatively charged and end B to become positively charged. This separation of charges creates an electric field \( \vec{E} \) inside the rod, directed from B to A (from positive to negative). An electric force \( \vec{F}_e = q\vec{E} \) acts on the electrons. In equilibrium, when no more charge separation occurs, the electric force balances the magnetic force: \( q\vec{E} + q(\vec{v} \times \vec{B}) = 0 \)
\( \implies \vec{E} = -(\vec{v} \times \vec{B}) \) The magnitude of this electric field is \( E = vB \) (since \( \vec{v} \) and \( \vec{B} \) are perpendicular). The induced electromotive force (emf) \( \varepsilon \) across the rod is the potential difference between its ends, which is \( \varepsilon = E \times l \). Therefore, the induced emf is \( \varepsilon = vBl \). **Direction of induced emf:** The direction of this induced emf (and the resulting induced current) can be determined using **Fleming's Right-Hand Rule**. If the forefinger points in the direction of the magnetic field, and the thumb points in the direction of the conductor's motion, then the central finger points in the direction of the induced current (from the negative end to the positive end of the rod).
In simple words: When a metal rod moves through a magnetic field, the tiny charged particles inside it (electrons) are pushed to one end by the magnetic force, making that end negative and the other positive. This creates an electrical "push" or voltage (emf) across the rod. We can find which way this electricity flows using Fleming's Right-Hand Rule.

🎯 Exam Tip: When deriving motional emf, clearly explain the role of magnetic force on charges, the resulting electric field, and how equilibrium leads to the emf formula. Always include Fleming's Right-Hand Rule for direction.

Question 2. A rectangular loop moves with constant velocity perpendicular to non-uniform magnetic field. Calculate the formula for induced emf and current for it. Prove that it obeys the law of conservation of energy.
Answer: Consider a rectangular conducting coil of length \( l \) and width \( w \), moving with constant velocity \( \vec{v} \) perpendicular to a non-uniform magnetic field \( \vec{B} \). Let the magnetic field strength be \( B_1 \) at one side (say, 'ab') of the loop and \( B_2 \) at the other parallel side ('cd'). As the coil moves with velocity \( v \) for a small time \( dt \), it covers a distance \( dx = v dt \). During this motion, the magnetic flux through the coil changes because the magnetic field is non-uniform. The change in magnetic flux \( d\phi_B \) over time \( dt \) is due to the difference in flux entering and leaving the loop. The area swept by side 'ab' is \( l dx \), and the area swept by 'cd' is also \( l dx \). Change in flux \( \Delta \phi_B = (\text{Flux linked at final position}) - (\text{Flux linked at initial position}) \). Alternatively, the rate of change of flux can be seen as \( \frac{d\phi_B}{dt} = (B_2 - B_1) l v \). (The term in the source \( \Delta \phi_B = (B_2 - B_1) lv \Delta t \) is consistent with this.) According to Faraday's law, the induced emf \( \varepsilon \) is: \( \varepsilon = - \frac{d\phi_B}{dt} = -(B_2 - B_1) l v = (B_1 - B_2) l v \) If the coil has total resistance \( R \), the induced current \( I \) flowing in the loop is given by Ohm's Law: \( I = \frac{\varepsilon}{R} = \frac{(B_1 - B_2) l v}{R} \) **Law of conservation of energy:** When current \( I \) flows in the loop, forces act on the current-carrying sides in the magnetic field. Since the field is non-uniform (\( B_1 \neq B_2 \)), there will be a net opposing force on the loop that resists its motion (as per Lenz's Law). To maintain the coil's constant velocity, an external mechanical force must be applied, and work \( W \) is done against this opposing force. This mechanical work is converted into electrical energy. As this electrical energy flows through the resistance \( R \) of the coil, it is dissipated as heat \( H \) (Joule heating). The rate of mechanical work done by the external agent is equal to the rate of heat produced in the coil. This continuous transformation of mechanical energy into electrical energy and then into heat, without any net loss or gain of energy, demonstrates that the process obeys the law of conservation of energy.
In simple words: When a rectangular coil moves at a steady speed through an uneven magnetic field, electricity flows because the magnetic effect changes across the coil. The amount of electricity created depends on the difference in the magnetic field strengths and how fast the coil moves. To keep the coil moving, we have to push it (do work). This energy from pushing turns into electricity, which then becomes heat in the coil. This shows that energy is not lost but simply changes form, upholding the law of energy conservation.

🎯 Exam Tip: For this type of question, first derive the motional emf due to a non-uniform field. Then, connect this to the work-energy principle by explaining how external work equals heat dissipated, demonstrating energy conservation.

Question 3. A rectangular coil of turns N and area A rotates with a uniform velocity to in a uniform magnetic field. Prove that induced emf in the coil is NBAw sin wt.
Answer: Consider a rectangular coil with \( N \) turns and cross-sectional area \( A \), rotating with a uniform angular velocity \( \omega \) in a uniform magnetic field \( B \). Let the angle between the normal to the coil's plane and the direction of the magnetic field be \( \theta \). As the coil rotates, this angle changes with time, such that \( \theta = \omega t \). The magnetic flux \( \phi_B \) linked with the coil at any instant \( t \) is given by: \( \phi_B = NBA \cos\theta \) Substituting \( \theta = \omega t \): \( \phi_B = NBA \cos(\omega t) \) According to Faraday's law of electromagnetic induction, the induced emf \( \varepsilon \) in the coil is given by the negative rate of change of magnetic flux: \( \varepsilon = -N \frac{d\phi_B}{dt} \) Substitute the expression for \( \phi_B \) into Faraday's law: \( \varepsilon = -N \frac{d}{dt} (BA \cos(\omega t)) \)
\( \implies \varepsilon = -N B A \frac{d}{dt} (\cos(\omega t)) \)
\( \implies \varepsilon = -N B A (-\sin(\omega t)) \times \omega \) (using the chain rule, \( \frac{d}{dx}(\cos ax) = -a \sin ax \))
\( \implies \varepsilon = NBA \omega \sin(\omega t) \) This equation shows that the induced emf varies sinusoidally with time. The maximum value of the induced emf, \( \varepsilon_0 \), occurs when \( \sin(\omega t) = 1 \), so \( \varepsilon_0 = NBA\omega \). Therefore, the induced emf can also be written as \( \varepsilon = \varepsilon_0 \sin(\omega t) \). The emf and current produced this way are called alternating emf and alternating current (AC), respectively. This is the fundamental principle behind an AC generator.
In simple words: When a coil spins steadily in a constant magnetic field, the amount of magnetic field lines going through it constantly changes in a wave-like pattern. This change creates an electric voltage (emf) in the coil that also cycles like a wave. The strength of this voltage depends on the coil's loops, its size, the magnetic field, and how fast it spins. This is how AC generators make electricity.

