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Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 6 Electromagnetic Induction. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 6 Important Extra Questions Electromagnetic Induction

Electromagnetic Induction Important Extra Questions Very Short Answer Type

Question 1.
What is the function of a step-up transformer? (CBSE AI 2011C)
Answer:
The function of a step-up transformer is to step-up the alternating voltage.

Question 2.
State Lenz’s law. (CBSE AI 2012C)
Answer:
It states that the direction of induced emf is such that it opposes the cause of its production.

Question 3.
How can the self-inductance of a given coil having ‘N’ number of turns, area of cross-section of ‘A’ and length T be increased? (CBSE AI 2012C)
Answer:
By inserting a core of high permeability inside the coil.

Question 4.
How does the mutual inductance of a pair of coils change when
(a) the distance between the coils is increased and
(b) the number of turns in the coils is increased? (CBSE AI 2013)
Answer:
(a) decreases
(b) increases.

Question 5.
The motion of the copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? (CBSE AI 2013)
Answer:
Production of eddy current.

Question 6.
Why is the core of a transformer laminated? (CBSE Delhi 2013C)
Answer:
To reduce the effects of eddy currents.

Question 7.
A metallic piece gets hot when surrounded by a coil carrying a high-frequency alternating current. Why? (CBSE Delhi 2014C)
Answer:
Due to the production of eddy current which generates heat.

Question 8.
Name any two applications where eddy currents are used to advantage. (CBSE Delhi 2016C)
Answer:

1. Electromagnetic damping
2. Induction furnace.

Question 9.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 10.
Predict the polarity of the capacitor in the situation described below. (CBSE AI 2017)

Answer:
The upper plate will be positive with respect to the lower plate in the capacitor.

Question 11.
In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. (CBSE AI 2017C)

Answer:
Plate A will be positive with respect to plate B in the capacitor.

Question 12.
A long straight current-carrying wire passes normally through the centre of the circular loop. If the current through the wire increases, will there be an induced emf in the loop? Justify. (CBSE Delhi 2017)
Answer:
Yes, as there will be a change in magnetic flux.

Question 13.
An air-cored solenoid has self-inductance 2.8 H. When the core is removed, the self-inductance becomes 2 mH. What is the relative permeability of the core used? (CBSE Delhi 2017C)
Answer:
μ_r = 2.8 / 2 × 10^-3 = 1.4 × 10³

Question 14.
A choke and a bulb are in series to a dc source. The bulb shines brightly. How does its brightness change when an iron core is inserted inside the choke coil?
Answer:
There is no change in the brightness as the inductive reactance is zero for dc.

Question 15.
Why does the acceleration of a magnet falling through a long solenoid decrease?
Answer:
It decreases because of the opposing induced emf produced in the soLenoid due to the rate of change of magnetic flux.

Question 16.
Why is the core of a transformer laminated?
Answer:
It is done to reduce the effect of Eddy Currents.

Question 17.
A vertical metallic pole falls down through the plane of the magnetic meridian. Will any emf be Induced between Its ends?
Answer:
No emf will be induced because the pote neither intercepts the vertical component nor the horizontal component of the earth’s magnetic field.

Question 18.
A magnet Is moved towards a coil and an electric charge is induced in it. If the resistance of the coil is increased, how will the induced charge change?
Answer:
On increasing the resistance of the colt, the magnitude of induced charge decreases.

Question 19.
Can a transformer be used In a dc circuit?
Answer:
No, because there is no change in magnetic flux.

Question 20.
Why does a metallic piece become very hot when it is surrounded by a coil carrying high-frequency alternating current?
Answer:
The high-frequency coil induces eddy currents in the metallic piece. These eddy currents produce heat hence the metalLic piece becomes hot.

Question 21.
The figure shows a horizontal solenoid PQ connected to a battery and a switch. A copper ring R is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. What would happen to the ring as the switch S is closed.

Answer:
The ring WilL be repelled due to opposing induced emf produced in it.

Question 22.
An air-core solenoid is connected to an ac source and a bulb. If an iron core Is inserted in the solenoid, how does the brightness of the bulb change? Give reasons for your answer.
Answer:
Insertion of an iron core In the solenoid increases its inductance. This in turn increases the value of inductive reactance. This decreases the current and hence the brightness of the bulb.

Question 23.
A magnet is moved in the direction indicated by an arrow between two coils AB and CD as shown in the figure. Suggest the direction of current in each coil,

Answer:
In coil AB induced current flows from A to B and in colt CD current flows from C to D.

Question 24.
Predict the directions of induced currents in metal rings 1 and 2 lying in the same place where current I in the wire is increasing steadily. (CBSE Delhi 2012)

Answer:
1 -clockwise, 2-anticlockwise.

Question 25.
The electric current flowing in a wire In the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown. (CBSE AI 2014)
Answer:
Clockwise.

Question 26.
Predict the polarity of plate A of the capacitor, when a magnet is moved towards it, as is shown In the figure. (CBSE AI 2014C)

Answer:
Positive.

Question 27.
The figure below shows two positions of a loop PQR in a perpendicular uniform magnetic field. In which position of the coil is there induced emf?

Answer:
In position b.

Question 28.
If the self-inductance of an air-core inductor increases from 0.01 mH to 10 mH on introducing an iron core into it, what is the relative permeability of the core used?
Answer:
We know that µ = \(\frac{L}{L_{0}}=\frac{10}{0.01}\) = 1000

Electromagnetic Induction Important Extra Questions Short Answer Type

Question 1.
An induced current has no direction of its own, comment.
Answer:
Yes, it is perfectly correct to say that an induced current has no fixed direction of its own. The direction of induced current depends upon the change in magnetic flux because in accordance with Lenz’s law the induced current always opposes the change in magnetic flux.

Question 2.
How are eddy currents produced? Mention two applications of eddy currents?
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with the conductor changes. When a thick conductor is moved in a magnetic field, magnetic flux linked with it changes. In situations like these, we can have induced currents that circulate throughout the volume of a material.

Because their flow patterns resemble swirling eddies in a river, therefore they are called eddy currents.

• Electromagnetic braking, and
• Induction furnace.

Question 3.
Name and define the unit used for measuring the coefficient of mutual inductance. State the relation of this unit with the units of magnetic flux and electric current.
Answer:
In SI the unit of mutual inductance is henry (H). Now from the expression
ε = – \(\frac{d \phi}{d t}\) = – M \(\frac{d l}{d t}\)
we have M = ε l \(\frac{d l}{d t}\).

Let ε = 1 volt and dl/dt = 1 As^-1, then
M = 1 volt/1 As^-1 = 1 henry.

The mutual-inductance of a coil is said to be 1 henry if a rate of change of current of 1 ampere per sec in the neighbouring coil induces in at an emf of 1 volt.

Question 4.
What are eddy currents? Write any two applications of eddy currents. (CBSE A! 2011)
Answer:
Eddy currents are the currents induced in the body of a thick conductor when the magnetic flux linked with a bulk piece of conductor changes.

1. Dead Beat Galvanometer, and
2. Induction furnace.

Question 5.
(a) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length l of the solenoid.
(b) How is this magnetic energy per unit volume compared with the electrostatic energy per unit volume stored in a parallel plate capacitor? (CBSE Delhi 2011C)
Answer:
The magnetic field stored in a solenoid is given by the expression U = – \(\frac{1}{2}\)Ll².