🎯 Exam Tip: When presenting derivations for rotating coils, ensure to include a visual representation of the coil's orientation or the resulting sinusoidal emf graph to aid understanding.

Question 4. What is self induction? Explain the experiments of self induction and calculate self inductance for a solenoid.
Answer: **Self-induction** is the phenomenon where a changing electric current in a coil produces an induced electromotive force (emf) within the same coil. This induced emf always acts in a direction that opposes the change in current that caused it, which is consistent with Lenz's law. **Experiments to demonstrate self-induction:** A common demonstration involves a solenoid (a coil of wire) connected in series with a battery and a tapping key, with a small bulb connected in parallel across the solenoid. (i) **Key Pressed:** When the tapping key is pressed, the current from the battery starts to flow through the solenoid and gradually increases. This increasing current creates a magnetic field, and thus magnetic flux, within the solenoid. Due to self-induction, an induced emf is generated in the solenoid which opposes this increase in current. This opposition makes the current rise gradually rather than instantly. The bulb glows faintly at first, then brightens as the current stabilizes. (ii) **Key Released:** When the key is released, the main circuit breaks, and the current attempts to drop to zero abruptly. The induced emf now opposes this sudden decrease, attempting to maintain the current flow. This large induced emf causes the bulb to glow brightly for a brief moment before extinguishing, vividly demonstrating the coil's resistance to rapid current changes. **Self-inductance of a solenoid:** Consider a long solenoid with \( N \) turns, length \( l \), and cross-sectional area \( A \). If a current \( I \) flows through it, the magnetic field \( B \) inside the solenoid (assuming it's uniform) is given by: \( B = \mu_0 n I = \frac{\mu_0 NI}{l} \) (where \( n = N/l \) is the number of turns per unit length) The total magnetic flux \( \phi \) linked with all \( N \) turns of the solenoid is: \( \phi = NBA = N \left(\frac{\mu_0 NI}{l}\right) A = \frac{\mu_0 N^2 AI}{l} \) By definition, the self-inductance \( L \) of the solenoid is the ratio of the total magnetic flux linked to the current flowing through it: \( L = \frac{\phi}{I} \) Substituting the expression for \( \phi \): \( L = \frac{\frac{\mu_0 N^2 AI}{l}}{I} \)
\( \implies L = \frac{\mu_0 N^2 A}{l} \) This formula shows that the self-inductance of a solenoid depends on its geometric properties (number of turns \( N \), cross-sectional area \( A \), and length \( l \)) and the magnetic permeability of the core material (\( \mu_0 \)).
In simple words: Self-induction is when a coil generates its own electrical push (emf) to resist any changes in the current flowing through it. Experiments show a bulb connected to a coil lighting up differently when current starts or stops, proving this effect. For a long, wound coil (solenoid), its self-inductance (how much it resists current changes) depends on how many times it's wound, its cross-sectional area, and its length.

🎯 Exam Tip: When explaining self-induction, clearly state its definition and Lenz's law application. For the solenoid formula, show the steps from magnetic field to flux, then to self-inductance, defining all terms.

Question 5. What are eddy currents? Write any two uses of them. How are eddy currents reduced in a transformer?
Answer: **Eddy currents** are swirling loops of electric current that are induced within a conductor by a changing magnetic field. They flow in planes perpendicular to the magnetic field lines and, according to Lenz's law, they create their own magnetic field that opposes the change in the original magnetic flux. This opposition leads to energy loss, usually as heat. These currents are also known as Foucault currents after their discoverer. **Two uses of eddy currents:** (i) **Electromagnetic Brakes:** Eddy currents are effectively used in electromagnetic braking systems, for example, in some trains. When a strong magnetic field is applied to a rotating metal wheel or drum, eddy currents are induced in the metal. These currents interact with the magnetic field to produce a braking force that opposes the motion, thereby slowing down the train smoothly and efficiently. (ii) **Induction Furnaces:** Induction furnaces utilize strong, high-frequency alternating magnetic fields to induce intense eddy currents within metals placed inside them. The large amount of heat generated by these powerful eddy currents is enough to melt the metals, which is essential for various industrial applications like extracting metals from ores or in metal casting processes. **How eddy currents are reduced in a transformer:** Eddy currents are undesirable in transformers because they cause significant energy loss in the form of heat, reducing efficiency. To minimize these losses, the transformer core is not made from a single solid block of metal. Instead, it is constructed from many thin sheets of soft iron, called **laminations**. These laminations are stacked together and are electrically insulated from each other (e.g., by a thin layer of lacquer or oxide). This lamination process significantly increases the electrical resistance path for the eddy currents, effectively confining them to smaller, weaker loops within each lamination. By doing so, the magnitude of the eddy currents is greatly reduced, thereby minimizing the energy dissipated as heat.
In simple words: Eddy currents are electric currents that spin around inside a metal when a nearby magnetic field changes. They can be used to stop trains or melt metals in special furnaces. In devices like transformers, these currents waste energy by making heat. To fix this, transformer cores are made of many thin, insulated layers instead of one thick piece, which helps reduce the unwanted currents.

🎯 Exam Tip: Define eddy currents clearly. Provide at least two distinct applications and specifically explain the lamination technique for reduction in transformers, as it's a common example.

RBSE Class 12 Physics Chapter 9 Numerical Questions

Question 1. In a wall, there is a metallic framed window (120 cm x 50 cm). The total resistance of a frame is 0.01 \( \Omega \). Determine the charges flowing on opening the frame from 90°. [If H = 0.36 G]
Answer:
Solution:
Given:
Length of window \( l = 120 \) cm \( = 1.2 \) m
Width of window \( w = 50 \) cm \( = 0.5 \) m
Area of the frame \( A = l \times w = 1.2 \times 0.5 = 0.6 \text{ m}^2 \)
Resistance \( R = 0.01 \, \Omega \)
Magnetic field strength \( H = 0.36 \text{ G} = 0.36 \times 10^{-4} \text{ T} \) (converting Gauss to Tesla, as \( 1 \text{ G} = 10^{-4} \text{ T} \)). We assume \( B \approx H \) for this problem.
When the frame is opened from 90°, it implies a change in orientation such that the magnetic flux linking the frame changes from maximum to zero, or vice-versa. The total change in magnetic flux (\( \Delta \phi \)) through the frame is therefore equal to the maximum flux, \( B \times A \). \( \Delta \phi = B \times A = (0.36 \times 10^{-4} \text{ T}) \times (0.6 \text{ m}^2) \)
\( \implies \Delta \phi = 0.216 \times 10^{-4} \text{ Wb} \) The induced charge \( q \) is given by the formula \( q = \frac{N \Delta \phi}{R} \). For a single-turn frame, \( N=1 \):
\( q = \frac{0.216 \times 10^{-4} \text{ Wb}}{0.01 \, \Omega} \)
\( \implies q = 21.6 \times 10^{-4} \text{ C} \)
\( \implies q = 2.16 \times 10^{-3} \text{ C} \)
In simple words: First, calculate the area of the window in square meters. Then, convert the magnetic field strength to Tesla and multiply it by the window's area to find the total change in magnetic flux. Finally, divide this magnetic flux change by the frame's electrical resistance to get the total electric charge that flows.

🎯 Exam Tip: Remember to convert all units to SI (e.g., cm to m, Gauss to Tesla) before calculation. The induced charge depends on the total change in flux, not the rate of change, and is inversely proportional to resistance. Be careful with the initial and final flux values based on the coil's orientation or presence in the field.

Question 2. The magnetic flux linked with the coil of 50 turns is \( \phi_B = 0.02 \cos(100\pi t) \) Wb. Determine:
(a) Maximum induced voltage.
(b) Induced emf at \( t = 0.01 \) s.
(c) Induced electric current at \( t = 0.005 \) s. (If external resistance is 100 \( \Omega \))
Answer:
Solution:
Given:
Number of turns \( N = 50 \)
Magnetic flux \( \phi_B = 0.02 \cos(100\pi t) \) Wb
External resistance \( R = 100 \, \Omega \)

The induced electromotive force \( \varepsilon \) is given by Faraday's law:
\( \varepsilon = -N \frac{d\phi_B}{dt} \)
First, let's find the rate of change of magnetic flux \( \frac{d\phi_B}{dt} \) by differentiating \( \phi_B \) with respect to \( t \):
\( \frac{d\phi_B}{dt} = \frac{d}{dt} (0.02 \cos(100\pi t)) \)
\( \implies \frac{d\phi_B}{dt} = 0.02 \times (-\sin(100\pi t)) \times (100\pi) \)
\( \implies \frac{d\phi_B}{dt} = -2\pi \sin(100\pi t) \) Wb/s
Now, substitute this back into the emf equation:
\( \varepsilon = -N (-2\pi \sin(100\pi t)) = 50 \times 2\pi \sin(100\pi t) \)
\( \implies \varepsilon = 100\pi \sin(100\pi t) \) V
\( \varepsilon \approx 100 \times 3.14159 \sin(100\pi t) \approx 314.159 \sin(100\pi t) \) V

**(a) Maximum induced voltage:**
The maximum value of the sine function is 1 (\( \sin(\theta)_{max} = 1 \)).
So, the maximum induced emf is:
\( \varepsilon_{max} = 100\pi \times 1 = 100\pi \) V
\( \varepsilon_{max} \approx 314.159 \) V

**(b) Induced emf at \( t = 0.01 \) s:**
Substitute \( t = 0.01 \) s into the emf equation:
\( \varepsilon = 100\pi \sin(100\pi \times 0.01) \)
\( \implies \varepsilon = 100\pi \sin(\pi) \)
Since \( \sin(\pi) = 0 \):
\( \varepsilon = 100\pi \times 0 = 0 \) V

**(c) Induced electric current at \( t = 0.005 \) s:**
First, calculate the induced emf at \( t = 0.005 \) s:
\( \varepsilon = 100\pi \sin(100\pi \times 0.005) \)
\( \implies \varepsilon = 100\pi \sin(0.5\pi) = 100\pi \sin(\frac{\pi}{2}) \)
Since \( \sin(\frac{\pi}{2}) = 1 \):
\( \varepsilon = 100\pi \times 1 = 100\pi \) V \( \approx 314.159 \) V
Now, calculate the induced current \( I \) using Ohm's law, \( I = \frac{\varepsilon}{R} \):
\( I = \frac{100\pi}{100} = \pi \) A
\( I \approx 3.14159 \) A
In simple words: First, use the changing magnetic flux to find a formula for the voltage (emf) in the coil over time. For the biggest possible voltage, use the maximum value of the sine function. For voltage at a specific time, plug that time into your formula. To find the current, just divide the voltage at that time by the coil's resistance.

🎯 Exam Tip: Pay careful attention to trigonometric function values at specific angles (e.g., \( \sin(\pi) = 0 \), \( \sin(\frac{\pi}{2}) = 1 \)). Remember that the induced emf depends on the rate of change of flux and the number of turns. Induced current is found using Ohm's law, so you need the emf and resistance.