But for a solenoid B = μ₀nl
or
l = B / μ₀ n

Substituting in the above expression we have
U = \(\frac{1}{2}\) × (μ₀n²Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) as L = μ₀ n² A l

U = \(\frac{1}{2}\)\(\frac{B^{2} A l}{\mu_{0}}\)

We know that the energy stored per unit volume in a parallel plate capacitor is
U_E = \(\frac{1}{2}\)ε₀E²

It is clear that in both cases the energy stored per unit volume is proportional to the square of the field intensity.

Question 6.
State Lenz’s Law.
A metallic rod held horizontally along the east-west direction is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer. (CBSE Delhi 2013)
Answer:
Lenz’s law states that the polarity of the induced emf is such that it tends to oppose the cause of its production.

Yes, as it will cut the horizontal component of the earth’s magnetic field.

Question 7.
Starting from the expression for the energy W = 1/2Ll², stored in a solenoid of self-inductance L to build up the current l, obtain the expression for the magnetic energy in terms of the magnetic field B, area A and length l of the solenoid having n number of turns per unit length. Hence show that the energy density is given by 8z/2m0. (CBSE Delhi 2013C)
(i) The magnetic energy is
U = \(\frac{1}{2}\)LI² = \(\frac{1}{2}\)L\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) since B = μ₀nl

Now L = μ₀n² Al, therefore we have
U_B = \(\frac{1}{2}\)(μ0n2 Al)\(\left(\frac{B}{\mu_{0} n}\right)^{2}\) = \(\frac{1}{2 \mu_{0}} B^{2} A l\)

(ii) The magnetic energy per unit volume is
U_B = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 8.
Define mutual inductance. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (CBSE Delhi 2016)
Answer:
Mutual inductance is numerically equal to the magnetic flux linked with a coil when the unit current passes through the neighbouring coil.
Given M = 1.5 H, dl = 20 – 0 = 20 A,
dt = 0.5 s, Φ = ?
Φ = – M\(\frac{d l}{d t}\)
or
Φ = – 1.5 × \(\frac{20}{0.5}\) = – 60 Wb

Question 9.
Explain the principle on which the metal detector is used at airports for security reasons.
Answer:
The metal detectors used in airport security checkpoints operate by detecting eddy currents induced in metallic objects. The detector generates an alternating magnetic field. This induces eddy currents in the conduction object carried through the detector. The eddy currents in turn produce an alternating magnetic field. This field induces a current in the detectors receiver coil.

Question 10.
A current is induced in coil C₁, due to the motion of current-carrying coil C₂.
(i) Write any two ways by which a large deflection can be obtained in the galvanometer G.
(ii) Suggest an alternative device to demonstrate the induced current in place of a galvanometer. (CBSE Delhi 2011)

Answer:
(ii) The two ways are
(a) Passing a large current through coil C₂ and
(b) Moving coil C₂ quickly towards the coil.
(ii) A magnetic compass can be placed at the centre of coil C₁. Whenever current will be induced it will show a deflection.

Question 11.
Consider a cube EFGHIJKL of side ‘a’ placed in a magnetic field B acting perpendicular to the face FJKG as shown in the figure. Write magnetic flux through the following faces (a) EFGH (b) EFJI (c) EILH (d) FJKG

Answer:
The magnetic flux depends upon the angle (q) between the magnetic field and area vector.
(a) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(b) Here θ = 90°, therefore magnetic flux Φ = BA cos θ = BA cos 90° = 0
(c) Here θ = 180°, therefore magnetic flux Φ = BA cos 180° = -BA
(d) Here θ = 0°, therefore magnetic flux Φ = BA cos 0° = BA

Question 12.
A metallic rod of ‘L’ length is rotated with an angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. (CBSE Delhi 2012, 2013)
Answer:
Let the rod move from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA= πL² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L²θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)

ε = – \(\frac{1}{2}\)L2B\(\frac{d \theta}{d t}=\frac{1}{2}\)BωL²

Question 13.
A metallic rod of length l length is rotated with a frequency ω, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R = l, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf induced in the rod. If r is the resistance of the rod and the metallic ring has negligible resistance, obtain the expression for the power generated. (CBSE AI 2013C)
Answer:
For the first part refer above question.
The power generated is given by
P = \(\frac{\varepsilon^{2}}{r}=\frac{B^{2} \omega^{2} l^{4}}{2 r}\)

Question 14.
A rectangular loop PQMN with movable ‘ arm PQ of length 10 cm and resistance 4 Ω is placed in a uniform magnetic field of 0.25 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQare negligible.
Calculate the
(i) emf induced in the arm PQand
(ii) current induced in the loop when arm PQ is moved with velocity 20 m s^-1. (CBSE Delhi 2014C)

Answer:
Given L = 10 cm = 0.1 m, B = 0.25 T, v = 20 m s^-1
(a) ε = BLv = 0.25 × 0.1 × 20 = 0.5 V
(b) l = ε/R = 0.5/4 = 0.125 A

Question 15.
In the given diagram a coil B is connected to a low voltage bulb L and placed parallel to another coil A as shown. Explain the following observations
(i) Bulb lights, and
(ii) Bulb gets dimmer if coil B is moved upwards.

Answer:
(a) When ac is applied across coil A an induced emf is produced in coil B due to mutual induction between the two coils. This makes the lamp light up.
(b) When coil B is moved upwards the mutual induction and hence induced emf in coil B decreases. This makes the lamp dimmer.

Question 16.
A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B.
(a) Write the expression for the instantaneous value of the emf induced in the wire.
Answer:
ε = BLv

(b) What is the direction of the emf?
Answer:
west to east

(c) Which end of the wire is at the higher potential? (CBSEAI2011)
Answer:
East.

Question 17.
Two concentric circular coils one of small radius r₁ and the other of large radius r₂, such that r₁ << r₂, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. (NCERT)
Answer:
Let a current l₂ flow through the outer circular coil. The field at the centre of the coil is B₂ = \(\frac{\mu_{0} I_{2}}{2 r_{2}}\). Since the other coaxially placed coil has a very small radius, B2 may be considered constant over its cross¬sectional area. Hence,
Φ = πr₁²B2 = \(\frac{\mu_{0} \pi r_{1}^{2} l_{2}}{2 r_{2}}\) = M₁₂l₂

Thus M₁₂ = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

But M₁₂ = M₂₁, therefore M₁₂ = M₂₁ = \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)

Note that we calculated M₁₂ from an approximate value of Φ₁ assuming the magnetic field B₂ to be uniform over the area πr₁². However, we can accept this value because r₁ << r₂.

Question 18.
Consider a magnet surrounded by a wire with an on/off switch S (figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain. (NCERT Exemplar)

Answer:
There is no relative motion between the magnet and the coil. This means that there is no change in magnetic flux, hence no electromotive force is produced and hence no current will flow in the circuit.

Question 19.
A wire in the form of a tightly wound solenoid is connected to a DC source and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced e.m.f. resist this decrease, which can be done by an increase in current.

Question 20.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced e.m.f. should resist this increase, which can be achieved by a decrease in current.

Electromagnetic Induction Important Extra Questions Long Answer Type

Question 1.
11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22,000 V. Which of the two modes of transmission should be preferred and why? Support your answer with possible calculations.
Answer:
1. Consider that 11000 watt of energy has to be transmitted. First at 220 V and then at 22000 V. When the power is transmitted at 220 V then the current flowing through the wires is 11000/220 = 50 A

2. When power is transmitted at 22000 V then the current through the wires is 11000 / 22000 = 0.5 A. If R is the resistance of the line wire then the energy dissipated in the two cases is 2500R joule per sec and 0.25R. joule per sec.