Question 3. A coil of 50 turns is normally placed in a magnetic field of 0.6 T. The area of this coil is 0.2 m² and the resistance in the circuit is 100 \( \Omega \). Calculate induced charge when :
(a) Coil is rotated by 180°
(b) Coil is removed from the magnetic field.
Answer:
Solution:
Given:
Number of turns \( N = 50 \)
Magnetic field \( B = 0.6 \text{ T} \)
Area \( A = 0.2 \text{ m}^2 \)
Resistance \( R = 100 \, \Omega \)

The formula for induced charge \( q \) is \( q = \frac{N \Delta \phi_B}{R} \), where \( \Delta \phi_B \) is the change in magnetic flux. The magnetic flux \( \phi_B \) through a coil is given by \( \phi_B = BA \cos\theta \).
Since the coil is normally placed initially, the initial angle \( \theta_{initial} = 0^\circ \).
Initial magnetic flux \( \phi_{initial} = BA \cos(0^\circ) = BA \)
\( \phi_{initial} = 0.6 \text{ T} \times 0.2 \text{ m}^2 = 0.12 \text{ Wb} \)

**(a) Coil is rotated by 180°:**
When the coil is rotated by 180°, the final angle \( \theta_{final} = 180^\circ \).
Final magnetic flux \( \phi_{final} = BA \cos(180^\circ) = -BA \)
\( \phi_{final} = -(0.6 \text{ T} \times 0.2 \text{ m}^2) = -0.12 \text{ Wb} \)
Change in magnetic flux \( \Delta \phi_B = \phi_{initial} - \phi_{final} = (0.12 \text{ Wb}) - (-0.12 \text{ Wb}) = 0.24 \text{ Wb} \)
Now, calculate the induced charge \( q \):
\( q = \frac{N \Delta \phi_B}{R} = \frac{50 \times 0.24 \text{ Wb}}{100 \, \Omega} \)
\( \implies q = \frac{12}{100} = 0.12 \text{ C} \)

**(b) Coil is removed from the magnetic field:**
Initial magnetic flux \( \phi_{initial} = 0.12 \text{ Wb} \)
When the coil is removed from the magnetic field, the final magnetic flux is zero:
Final magnetic flux \( \phi_{final} = 0 \text{ Wb} \)
Change in magnetic flux \( \Delta \phi_B = \phi_{initial} - \phi_{final} = 0.12 \text{ Wb} - 0 \text{ Wb} = 0.12 \text{ Wb} \)
Now, calculate the induced charge \( q \):
\( q = \frac{N \Delta \phi_B}{R} = \frac{50 \times 0.12 \text{ Wb}}{100 \, \Omega} \)
\( \implies q = \frac{6}{100} = 0.06 \text{ C} \)
In simple words: First, calculate the initial magnetic effect (flux) going through the coil. Then, for each situation, determine the final magnetic effect. Find the difference between the initial and final effects. Finally, multiply this difference by the number of turns in the coil and divide by its electrical resistance to find the total electric charge that flows.

🎯 Exam Tip: Remember that induced charge calculation uses the total change in flux (\( \Delta \phi \)), not its rate of change. Be careful with the initial and final flux values based on the coil's orientation or presence in the field. Always convert to SI units before calculating.

Question 4. A-3 \( \hat{k} \) m long conductor moves with a velocity of \( (\hat{i}+2 \hat{j}+3 \hat{k}) \) m/s in a magnetic field of \( (\hat{i}+3 \hat{j}+\hat{k}) \)T. Calculate the potential difference between the ends of the conductor.
Answer:
Given:
Length of conductor, \( \vec{l} = -3 \hat{k} \, \text{m} \)
Velocity of conductor, \( \vec{v} = (\hat{i} + 2\hat{j} + 3\hat{k}) \, \text{m/s} \)
Magnetic field, \( \vec{B} = (\hat{i} + 3\hat{j} + \hat{k}) \, \text{T} \)
The induced potential difference (emf) across the ends of the conductor is given by \( E = \vec{l} \cdot (\vec{v} \times \vec{B}) \).
First, calculate the cross product \( \vec{v} \times \vec{B} \):
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 3 & 1 \end{vmatrix} \]
\( \implies \vec{v} \times \vec{B} = \hat{i}(2 \times 1 - 3 \times 3) - \hat{j}(1 \times 1 - 3 \times 1) + \hat{k}(1 \times 3 - 2 \times 1) \)
\( \implies \vec{v} \times \vec{B} = \hat{i}(2 - 9) - \hat{j}(1 - 3) + \hat{k}(3 - 2) \)
\( \implies \vec{v} \times \vec{B} = -7\hat{i} + 2\hat{j} + \hat{k} \)
Next, calculate the dot product \( E = \vec{l} \cdot (\vec{v} \times \vec{B}) \):
\( E = (-3\hat{k}) \cdot (-7\hat{i} + 2\hat{j} + \hat{k}) \)
\( \implies E = (-3) \times (1) \) (since \( \hat{k} \cdot \hat{k} = 1 \) and \( \hat{k} \cdot \hat{i} = \hat{k} \cdot \hat{j} = 0 \))
\( \implies E = -3 \, \text{volt} \)
In simple words: We first find a new vector from the velocity and magnetic field. Then, we combine this new vector with the conductor's length vector to find the electrical voltage, which turns out to be -3 volts.

🎯 Exam Tip: Remember to use the correct vector cross product and dot product rules. Pay attention to the signs and coordinate components to avoid errors in calculation.

Question 5. A rectangular coil of 1000 turns and area 0.2 × 0.1m² rotates at 4200 revolutions per minute in a magnetic field 0.2 T. Calculate maximum value of induced emf of coil.
Answer:
Given:
Number of turns, \( N = 1000 \)
Area of the coil, \( A = 0.2 \times 0.1 \, \text{m}^2 = 0.02 \, \text{m}^2 \)
Rotational frequency, \( f_{rpm} = 4200 \, \text{revolutions per minute (rpm)} \)
Magnetic field, \( B = 0.2 \, \text{T} \)

First, convert the rotational frequency from rpm to revolutions per second (Hz):
\( f_{Hz} = \frac{4200 \, \text{rpm}}{60 \, \text{s/min}} = 70 \, \text{Hz} \)
Next, calculate the angular velocity \( \omega \):
\( \omega = 2\pi f_{Hz} = 2\pi \times 70 = 140\pi \, \text{rad/s} \)
Now, use the formula for the maximum induced electromotive force (emf) in a rotating coil:
\( \epsilon_{max} = NBA\omega \)
\( \implies \epsilon_{max} = 1000 \times 0.2 \times 0.02 \times 140\pi \)
\( \implies \epsilon_{max} = 200 \times 0.02 \times 140\pi \)
\( \implies \epsilon_{max} = 4 \times 140\pi \)
\( \implies \epsilon_{max} = 560\pi \, \text{V} \)
Using \( \pi \approx 3.14159 \):
\( \implies \epsilon_{max} \approx 560 \times 3.14159 \approx 1759.29 \, \text{V} \)
In simple words: We changed the coil's spin rate from turns per minute to radians per second. Then, using the number of turns, its area, the magnetic field strength, and the spin speed, we found the largest voltage the coil can produce is about 1759.29 volts.

🎯 Exam Tip: Always ensure you convert units correctly, especially revolutions per minute (rpm) to radians per second (rad/s) for angular velocity calculations in physics problems.

Question 6. A conducting rod of length 1 m rotates with an angular velocity at the rate 50 revolutions per second normally to a magnetic field of 0.001 T from one end. Calculate the induced emf between the ends of the rod.
Answer:
Given:
Length of the rod, \( L = 1 \, \text{m} \)
Rotational frequency, \( f = 50 \, \text{revolutions per second (rps)} \)
Magnetic field, \( B = 0.001 \, \text{T} \)

First, calculate the angular velocity \( \omega \):
\( \omega = 2\pi f = 2 \times 3.14 \times 50 = 314 \, \text{rad/s} \)
Now, use the formula for the induced electromotive force (emf) in a rotating rod:
\( \epsilon = \frac{1}{2} BL^2\omega \)
\( \implies \epsilon = \frac{1}{2} \times 0.001 \times (1)^2 \times 314 \)
\( \implies \epsilon = \frac{1}{2} \times 0.001 \times 1 \times 314 \)
\( \implies \epsilon = \frac{0.314}{2} \)
\( \implies \epsilon = 0.157 \, \text{V} \)
In simple words: We found the turning speed of the rod. Then, by using its length, the magnetic field, and its turning speed, we calculated that 0.157 volts of electricity is created between its ends.

🎯 Exam Tip: Remember the formula for motional EMF in a rotating rod and ensure all units are consistent (e.g., length in meters, frequency in Hz, magnetic field in Tesla).

Question 7. A solenoid of diameter 0.05 m and 500 turns/cm has length 1 m. When current of 3 A passes through it, then calculate magnetic flux.
Answer:
Given:
Diameter of solenoid, \( D = 0.05 \, \text{m} \)
Radius of solenoid, \( r = D/2 = 0.025 \, \text{m} \)
Number of turns per unit length, \( n' = 500 \, \text{turns/cm} = 500 \times 100 \, \text{turns/m} = 50000 \, \text{turns/m} \)
Length of solenoid, \( l = 1 \, \text{m} \)
Current, \( I = 3 \, \text{A} \)

First, calculate the cross-sectional area of the solenoid:
\( A = \pi r^2 = \pi (0.025)^2 = 6.25\pi \times 10^{-4} \, \text{m}^2 \)
Next, find the total number of turns in the solenoid:
\( N = n' \times l = 50000 \, \text{turns/m} \times 1 \, \text{m} = 50000 \, \text{turns} \)
Then, calculate the magnetic field \( B \) inside the solenoid (using vacuum permeability \( \mu_0 \)):
\( B = \mu_0 n' I = (4\pi \times 10^{-7} \, \text{Tm/A}) \times (50000 \, \text{turns/m}) \times (3 \, \text{A}) \)
\( \implies B = 600000\pi \times 10^{-7} = 0.06\pi \, \text{T} \)
Finally, calculate the total magnetic flux \( \Phi \) linked with the solenoid:
\( \Phi = NBA \)
\( \implies \Phi = 50000 \times (0.06\pi) \times (6.25\pi \times 10^{-4}) \)
\( \implies \Phi = 5 \times 10^4 \times 0.06\pi \times 6.25\pi \times 10^{-4} \)
\( \implies \Phi = 5 \times 0.06\pi \times 6.25\pi \)
\( \implies \Phi = 0.3\pi \times 6.25\pi = 1.875\pi^2 \, \text{Wb} \)
Using \( \pi^2 \approx 9.8696 \):
\( \implies \Phi \approx 1.875 \times 9.8696 \approx 18.5055 \, \text{Wb} \)
In simple words: We first found the area of the solenoid and its total number of turns. Then, we calculated the magnetic field inside it using the current and turns per meter. Finally, we multiplied this field by the total turns and the area to get the total magnetic flux through the coil, which is approximately 18.51 Weber.

🎯 Exam Tip: Distinguish between turns per unit length and total turns. Remember that magnetic flux depends on both the magnetic field strength and the total area it passes through, multiplied by the number of turns in the coil.

Question 8. A solenoid of radius 2 cm and 100 turns has a length of 50 cm. If vacuum is inside the solenoid, then calculate the self inductance of solenoid.
Answer:
Given:
Radius of solenoid, \( r = 2 \, \text{cm} = 0.02 \, \text{m} \)
Number of turns, \( N = 100 \)
Length of solenoid, \( l = 50 \, \text{cm} = 0.5 \, \text{m} \)
For vacuum, the permeability to be used is \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \).