This shows that if energy is transmitted at low voltages there is more loss in energy than when it is transmitted at high voltages. Furthermore, if power is to be transmitted at low voltage then the resistance of the line wire should be low, as such thick wires will be required. To support these thick wires strong poles situated close to each other will be needed. This will increase the cost of transmission. But at high voltages, even thin wires will do.

Question 2.
A coil A is connected to a voltmeter V and the other coil B to an alternating current source D. If a large copper sheet C is placed between the two coils, how does the induced emf in coil A change due to current in coil B. Justify your answer.

Answer:
In the absence of sheet C, an induced emf is set up in coil due to mutual induction phenomenon when an alternating current is passed through coil B.

However, when induced copper sheet C is placed, eddy currents are set up in the sheet due to a change in flux.

Thus, now coil A has a positive effect due to coil B and a negative effect due to eddy currents in C. Consequently, the flux of coil A and hence the induced emf in coil A is decreased, i.e. the reading of voltmeter V is reduced.

Question 3.
A bar magnet is dropped so that it falls vertically through coil C. The graph obtained for the voltage produced across the coil versus time is as shown in figure (b).
(i) Explain the shape of the graph and
(ii) why is the negative peak longer than the positive peak?

Answer:
(a) As the magnet approaches the coil, an emf is induced in it. As the magnet approaches the coil the magnetic flux linked with the coil increases. As a result the induced emf increases. When the magnet enters the coil, the change in magnetic flux linked with the coil begins to decrease and becomes zero when the magnet is completely inside the coil. This starts decreasing the emf and makes it zero.

When the magnet comes out of the coil the direction of induced emf changes direction and begins to increase in the opposite direction. When the magnet moves far away from the coil the induced emf becomes zero.

As the magnet comes out of the coil with a speed greater than the speed at which it approaches the coil, therefore the induced emf is more in the second case. Hence the longer negative peak.

Question 4.
(a) State Lenz’s law: Using this law indicate the direction of the current in a closed loop when a bar magnet with the North Pole is brought close to it. Explain briefly how the direction of the current predicted wrongly results in the violation of the law of conservation of energy.
(b) A rectangular loop and circular loop are moving out of a uniform magnetic field region with a constant velocity v as shown in the figure. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops. (CBSE AI 2011C)

Answer:
(a) Lenz’s law states that the direction of the induced emf is such that it opposes the cause of its production. When the north pole of the magnet is brought near the coil, the upper end of the coil will acquire north polarity so as to oppose the approaching North Pole (by repelling). This means that the direction of current must be anticlockwise as seen from the side of the magnet.

If the current is wrongly predicted as clockwise, then the upper face will acquire south polarity which will attract the North Pole. This means the current is being produced without doing any work. This leads to the violation of the law of conservation of energy.

(b) It is expected that the induced emf will be constant in the rectangular coil. In the case of the rectangular coil, when pulled out of the magnetic field, the rate of change of magnetic flux will be constant because the rate of change of area is constant. This is not so in the case of the circular coil.

Question 5.
The current through two inductors of self-inductance 15 mH and 25 mH is increasing with time at the same rate. Draw graphs showing the variation of the
(a) emf induced with the rate of change of current
(b) energy stored in each inductor with the current flowing through it. Compare the energy stored in the coils, if the power dissipated in the coils is the same. (CBSE AI 2017C)
Answer:
Given L₁ = 15 mH, L₂ = 25 mH.
(a) The emf induced across an inductor is given by the expression ε = – L\(\frac{d i}{d t}\) since di/dt is the same for both coils therefore induced emf will depend upon the value of inductance. Hence the graphs are as shown

(b) The energy stored in an inductor is given by U = \(\frac{1}{2}\)Ll²

(c) The power dissipated P = l₂X_L
or
l₂ = Pl XL and U = \(\frac{1}{2}\)Ll¹ = \(\frac{1}{2} \frac{L P}{X_{L}}=\frac{1}{2} \frac{P L}{\omega L}=\frac{P}{2 \omega}\)

Since power dissipated is same and co is also same therefore energy stored in the coils will also be the same.

Question 6.
A rectangular frame of wire is placed in a uniform magnetic field directed outwards, normal to the paper. AB is connected to a spring which is stretched to AB and then released at time t = 0. Explain qualitatively how induced e.m.f. in the coil would vary with time. (Neglect damping of oscillations of spring) (CBSE AI 2018C)

Answer:
As the spring is released AB is pulled out of the field. This increases the area of the loop inside the magnetic field. This increases flux and hence an induced emf is produced. The portion AB does not stop at Ab but moves outwards. Now the spring will push AB inwards. This will decrease the area of the loop thereby decreasing the induced emf. This continues and hence the emf increases and decreases periodically.

Question 7.
Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of the same length wound one over the other. (CBSE AI 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ0 n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N₁ N₂ of their number of turns. This is termed the reciprocity theorem.

Question 8.
Define mutual inductance and write its SI unit.
A square loop of side ‘a’ carrying a current l₂ is kept at distance x from an infinitely long straight wire carrying a current l1 as shown in the figure. Obtain the expression for the resultant force acting on the loop. (CBSE Delhi 2019)

Answer:
(a) Mutual inductance equals the magnetic flux associated with a coil when unit current flows in its neighbouring coil. (b) Force per unit length between two parallel straight conductors
\(\frac{F}{L}=\frac{\mu_{o} I_{1} I_{2}}{2 \pi r}\)

Force on the part of the loop which is parallel to infinite straight wire and at a distance x from it F = \(\frac{\mu_{o} l_{1} l_{2} a}{2 \pi x}\)(away from the infinitely 2KX long wire)

Force on the part of the loop which is at a distance (x + a) from it
F₂ = \(\frac{\mu_{0} I_{1} I_{2} a}{2 \pi(x+a)}\) (towards the infinite long wire)

Net force F = F₁ – F₂
F = \(\frac{\mu_{o} l_{1} l_{2} a^{2}}{2 \pi(x+a)}\) (away from the infinite straight line)

Question 9.
What are the possible causes of energy loss in a transformer? How are these minimised?
Answer:
The possible causes of energy losses in transformers are:
(a) Flux leakage: There is always some flux leakage. It can be reduced by winding the primary and secondary coils one over the other.
(b) Copper: The copper wires used for the windings have some resistance and hence some energy is lost due to heat produced in the wire. It can be minimised by taking thick wire.
(c) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core, which results in loss of electrical energy. To minimise it we use a laminated iron core.
(d) Hysteresis loss: As the magnetisation cycle of the iron core is repeated again and again some loss of energy takes place due to magnetic hysteresis. To minimise it we prefer a soft iron core for which hysteresis loss is less.

Question 10.
Discuss how Faraday’s law of e.m.f induction is applied in an ac generator for converting mechanical energy into electrical energy. Draw graphs to show the ’phase relationship’ between the instantaneous
(a) magnetic flux (Φ) linked with the coil and
(b) induced emf (ε) in the coil. (CBSE Delhi 2016C)
Answer:
In an ac generator, the change in magnetic flux is brought about by rotating the coil in a magnetic field. According to Faraday’s law, induced emf is set up in the coil on changing the magnetic flux linked with it. Hence mechanical energy, which is supplied to rotate the coil, gets converted into electrical energy.

Let the coil be rotated with a constant angular speed, co, the angle 0 between magnetic field 8, and area vector A of the coil at any instant is θ = ωt

Therefore magnetic flux at any instant is Φ = BA cos ωt

From Faraday’s law, induced emf is
ε = – n\(\frac{d \phi}{d t}\) = -nBA\(\frac{d}{d t}\)cos ωt
ε = nBAa sin ωt

Magnetic flux is given by Φ = BA cos ωt, therefore the graph will be a cosine curve as shown

The induced emf is given by ε = nBA sin ωt, therefore the graph will be a sine curve as shown.