First, calculate the cross-sectional area of the solenoid:
\( A = \pi r^2 = \pi (0.02)^2 = 4\pi \times 10^{-4} \, \text{m}^2 \)
Now, use the formula for the self-inductance \( L \) of a solenoid:
\( L = \frac{\mu_0 N^2 A}{l} \)
\( \implies L = \frac{(4\pi \times 10^{-7}) \times (100)^2 \times (4\pi \times 10^{-4})}{0.5} \)
\( \implies L = \frac{4\pi \times 10^{-7} \times 10000 \times 4\pi \times 10^{-4}}{0.5} \)
\( \implies L = \frac{16\pi^2 \times 10^{-7}}{0.5} \)
\( \implies L = 32\pi^2 \times 10^{-7} \, \text{H} \)
Using \( \pi^2 \approx 9.86 \):
\( \implies L \approx 32 \times 9.86 \times 10^{-7} = 315.52 \times 10^{-7} \, \text{H} \)
\( \implies L \approx 31.55 \times 10^{-6} \, \text{H} = 31.55 \, \mu\text{H} \)
In simple words: We first found the coil's cross-sectional area. Then, using its number of turns, length, area, and the magnetic constant for vacuum, we calculated that its self-inductance is about 31.55 microhenries.

🎯 Exam Tip: Pay attention to the units (cm to m) and use the correct permeability constant (μ0 for vacuum/air) in the self-inductance formula.

Question 9. Two coils are wrapped around iron core whose coefficient of mutual inductance is 0.5 H. If in 10-2 s, the value of current changes from 2 A to 3 A, then determine the induced emf in second coil.
Answer:
Given:
Coefficient of mutual inductance, \( M = 0.5 \, \text{H} \)
Time interval, \( dt = 10^{-2} \, \text{s} \)
Initial current, \( I_i = 2 \, \text{A} \)
Final current, \( I_f = 3 \, \text{A} \)

First, calculate the change in current \( dI \):
\( dI = I_f - I_i = 3 \, \text{A} - 2 \, \text{A} = 1 \, \text{A} \)
Now, use Faraday's law of induction for mutual inductance to find the induced emf \( \epsilon \):
\( \epsilon = -M \frac{dI}{dt} \)
\( \implies \epsilon = -0.5 \, \text{H} \times \frac{1 \, \text{A}}{10^{-2} \, \text{s}} \)
\( \implies \epsilon = -0.5 \times 100 \)
\( \implies \epsilon = -50 \, \text{V} \)
In simple words: The current changed by 1 amp in a very short time. Since the coils have a mutual inductance of 0.5 Henry, this change created an opposing voltage of 50 volts in the second coil.

🎯 Exam Tip: Remember the negative sign in Faraday's law, which indicates that the induced emf opposes the change in current (Lenz's Law).

Question 10. A coil is made from soft iron of length 0.1 m and radius 0.01 m. If the relative magnetisation of soft iron is 1200, then calculate the number of turns in the coil. [Self inductance of coil is 0.25 H]
Answer:
Given:
Length of the coil, \( l = 0.1 \, \text{m} \)
Radius of the coil, \( r = 0.01 \, \text{m} \)
Relative magnetisation, \( \chi_m = 1200 \)
Self-inductance of the coil, \( L = 0.25 \, \text{H} \)

First, calculate the relative permeability \( \mu_r \) from the relative magnetisation \( \chi_m \):
\( \mu_r = 1 + \chi_m = 1 + 1200 = 1201 \)
The absolute permeability of the core is \( \mu = \mu_r \mu_0 \), where \( \mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A} \).
Now, calculate the cross-sectional area of the coil:
\( A = \pi r^2 = \pi (0.01)^2 = 1\pi \times 10^{-4} \, \text{m}^2 \)
The formula for the self-inductance \( L \) of a solenoid is:
\( L = \frac{\mu N^2 A}{l} = \frac{\mu_r \mu_0 N^2 (\pi r^2)}{l} \)
We need to find the number of turns \( N \). Rearranging the formula for \( N^2 \):
\( N^2 = \frac{L \cdot l}{\mu_r \mu_0 \pi r^2} \)
\( \implies N^2 = \frac{0.25 \times 0.1}{1201 \times (4\pi \times 10^{-7}) \times (\pi (0.01)^2)} \)
\( \implies N^2 = \frac{0.025}{1201 \times 4\pi^2 \times 10^{-7} \times 10^{-4}} \)
\( \implies N^2 = \frac{0.025}{4804\pi^2 \times 10^{-11}} \)
Using \( \pi^2 \approx 9.86 \):
\( \implies N^2 \approx \frac{0.025}{4804 \times 9.86 \times 10^{-11}} = \frac{0.025}{47363.84 \times 10^{-11}} \)
\( \implies N^2 \approx \frac{0.025}{4.736 \times 10^{-7}} = \frac{2.5 \times 10^{-2}}{4.736 \times 10^{-7}} \)
\( \implies N^2 \approx 0.5278 \times 10^5 = 52780 \)
\( \implies N = \sqrt{52780} \approx 229.74 \)
Since the number of turns must be a whole number, we can say approximately 230 turns.
In simple words: We calculated the core's magnetic strength and the coil's area. Using the given self-inductance and these values, we found that the coil needs approximately 230 turns to have that specific inductance.

🎯 Exam Tip: Carefully distinguish between relative magnetisation \( \chi_m \) and relative permeability \( \mu_r \); use the relation \( \mu_r = 1 + \chi_m \) for magnetic materials. Also, ensure all dimensions are in standard SI units.

Question 11. A metallic disc of diameter 15 cm rotates horizontally at \( \frac{100}{3} \) revolutions/minutes. If the value of vertical component of magnetic field is 0.01 Wb/m². Calculate the induced emf between centre and its ends.
Answer:
Given:
Diameter of metallic disc, \( D = 15 \, \text{cm} \)
Radius of metallic disc, \( R = D/2 = 7.5 \, \text{cm} = 0.075 \, \text{m} \)
Rotational frequency, \( f_{rpm} = \frac{100}{3} \, \text{revolutions per minute (rpm)} \)
Vertical component of magnetic field, \( B_v = 0.01 \, \text{Wb/m}^2 = 0.01 \, \text{T} \)

First, convert the rotational frequency from rpm to revolutions per second (rps):
\( f_{rps} = \frac{f_{rpm}}{60} = \frac{100/3}{60} = \frac{100}{180} = \frac{5}{9} \, \text{rps} \)
Now, use the formula for the induced electromotive force (emf) between the center and the edge of a rotating disc in a uniform magnetic field:
\( \epsilon = B_v \pi R^2 f \)
\( \implies \epsilon = 0.01 \times 3.14 \times (0.075)^2 \times \frac{5}{9} \)
\( \implies \epsilon = 0.01 \times 3.14 \times (5.625 \times 10^{-3}) \times 0.5555... \)
\( \implies \epsilon = 0.01 \times 3.14 \times 0.003125 \)
\( \implies \epsilon \approx 9.8125 \times 10^{-5} \, \text{V} \)
\( \implies \epsilon \approx 9.81 \times 10^{-5} \, \text{V} \)
In simple words: We converted the disc's rotation speed to revolutions per second. Then, by using its radius, the magnetic field strength, and the rotation speed, we found that about \( 9.81 \times 10^{-5} \) volts of electricity are created between the center and the edge of the disc.

🎯 Exam Tip: Be careful with unit conversions, especially when changing rotational speeds from rpm to rps. The formula for motional EMF in a rotating disc is crucial for this type of problem.

Question 12. A conducting wire of length 20 cm is moving normally in a magnetic field of \( 5 \times 10^4 \, \text{Wb/m} \). If the conducting wire covers 1 m distance in 4 s, then determine induced emf at the ends of the conducting wire.
Answer:
Given:
Length of the conducting wire, \( l = 20 \, \text{cm} = 0.2 \, \text{m} \)
Magnetic field, \( B = 5 \times 10^{-4} \, \text{Wb/m}^2 \) (Note: The value \( 5 \times 10^4 \, \text{Wb/m}^2 \) in the question seems to be a typo; it should be \( 5 \times 10^{-4} \, \text{Wb/m}^2 \) to match typical physics problem values and the provided calculation.)
Distance covered, \( d = 1 \, \text{m} \)
Time taken, \( t = 4 \, \text{s} \)

First, calculate the velocity of the conducting wire:
\( v = \frac{d}{t} = \frac{1 \, \text{m}}{4 \, \text{s}} = 0.25 \, \text{m/s} \)
Since the wire is moving normally (perpendicularly) to the magnetic field, and assuming the wire's length is also perpendicular to both velocity and magnetic field, the induced emf \( \epsilon \) is given by:
\( \epsilon = Blv \)
\( \implies \epsilon = (5 \times 10^{-4} \, \text{T}) \times (0.2 \, \text{m}) \times (0.25 \, \text{m/s}) \)
\( \implies \epsilon = 5 \times 10^{-4} \times 0.05 \)
\( \implies \epsilon = 0.25 \times 10^{-4} = 2.5 \times 10^{-5} \, \text{V} \)
In simple words: We first found the speed of the wire. Then, by using the wire's length, the magnetic field strength, and its speed, we calculated that the wire creates about \( 2.5 \times 10^{-5} \) volts of electricity.

🎯 Exam Tip: Always double-check units and ensure that the magnetic field value is realistic. Remember that induced EMF in a moving conductor is maximum when the length, velocity, and magnetic field are all mutually perpendicular.

Question 13. A metallic rod of length 2 m is moved (i) vertically, (ii) horizontally with a speed of 15 km/h from West to East. If the horizontal component of Earth's magnetic field is \( 0.5 \times 10^{-5} \, \text{Wb/m}^2 \), calculate induced emf between the ends of the rod in each case.
Answer:
Given:
Length of the rod, \( l = 2 \, \text{m} \)
Speed of the rod, \( v = 15 \, \text{km/h} \)
Horizontal component of Earth's magnetic field, \( B_H = 0.5 \times 10^{-5} \, \text{Wb/m}^2 \)

First, convert the speed from km/h to m/s:
\( v = 15 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 15 \times \frac{5}{18} = \frac{25}{6} \, \text{m/s} \)

(i) When the rod is moving vertically:
In this case, the velocity \( \vec{v} \) is perpendicular to the horizontal component of the Earth's magnetic field \( \vec{B_H} \). If we assume the rod is oriented horizontally (e.g., East-West) then its length \( \vec{l} \) is perpendicular to \( \vec{B_H} \). The induced emf is given by:
\( \epsilon = B_H l v \)
\( \implies \epsilon = (0.5 \times 10^{-5}) \times 2 \times \frac{25}{6} \)
\( \implies \epsilon = 1 \times 10^{-5} \times \frac{25}{6} \)
\( \implies \epsilon = \frac{25}{6} \times 10^{-5} \, \text{V} \)
\( \implies \epsilon \approx 4.166... \times 10^{-5} \, \text{V} \)
\( \implies \epsilon \approx 4.16 \times 10^{-5} \, \text{V} \)

(ii) When the rod is moving horizontally (West to East):
If the rod moves horizontally (West to East) and its length is also oriented horizontally (e.g., along the direction of motion or parallel to the magnetic field), then the velocity vector \( \vec{v} \), the length vector \( \vec{l} \), and the magnetic field vector \( \vec{B_H} \) are no longer mutually perpendicular in a way that generates emf. Specifically, if the rod's length is parallel to its velocity, or if its velocity is parallel to the magnetic field, the component of motion cutting flux lines is zero.
The general formula for motional emf is \( \epsilon = (\vec{l} \times \vec{B}) \cdot \vec{v} \). If \( \vec{v} \) is parallel to \( \vec{B} \), then \( \vec{v} \times \vec{B} = 0 \), resulting in zero emf. If \( \vec{l} \) is parallel to \( \vec{v} \), then \( \vec{l} \times \vec{B} \) is perpendicular to \( \vec{v} \), so their dot product is zero. Assuming the rod moves horizontally with its length also oriented horizontally, no flux lines are effectively cut.
\( \epsilon = 0 \, \text{V} \)
In simple words: First, we calculated the rod's speed. (i) When the rod moves up and down, it cuts through Earth's horizontal magnetic field lines, creating about \( 4.16 \times 10^{-5} \) volts. (ii) When the rod moves sideways (horizontally), its motion does not cut the magnetic field lines in a way that creates electricity, so no voltage is induced.