Question 11.
(a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of self-inductance ‘L’ when the current through it grows from zero to l.
(b) A square loop MNOP of the side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cm s^-1 till it goes out of the field.

(c) Depict the direction of the induced current in the loop as it goes out of the field. For how long would the current in the loop persist?
(d) Plot a graph showing the variation of magnetic flux and induced emf as a function of time. (CBSE AI 2015)
Answer:
(a) Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = V_ab i = Li \(\frac{d i}{d t}\)

The energy dU supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is
U = L ∫_o¹ i di = \(\frac{1}{2}\)L l²

(b) (i) Direction of induced current – clockwise (MNOP), the duration of induced current is 1s.
(ii) The graph is as shown.

Question 12.
(a) Define mutual inductance and write its SI units.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil. It is measured in henry.

(b) Derive an expression for the mutual inductance of two long co-axial solenoids of the same length wound one over the other.
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ_o n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ0n2AN1 = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N1 N2 of their number of turns. This is termed the reciprocity theorem.

(c) In an experiment, two coils c₁ and c₂ are placed close to each other. Find out the expression for the emf induced in the coil c₁ due to a change in the current through the coil c₂. (CBSE Delhi 2015)
Answer:
Consider the experimental set up as shown.

When the key K is pressed in coil C₁ a magnetic flux linked with C₂ changes.

The magnetic flux linked with coil C₂ is given by
f₂ = M l₁ where M is the mutual inductance between the two coils.

But by Faraday’s flux rule we have
ε₂ = – \(\frac{d \phi}{d t}\) therefore we have
ε₂ = – \(\frac{d M l_{1}}{d t}\) = – M \(\frac{d l_{1}}{d t}\)

Question 13.
(a) When a bar magnet is pushed towards (or away) from the coil connected to a galvanometer, the pointer in the galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depend. State the laws describing this phenomenon.
(b) Sketch the change In flux, emf and force when a conducting rod PQ of resistance R and length I moves freely to and fro between A and C with speed v on a rectangular conductor placed in the uniform magnetic field as shown in the figure. (CBSE AI 2016)

Answer:
(a) Phenomenon: electromagnetic induction

Factors:

• Strength of the magnetic field of the magnet
• speed of motion of bar magnet.

Direction depends upon

• the motion of magnet whether inward or outward
• the direction of the north/south pole.

Law: The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit.

(b) The sketch is as shown below.

Question 14.
What is induced emf? Write Faraday’s laws of electromagnetic induction. Express it mathematically. A conducting rod of length ‘L’ with one end pivoted is rotated with a uniform angular velocity ‘ω’ in a vertical plane, normal to a uniform magnetic field ‘B’. Deduce an expression for the emf induced in this rod.
Answer:
It is the emf induced when the magnetic flux linked with a coil changes.

Faraday put forward the following laws called Faraday’s laws of electromagnetic induction.

• Law 1. Whenever the magnetic flux linked with a coil changes an induced emf is produced.
• Law 2. The induced emf lasts as long as the change in magnetic flux continues.

The expression for induced emf in rotating rod: Let the rod move from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA = πL² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)L2θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is
ε = \(\frac{d \phi}{d t}=\frac{d B A}{d t}=\frac{B d A}{d t}=B \frac{d}{d t}\left[\frac{1}{2} L^{2} \theta\right]\)
Or
ε = \(\frac{1}{2} L^{2} B \frac{d \theta}{d t}=\frac{1}{2}\)BωL²

Question 15.
How is the mutual inductance of a pair of coils affected when:
(a) the separation between the coils is increased?
Answer:
When the separation between the two coils is increased the mutual inductance decreases. It is because the flux linked with the secondary due to a current in the primary decreases.

(b) the number of turns of each coil is increased?
Answer:
When the number of turns of each coil is increased the mutual inductance increases because M₁₂ (= M₂₁) ∝ N₁N₂,

i.e. mutual inductance is directly proportional to the number of turns. The linkage of flux increases with an increase in the number of turns in the coils.

(c) a thin iron sheet is placed between the two coils, other factors remaining the same? Explain your answer in each case.
Answer:
When a thin iron sheet is placed between the coils the permeability of the medium between the coil increases. As mutual inductance is directly proportional to the permeability, therefore the mutual inductance increases.

Question 16.
A rectangular coil of area A, having a number of turns N, is rotated at f revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that the maximum emf induced in the coil is 2π f N B A.
Answer:
Consider an armature of the ac generator having n turns and placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle 0 with the direction of the magnetic field. If r_o is the uniform angular velocity with which, the coil rotates then θ = ωt.

The flux through the loop equals its area A multiplied by B_⊥ = Bcos θ, the component of magnetic field B perpendicular to the area, hence

Φ = n B A cos Φ = n B A cos ω t …(1)
where is the number of turns in the armature?

By Faraday’s flux rule,
ε = – \(\frac{d \phi}{d t}=-\frac{d}{d t}\) n B A cos ω t
= – n B A \(\frac{d}{d \phi}\)cos ω t = – n B A (- ω sin ω t) ….(2)
Or
ε = n B A ω sin ω t

The induced emf is maximum when sin ωt = maximum = 1, therefore the maximum induced emf is given by
ε₀ = n B A ω sin ω t …(3)

Since ω = 2πf, therefore equation (3) becomes ε₀ = 2πf nBA

Question 17.
State Lenz’s law. Explain by giving examples that Lenz’s law is a consequence of the law of conservation of energy. (CBSE Delhi 2017C)
Answer:
It states that the direction of induced emf in a coil is such that it opposes the cause of its production. Lenz’s law is a consequence of the law of conservation of energy. To show it, let us consider a bar magnet pushed towards a conducting loop. When N-pole moves towards the loop, the face of the loop facing the North Pole develops north polarity as per Lenz’s law so as to oppose the motion of the magnet.

Again, when N-pole moves away from the loop, the nearby face develops south polarity, thus opposing the motion of the magnet away from the loop. It means that the motion of the magnet is automatically opposed every time. Hence, some work is to be done on the magnet to move it and this mechanical work is transformed into electrical energy. Thus, the conservation law of energy is followed.

Question 18.
Define the term ‘self-inductance’ and write its SI unit.
Obtain the expression for the mutual inductance of two long co-axial solenoids S₁ and S₂ wound one over the other, each of length L and radii r₁ and r₂ and n₁ and n₂ number of turns per unit length when a current I is set up in the outer solenoid S₂. (CBSE Delhi 2017)
Answer:
The mutual inductance of two coils is numerically equal to the magnetic flux linked with one coil when a unit current flows through the neighbouring coil.

As shown in the figure consider two long co-axial solenoids S₁ and S₂ with S₂ wound over S₁.

Let l = length of each solenoid r₁, r₂ = radii of the two solenoids

A = πr₁² = area of cross-section of inner solenoid S₁
N₁, N₂ = number of turns in the two solenoids First we pass a time-varying current l₂ through S₂. The magnet field set up inside S₂ due to l₂ is
B₂ = μ_o n₂ l₂

where n₂ = N₂/l = the number of turns per unit length of S₂.