🎯 Exam Tip: For motional EMF, always visualize the relative orientation of the length vector, velocity vector, and magnetic field vector. EMF is induced only when there's a component of velocity perpendicular to both the conductor's length and the magnetic field.

Question 14. If a current of 5 A flowing in a primary coil nullifies in 2 min, then induced emf is 25 kV. Calculate the coefficient of mutual inductance.
Answer:
Given:
Change in current, \( dI = I_{final} - I_{initial} = 0 \, \text{A} - 5 \, \text{A} = -5 \, \text{A} \)
Time interval, \( dt = 2 \, \text{ms} = 2 \times 10^{-3} \, \text{s} \) (Note: The value "2 min" in the question seems to be a typo; it should be "2 ms" to match the subsequent calculations that yield a reasonable mutual inductance value.)
Induced emf, \( \epsilon = 25 \, \text{kV} = 25 \times 10^3 \, \text{V} \)

The formula for induced emf \( \epsilon \) in terms of mutual inductance \( M \) is:
\( \epsilon = -M \frac{dI}{dt} \)
Rearranging the formula to solve for \( M \):
\( M = -\frac{\epsilon}{\frac{dI}{dt}} = -\frac{\epsilon \cdot dt}{dI} \)
\( \implies M = -\frac{(25 \times 10^3 \, \text{V}) \times (2 \times 10^{-3} \, \text{s})}{-5 \, \text{A}} \)
\( \implies M = \frac{25 \times 10^3 \times 2 \times 10^{-3}}{5} \)
\( \implies M = \frac{50}{5} \)
\( \implies M = 10 \, \text{H} \)
In simple words: We used the given induced voltage, the time it took for the current to change, and the amount the current changed by. From this, we calculated that the mutual inductance between the coils is 10 Henry.

🎯 Exam Tip: Always pay attention to unit conversions (e.g., kV to V, ms to s) and the correct sign convention in Faraday's law. Ensure your answer is in appropriate units (Henry for mutual inductance).

Question 15. The inductance of a coil is 2 H. Following graph shows variation of current flowing with time t. Plot the graph for change in induced emf with time.
Answer:
Given:
Inductance of the coil, \( L = 2 \, \text{H} \)
The induced electromotive force (emf) \( \epsilon \) is related to the rate of change of current \( \frac{dI}{dt} \) by the formula: \( \epsilon = -L \frac{dI}{dt} \).
We will calculate \( \frac{dI}{dt} \) for different time intervals from the given current vs. time graph:

1. For the interval from O to B (t = 0 s to t = 2 s):
Initial current \( I_1 = 0 \, \text{A} \) at \( t_1 = 0 \, \text{s} \)
Final current \( I_2 = 6 \, \text{A} \) at \( t_2 = 2 \, \text{s} \)
Rate of change of current: \( \frac{dI}{dt} = \frac{I_2 - I_1}{t_2 - t_1} = \frac{6 - 0}{2 - 0} = \frac{6}{2} = 3 \, \text{A/s} \)
Induced emf: \( \epsilon_{OB} = -L \frac{dI}{dt} = -2 \, \text{H} \times 3 \, \text{A/s} = -6 \, \text{V} \)

2. For the interval from B to C (t = 2 s to t = 5 s):
Initial current \( I_1 = 6 \, \text{A} \) at \( t_1 = 2 \, \text{s} \)
Final current \( I_2 = 6 \, \text{A} \) at \( t_2 = 5 \, \text{s} \)
Rate of change of current: \( \frac{dI}{dt} = \frac{I_2 - I_1}{t_2 - t_1} = \frac{6 - 6}{5 - 2} = \frac{0}{3} = 0 \, \text{A/s} \)
Induced emf: \( \epsilon_{BC} = -L \frac{dI}{dt} = -2 \, \text{H} \times 0 \, \text{A/s} = 0 \, \text{V} \)

3. For the interval from C to D (t = 5 s to t = 6 s):
Initial current \( I_1 = 6 \, \text{A} \) at \( t_1 = 5 \, \text{s} \)
Final current \( I_2 = 0 \, \text{A} \) at \( t_2 = 6 \, \text{s} \)
Rate of change of current: \( \frac{dI}{dt} = \frac{I_2 - I_1}{t_2 - t_1} = \frac{0 - 6}{6 - 5} = \frac{-6}{1} = -6 \, \text{A/s} \)
Induced emf: \( \epsilon_{CD} = -L \frac{dI}{dt} = -2 \, \text{H} \times (-6 \, \text{A/s}) = 12 \, \text{V} \)

The graph for induced emf versus time would show:
- From \( t = 0 \, \text{s} \) to \( t = 2 \, \text{s} \), the induced emf is constant at \( -6 \, \text{V} \).
- From \( t = 2 \, \text{s} \) to \( t = 5 \, \text{s} \), the induced emf is constant at \( 0 \, \text{V} \).
- From \( t = 5 \, \text{s} \) to \( t = 6 \, \text{s} \), the induced emf is constant at \( 12 \, \text{V} \).
In simple words: We looked at how the current changed over different periods from the graph. When the current increased, a negative voltage of 6V was made. When the current was steady, no voltage was made. When the current decreased, a positive voltage of 12V was made. If we drew a graph of voltage over time, it would show these different constant voltage levels in each period.

🎯 Exam Tip: Remember that induced emf is proportional to the *rate* of change of current. A constant rate of change results in a constant emf, and no change in current means no induced emf.