Total magnetic flux linked with the inner solenoid S₁ is
Φ₁ = B₂AN₁ = μ_o n₂ l₂ × AN₁

Mutual inductance of coil 1 with respect to coil 2 is
M₁₂ = \(\frac{\phi_{1}}{l_{2}}\) = μ_on₂AN₁ = \(\frac{\mu_{0} A N_{1} N_{2}}{l}=\frac{\mu_{0} \pi r^{2} N_{1} N_{2}}{l}\)

The mutual inductance of any two coils is always proportional to the product N₁ N₂ of their number of turns. This is termed the reciprocity theorem.

Question 19.
Obtain the expression for the magnetic energy stored in an ideal inductor of self-inductance L when a current l passes through it. Hence obtain the expression for the energy density of magnetic field B produced in the inductor. (CBSE Delhi 2016C)
Answer:
Let i be the current at some instant through a pure inductor. If di / dt is the rate of change of current through the inductor then the voltage between the terminals of the inductor at this instant is
V=L di / dt

Therefore instantaneous power in the inductor is
P = V_ab i = Li\(\frac{d i}{d t}\)

The energy di supplied to the inductor during an infinitesimal time interval dt is
dU = P dt, so dU = Li di

The total amount of energy supplied while the current increases from zero to a final value l is

U = L ∫₀¹ di = \(\frac{1}{2}\)L l²

This gives the expression for the energy stored in an inductor.

The magnetic energy is

The magnetic energy per unit volume is
U_B = \(\frac{U_{B}}{V}=\frac{U_{B}}{A l}=\frac{B^{2}}{2 \mu_{0}}\)

Question 20.
Derive the expression for the self-inductance of a solenoid.
Answer:
Consider a uniformly wound solenoid with N turns and length l. Let the length of the solenoid be large as compared to its radius and let the core of the solenoid have air. Due to this, we can take the interior field uniform. Therefore the magnetic field in the interior of the solenoid is given by

B = μ₀ n l = μ₀ \(\frac{N}{l}\) l …(1)

where n is the number of turns per unit length. The flux through each turn is given by
Φ_m = B A = μ₀ \(\frac{N A}{l}\) l …(2)

where A Is the area of cross-section of the solenoid. But
L = \(\frac{N \Phi_{\mathrm{m}}}{I}\) …(3)

Therefore from equations (2) and (3) we have
L = \(\frac{N \Phi_{m}}{l}=\frac{\mu_{0} N^{2} A}{l}\) ….(4)

This shows that L depends on the geometric factors and is proportional to the square of the number of turns. Since N N = n l, we can therefore express the above result as
L = µ₀\(\frac{(n l)^{2}}{l}\)A = µ₀ n² A l …(5)

Question 21.
(a) State Faraday’s laws of electromagnetic Induction.
Answer:
Faraday’s Laws of electromagnetic induction
First law: Whenever the magnetic flux (inked with a circuit (or coil) changes, an emf is induced in the circuit. The induced emf Lasts so Long as the change in the magnetic flux continues.

Second law:
The magnitude of induced emf in the circuit (or coil) is directly proportional to the rate of change of magnetic flux linked with the circuit.
i.e. e = – \(\frac{d \phi}{d t}\)

(b) Derive an expression for the emf induced across the ends of a straight conductor of length I moving at right angles to a uniform magnetic field B with a uniform speedy.
Answer:
Let a conductor ab of Length l be placed in a magnetic field \(\vec{B}\) shown by (x) directed towards the reader. When the conductor moves with velocity v perpendicular to B, the force on any free electron of this conductor is given

F = e vB sin 90° = evß

When this electron moves across the Length of the conductor, the work done by the electron
W = F × l = evBl

After some time when all the free electrons are shifted towards the end b, the end b becomes negative and end a becomes positively charged.

Hence the net induced emf in the conductor
e = \(\frac{W}{q}=\frac{e v B l}{e}=v B l\)
e = Blv

(c) Obtain the expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length I of the solenoid through which a current j is passed. (CBSE 2019C)
Answer:
Energy stored in an inductor Considers an inductor of inductance L. Let I will be instantaneous current in the inductor and dl/dt be the rate of growth of current.

Induced emf, E = L\(\frac{dl}{dt}\) (in magnitude)

If the source sends a charge dq in time dt, then
dq = ldt

Small amount of work done by the source
dW = edq = eldt = L\(\frac{dl}{dt}\). ldt
or
dW = Ll dl

Total work was done during the growth of current in an Inductor by the external source
W = ∫dW =L∫₀¹ ldl
= L \(\left[\frac{l^{2}}{2}\right]_{0}=\frac{1}{2}\)Ll²
W = \(\frac{1}{2}\) Ll² …(1)

This work done is stored in the form of magnetic energy.
In a solenoid,
L = μ₀n²Al
And B = μ₀nl
∴ l = \(\frac{B}{\mu_{0} n}\)

So Eq. (1) becomes
W = \(\frac{1}{2}\)μ0n2Al\(\left(\frac{B}{\mu_{0} n}\right)^{2}=\frac{B^{2} A l}{2 \mu_{0}}\)

Question 22.
(a) A metallic rod of length ‘l’ and resistance ‘R’ is rotated with a frequency ‘v’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius ‘l’, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field ‘B’ parallel to the axis is present everywhere.
(i) Derive the expression for the induced emf and the current in the rod.
(ii) Due to the presence of current in the rod and of the magnetic field, find the expression for the magnitude and direction of the force acting on this rod.
(iii) Hence, obtain an expression for the power required to rotate the rod.
(b) A copper coil l is taken out of a magnetic field with a fixed velocity. Will it be easy to remove it from the same field if its ohmic resistance is increased?
OR
(a) A rectangular coil rotates in a uniform magnetic field. Obtain an expression for induced emf and current at any instant. Also find their peak values. Show the variation of induced emf versus angle of rotation (ωt) on a graph.
(b) An iron bar falling through the hollow region of a thick cylindrical shell made of copper experiences a retarding force. What can you conclude about the nature of the iron bar? Explain. (CBSEAI 2019)
Answer:
(a) (/) Let the rod moves from OP to OQ through a small sector of angle θ.

The small area covered by the rod is
dA = πl² × \(\frac{\theta}{2 \pi}=\frac{1}{2}\)l²θ

where L is the radius of the circle in which the rod rotates. Hence the induced emf is

Now induced current is given by
i = \(\frac{\varepsilon}{R}=\frac{B \omega l^{2}}{2 R}\)

(ii) \(\vec{F}=i(\vec{l} \times \vec{B})\)
F = \(\frac{l^{3} \omega B}{2 R}\)

Direction of \(\vec{F}\) is perpendicular to both \(\vec{i}\) and \(\vec{B}\),

(iii) P = i²R
= \(\left(\frac{B l^{2} \omega}{2 R}\right)^{2}\)R = \(\frac{B^{2} l^{4} \omega^{2}}{4 R}\)

(b) Yes, since induced current will reduce, it will be a little easier to remove the coil.
Or
(a) Consider an armature of the ac generator having n turns placed in a uniform magnetic field B.

Suppose at any instant t the normal to the plane of the coil makes an angle θ with the direction of the magnetic field. If ω is the uniform angular velocity with which the coil rotates then θ = ωt.

The flux through the loop is Φ = n BA cos Φ = n BA cos ω t
where n is the number of turns in the armature.

By Faraday’s flux rule,

Now,i₀ = \(\frac{\varepsilon_{0}}{R}=\frac{n B A \omega}{R}\)

The graph is as shown.

(b) The bar is magnetic in nature. It experiences retardation in accordance with Lenz’s law.

Question 23.
State Lenz’s law. The energy f required to build up a steady current l, in a given coil, varies with l in the manner as shown. Calculate the self-inductance of the coil.

A circular coil of radius r, is placed coaxially with another coil of radius R (R >> r) with the centre of the two coils coinciding with each other. Obtain an expression for the mutual inductance of the two coils.
Answer:
Lenz’s law states that the direction of induced emf is such that it opposes the cause of its production.
The energy stored in an inductor is given by the relation
E = – \(\frac{1}{2}\)Ll²

From the graph for a current of 200 mA the energy is 4 mJ, therefore self-inductance of the coil.
\(\frac{2 E}{l^{2}}\) = L
or
\(\frac{2 \times 4 \times 10^{-3}}{\left(200 \times 10^{-3}\right)^{2}}\) = 0.2 H
or
L = 200 mH

Consider the two coils of radius r and R placed coaxially, as shown in the figure.

Let n₁ and n₂ be the number of turns per unit length in them.

If a current l₂ is passed through the outer coil, there will a magnetic field
B_o = μ_o n₁ l₁ produced in it.

The flux associated with the inner coil will undergo variation as the field B0 grows.
The effective magnetic flux
Φ₁ = B_o × A_eff = \(\frac{\mu_{0} l_{2}}{2 R}\) × πr²

Therefore mutual inductance of the coils is
M = \(\frac{\phi_{1}}{l_{2}}=\frac{\mu_{0} \pi r^{2}}{2 R}\)

Question 24.
(a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is always such that it tends to produce a current which opposes the change of magnetic flux that produces it.
(b) The current flowing through an inductor of self-inductance L is continuously increasing. Plot a graph showing the variation of
(i) Magnetic flux versus the current
(ii) induced emf versus dl/dt
(iii) Magnetic potential energy stored versus the current. (CBSE Delhi 2014)
Answer:
(a) Consider a coil C connected to a galvanometer. When the North Pole of a bar magnet is pushed towards the coil, the pointer in the galvanometer deflects, indicating the presence of electric current in the coil. When the magnet is pulled away from the coil, the galvanometer shows deflection in the opposite direction, which indicates a reversal of the current’s direction.

We see that the North Pole of a bar magnet is being pushed towards the closed coil. As the North Pole of the bar magnet moves towards the coil, the magnetic flux through the coil increases. Hence current is induced in the coil in such a direction that it opposes the increase in flux.

This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet. Similarly, if the North Pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux, the induced current in the coil flows in a clockwise direction and the South Pole faces the receding North Pole of the bar magnet. This would result in an attractive force that opposes the motion of the magnet and the corresponding decrease in flux.

(b) The graphs are as shown.
(i) Φ = Ll
Therefore the graph is

(ii) ε = – L\(\frac{dl}{dt}\)
Therefore the graph is

(iii) U = \(\frac{1}{2}\)Ll²
Therefore the graph is

Numerical Problems:

Formulae for solving numerical problems

• Induced emf ε = –\(\frac{d \phi}{d t}\)
  or
  ε = \(\frac{-\left(\phi_{2}-\phi_{1}\right)}{t_{2}-t_{1}}\)
• Emf induced in a rod of Length L is ε = Blv.
• When a conducting rod of length L kept perpendicular to a uniform magnetic field B is rotated about one of its ends with uniform angular velocity, the emf induced between its ends has a magnitude
  \(\frac{1}{2}\)BωL² = BπvL².
• When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by
  ε = – \(\frac{d \phi}{d t}\) = – L \(\frac{d l}{d t}\) inductance of the coil. It is a measure of the inertia of the coil against the change of current through it. Also Φ = L l
• ε = – M\(\frac{d l}{d t}\) . Also Φ = Ml for mutual induction.
• \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{l_{p}}{l_{s}}\) = k.
• Average power η = \(\frac{\varepsilon_{s} l_{s}}{\varepsilon_{p} l_{p}}\)

Question 1.
A square loop of side 10 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^-1 in the positive X-direction containing a magnetic field in the positive Z-direction. The field is non-uniform and has a gradient of 10^-3 T cm^-1 along the negative X-direction (i.e. it increases by 10^-3 T cm^-1 as one move in the negative X-direction). Calculate the emf induced. (CBSE 2019C)
Answer:

Question 2.
In a ceiling fan, each blade rotates in a circle of radius 0.5 m. If the fan makes 2 rotations per second and the vertical component of the earth’s magnetic field is 8 × 10^-5 T, calculate the emf induced between the inner and outer ends of each blade. (CBSE2019C)
Answer:

Question 3.
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω are placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and the current induced in the coil. The horizontal component of the earth’s magnetic field at the place is 3.0 × 10^-5 T. (NCERT)
Answer:
Initial flux through the coil
Φ_in = BA cos θ = 3.0 × 10^-5 × π × 10^-2 × cos 0°
= 3π × 10^-7 Wb

Final flux after the rotation
Φ_f = BA cos θ = 3.0 × 10^-5 × π × 10^-2 × cos 180° = – 3π × 10^-7 Wb

Therefore estimated value of induced emf is
ε = n\(\frac{d \phi}{d t}\) = 500 × (6π × 10^-7)/ 0.25 = 3.8 × 10^-3 V dt

Hence the current induced is
l = ε/R = 3.8 × 10^-3 / 2 = 1.9 × 10^-3 A

Question 4.
Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m². The coil rotates at half a revolution per second and is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? (NCERT)
Answer:
Given n = 100, A = 0.10 m2, B = 0.01 T, f = 0.5 rps
Using the expression ε₀ = nBAω

we have ε₀ = 100 × 0.01 × 0.10 × 2 × 3.14 × 0.5 = 0.314 V

Question 5.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? (NCERT)
Answer:
Given L = 8 cm = 8 × 10^-2 m, b = 2 cm = 2 × 10^-2 m, 8 = 0.3 T, v = 1 cm s^-1 = 0.01 m s^-1 ε = ?
(a) Using the expression ε = B L v
ε = 0.3 × 8 × 10^-2 × 0.01 = 2.4 × 10^-4 V

This emf will last till the loop comes out of the magnetic field. Since the shorter side is moving out, therefore it will take 2 s to cover 2 cm with a velocity of 1 cm s^-1.

(b) Using the expression e = B L v
ε = 0.3 × 2 × 10^-2 × 0.01 = 0.6 × 10^-4 V

This emf will last till the loop comes out of the magnetic field. Since the longer side is moving out, therefore it will take 8 s to cover 8 cm with a velocity of 1 cm s^-1.

Question 6.
A 1.0 m long metallic rod is rotated with an angular frequency 400 rad s-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. (NCERT)
Answer:
Given L = 1.0 m, ω = 400 rad s^-1, B = 0.5 T,

ε = ? Using the relation ε = \(\frac{1}{2}\)BωL²

we have
ε = \(\frac{1}{2}\)0.5 × 400 × 1² =100 V

Question 7.
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^-1 in a uniform horizontal magnetic field of magnitude 3.0 × 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? (NCERT)
Answer:
Given r = 8.0 cm = 8.0 × 10^-2 m, n = 20, ω = 50 rad s-1 B = 3.0 × 10^-2 T, ε₀ = ?, ε_a = ? R= 10 Ω l = ?, P = ?
Using the relation
ε0 = nBAω
= 20 × 3 × 10^-2 × 3.14 × (8 × 10^-2)² × 50 = 0.603 V

The average induced emf is given by
εa = 0 over one cycle.

Also l₀ = ε₀ / R
= 0.603 / l₀
= 0.0603 A

Now P = 1 /2 ε₀ × l₀ = 1 /2 × 0.603 × 0.0603
= 0.018 W

The induced current causes a torque opposing the rotation of the coil. An external agent (rotor) must supply torque to counter this torque in order to keep the coil rotating uniformly. Thus the source of the power dissipated as heat in the coil is the external rotor.

Question 8.
Current in a circuit falls from 5.0 A to 0. 0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit (NCERT)
Answer:
Given dl = 5.0 – 0.0 = 5.0 A, dt = 0.1 s, ε = 200 V, L = ?

Using the relation ε = – L \(\frac{dl}{dt}\)
or
L = \(\frac{\varepsilon}{d l / d t}\)

L = \(\frac{200}{5 / 0.1}=\frac{200 \times 0.1}{5}\) = 4 H

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? (NCERT)
Answer:
Given dl = 20- 0.0 = 20 A, dt = 0.5 s, ε = ?,
M = 1.5 H,
Using the relation ε = – \(\frac{d \phi}{d t}\) = -M\(\frac{dl}{dt}\)
or
dΦ = Mdl = 1.5 × 20 = 30 Wb

Question 10.
A jet plane is travelling towards the west at a speed of 1800 km h^-1. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the earth’s magnetic field at the location has a magnitude of 5 × 10^-4 T and the dip angle is 30°? (NCERT)
Answer:
Given v= 1800 km h-1, ε = ?, L = 25 m,
B = 5 × 10^-4 T, δ = 30°
Now vertical component of earth’s magnetic field is
B_V = B × sin 30°
= 5 × 10^-4 × 0.5
= 2.5 × 10^-4T
Now using the expression ε = B L v
ε = 2.5 × 10^-4 × 25 × 500 = 3.125 V

Question 11.
1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of a radius of 0.5 cm for this purpose. Calculate the fraction of ohmic losses to the power transmitted if
(a) power is transmitted at 220 V. Comment on the feasibility of doing this.
(b) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring the voltage to 220 V. ρ_cu = 1.7 × 10^-8 SI unit. (NCERT Exemplar)
Answer:
Given P = 1 MW, 2L = 20 km = 2 × 104 m, r = 0.5 cm = 0.5 × 10^-2 m
(a) Resistance of the Cu wire used
R = \(\frac{\rho L}{A}=\frac{1.7 \times 10^{-8} \times 2 \times 10^{4}}{\pi \times\left(0.5 \times 10^{-2}\right)^{2}}\) = 4 Ω

Now current at 220 V is

l = \(\frac{P}{V}=\frac{10^{6}}{220}\) = 0.45 × 10⁴ A

Therefore power loss
= l²R = (0.45 × 10⁴)² × 4 = 10⁶W

This is a huge loss; therefore this method is not feasible.

(b) Now P = Vl’ or
l’ = \(\frac{P}{V^{\prime}}=\frac{10^{6}}{11000}=\frac{1}{1.1}\) × 10² A

Hence power loss
= l’²R = \(\frac{1}{1.21}\) × 4 × 10⁴ = 3.3 × 10⁴ W

Hence fraction of power loss
\(\frac{3.3 \times 10^{4}}{10^{6}}\) = 3.3%

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 6 General Principles and Processes of Isolation of Elements. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 6 Important Extra Questions General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements Important Extra Questions Very Short Answer Type

Question 1.
What is the composition of copper matte? (CBSE Delhi 2013)
Answer:
Copper matte contains cuprous sulphide (Cu₂S) and iron sulphide (FeS).

Question 2.
Name the method used for refining of
(i) Nickel
(ii) Zirconium (CBSE Sample Paper 2007, 2014)
Answer:
(i) Mond’s process
(ii) Van Arkel method

Question 3.
Write the overall reaction taking place in the process used for the electrolysis of alumina by Hall-Heroult process. (CBSE Sample Paper 2011)
Answer:
2Al₂O₃ + 3C → 4Al + 3CO₂

Question 4.
Write a non-exothermic reaction taking place in the blast furnace during extraction of iron. (CBSE Sample Paper 2011)
Answer:

Question 5.
Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
Answer:
Zone refining.

Question 6.
In the extraction of Al, impure Al₂O₃ is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of the process? (CBSE Delhi 2016)
Answer:
Al₂O₃ + 2NaOH + 3H₂O → 2Na [Al(OH)₄]
This process is called leaching.

Question 7.
Name the method of refining which is based on the principle of adsorption.
(CBSE AI 2016)
Answer:
Chromatography.

Question 8.
Out of PbS and PbCO₃ (ores of lead), which one is concentrated by froth floatation process preferably? (CBSE Delhi 2017)
Answer:
PbS

Question 9.
What is the role of depressants in the froth floatation process? (CBSE Al 2017)
Answer:
The depressants in the froth floatation process selectively prevent one of the sulphide ores from forming the froth with air bubbles.

Question 10.
What is the role of collector and froth stabilizer in froth floatation process?
(CBSE AI 2012)
Answer:

• Collector enhances non-wettability of the mineral particles.
• Froth stabilisers stabilise the froth.

Question 11.
Name the method used for refining of copper. (CBSE AI 2013)
Answer:
Electrolytic refining

Question 12.
What is the role of graphite in the electrometallurgy of aluminium? (CBSE Delhi 2012)
Answer:
The graphite rod is useful in the electrometallurgy of aluminium for reduction of alumina to aluminium.
2Al₂O₃ + 3C → 4Al + 3CO₂

Question 13.
Why is it that only sulphide ores are concentrated by ‘froth floatation process’? (CBSE Delhi 2011)
Answer:
This is because the sulphide ore particles are preferentially wetted by oil and the gangue particles by water.

Question 14.
What is the role of collectors in Froth- Floatation process? (CBSE AI 2012, 2017)
Answer:
The collectors such as pine oil, eucalyptus oil and fatty acids enhance the non – wettability of the mineral particles in froth – floatation process.

Question 15.
Name the substance used as depressant in the separation of two sulphide ores in froth floatation method. (CBSE Sample Paper 2019)
Answer:
Sodium cyanide.

Question 16.
Which reducing agent is employed to get copper from the leached low grade copper ore? (CBSE Delhi 2014)
Answer:
Iron scraps.

Question 17.
What is the role of zinc metal in the extraction of silver? (CBSE 2014)
Answer:
Zinc acts as a reducing agent.

General Principles and Processes of Isolation of Elements Important Extra Questions Short Answer Type

Question 1.
Out of C and CO which is a better reducing agent for FeO?
(i) In the lower part of blast furnace (Higher temperature)
(ii) In the upper part of blast furnace (Lower temperature) (CBSE Sample Paper 2011)
Answer:
(i) C is better reducing agent at higher temperature (lower part of blast furnace).
(ii) CO is better reducing agent at lower temperature (higher part of blast furnace).

Question 2.
What is the role of limestone in the extraction of iron from its oxides? (CBSE Al 2016)
Answer:
Limestone decomposes to form lime (CaO) and carbon dioxide (CO₂). The lime thus produced acts as a flux and combines with silica (present as impurity in oxides of iron) to produce slag.

Question 3.
How is copper extracted from a low grade ore of it? (CBSE Al 2012)
Answer:
Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing copper ions (Cu²⁺) is treated with scrap iron or H₂ as:
Cu²⁺(aq) + H₂(g) → Cu(s) + 2H⁺(aq)
In this way, copper is obtained.

Question 4.
(i) Which solution is used for the leaching of silver metal in the presence of air in the metallurgy of silver?
(ii) Out of C and CO, which is a better reducing agent at the lower temperature range in the blast furnace to extract iron from the oxide ore? (CBSE Delhi 2013)
Answer:
(i) Dilute solution (0.5%) of NaCN or KCN in the presence of air.
(ii) CO

Question 5.
(i) Which of the following ores can be concentrated by froth floatation method and why?
Fe₂O₃, ZnS, Al₂O₃
(ii) What is the role of silica in the metallurgy of copper? (CBSE Delhi 2013)
Answer:
(i) ZnS, because in sulphide ores, the sulphide ore particles are preferentially wetted by oil and gangue particles by water.
(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu₂O.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.

The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

Question 6.
(i) Name the method used for removing gangue from sulphide ores.
(ii) How is wrought iron different from steel? (CBSE Al 2013)
Answer:
(i) Froth floatation method.
(ii) Wrought iron is the pure form of iron and contains carbon and other impurities not more than 0.5%.
Steel contains 0.5 to 1.5% carbon along with small amounts of other elements such as Mn, Cr, Ni, etc. and other impurities.

Question 7.
Outline the principles behind the refining of metals by the following methods: (CBSE Delhi 2014)
(i) Zone refining method
(ii) Chromatographic method
Answer:
(i) Zone refining method. It is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
(ii) Chromatographic method. This is based on the principle that different components of a mixture are differently adsorbed on an adsorbent.

Question 8.
Write the principle behind the froth floatation process. What is the role of collectors in this process? (CBSE 2014)
Answer:
The froth floatation process is based on the principle of difference in the wetting properties of the ore and gangue particles with water and oil. The collectors such as pine oil, eucalyptus oil, fatty acids, etc. are used to enhance the non-wettability of the mineral particles.

Question 9.
What is meant by vapour phase refining? Write any one example of the process which illustrates this technique, giving the chemical equations involved.
OR
Write and explain the reactions involved in the extraction of gold. (CBSE Sample Paper 2019)
Answer:
Vapour phase refining: This method is based on the fact that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. Thus, the two requirements are:

• The metal should form a suitable compound with a suitable reagent.
• The volatile compound should be easily decomposable so that the metal can be easily recovered.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a stream of carbon monoxide to form volatile nickel tetracarbonyl, Ni(CO)₄, complex.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.

OR
The extraction of gold involves leaching of metal present in the ore with CN^– ions. This is an oxidation-reaction because during the leaching process, Au is oxidised to Au⁺ which then combines with CN^– ions to form their respective soluble complexes.

The metal is then recovered from these complexes by reduction or displacement method using a more electropositive zinc metal.
2 [Au (CN)₂]^– (aq) + Zn(s) → 2 Au (s) + [Zn (CN)₄]^2- (aq)
Zinc acts as a reducing agent.
This process is called hydro metallurgy.

General Principles and Processes of Isolation of Elements Important Extra Questions Long Answer Type

Question 1.
(i) Give one example of each the following:
(a) Acidic flux
(b) Basic flux
(ii) What happens when
(a) Cu₂O undergoes self reduction in a silica line converter
(b) Haematite oxidises carbon to carbon monoxide? (CBSE Sample Paper 2012)
Answer:
(i) (a) Acidic flux: SiO₂
(b) Basic flux: CaO

(ii) (a) Cu₂O undergoes self reduction to form blister copper as:

(b) Haematite oxidises carbon to carbon monoxide forming iron.
Fe₂O₃ + 3C → 3CO + 2Fe

Question 2.
What is a flux? What is the role of flux in the metallurgy of iron and copper? (CBSE Sample Paper 2011)
Answer:
Flux is a substance which combines with gangue which may still be present in the calcined or roasted ore to form an easily fusible material called the slag.
Flux + Gangue → Slag (fusible)
In the metallurgy of copper, most of the ferrous sulphide present as impurity gets oxidised to ferrous oxide which combines with silica (flux) to form fusible slag.
FeO + SiO₂ → FeSiO₃
The slag being lighter floats and forms the upper layer which is removed from the slag hole from time to time.
In the blast furnace for metallurgy of iron, lime acts as a flux and combines with silica present as impurity to form slag.

Question 3.
Describe the principle involved in each of the following processes.
(i) Mond process for refining of Nickel.
(ii) Column chromatography for purification of rare elements. (CBSE 2012)
Answer:
(i) Mond’s process for refining of nickel. Impure nickel is heated in a stream of carbon monoxide forming volatile complex, tetra carbonylnickel (0).

The volatile complex is decomposed at high temperature to give pure metal

(ii) Column chromatography for purification of rare elements. This technique is based on the differences in the adsorbing capacities of the metal and its impurities. The metal and the impurities are differently adsorbed. Impure metal is put in a liquid or gaseous medium (called moving phase) which is moved through an adsorbent (stationary phase).
Metal and impurities are adsorbed at different levels in column.

Question 4.
Which methods are usually employed for purifying the following metals?
(i) Nickel
(ii) Germanium
Mention the principle behind each one of them. (CBSE 2012)
Answer:
(i) Nickel. By vapour phase refining. This method is based on the principle that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal.

For example, nickel is refined by this technique and the method is known as Mond process. In this method, nickel is heated in a steam of carbon monoxide to form volatile nickel carbonyl Ni(CO)₄.
The carbonyl vapours when subjected to higher temperature (450-470 K) undergo thermal decomposition giving pure nickel.

(ii) Germanium. By Zone refining method. This method is based on the principle that the impurities are more soluble in melt than in the solid state of the metal. Therefore, an impure metal on solidification will deposit pure metal and the impurities will remain behind in the molten part of the metal.

Question 5.
Explain the principle of the method of electrolytic refining of metals. Give one example. (CBSE 2014)
Answer:
In electrolytic refining method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. Both anode and cathode are placed in a suitable electrolytic bath containing soluble salt of the same metal. On passing the electrical current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal, while equivalent amount of metal dissolves from the anode into the elctrolyte in the form of metal ions. The impurities fall down below the anode as anode mud. The reactions occurring at the electrodes are:
At cathode: Mⁿ⁺ + ne^– → M
At anode: M → Mⁿ⁺ + ne^–
Copper is refined using electrolytic refining method.

Question 6.
(i) Name the method used for the refining of zirconium. (CBSE 2015)
(ii) What is the role of CO in the extraction of iron?
(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why?
Answer:
(i) Van Arkel method
(ii) CO acts as reducing agent. It reduces oxides of iron to iron.
FeO + CO → Fe + CO₂
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂

(iii) This is because if the metal is in liquid state, its entropy is higher than when it is in solid state. Therefore, the value of entropy change (ΔS) of the reduction process is more on +ve side when the metal formed is in the liquid state and the metal oxide being reduced is in solid state. As a result, the value of ΔG° becomes more on negative side and hence the reduction becomes easier.

Question 7.
(i) Name the method of refining which is based on the principle of adsorption. (CBSE 2008, 2016)
(ii) What is the role of depressant in froth floatation process?
(iii) What is the role of limestone in the extraction of iron from its oxides?
Answer:
(i) Chromatography

(ii) The depressants are used to prevent certain types of particles from forming the froth with bubbles in froth floatation process. This helps to separate two sulphide ores. For example, in case of an ore containing zinc sulphide (ZnS) and lead sulphide (PbS), sodium cyanide (NaCN) is used as a depressant. It forms a layer of zinc complex Na₂[Zn(CN)₄] with ZnS on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.

However, NaCN does not prevent PbS from forming the froth and allows it to come with the froth.

(iii) Limestone decomposes to form lime (CaO) and carbon dioxide (CO₂). The lime thus produced acts as a flux and combines with the silica (present as impurity in oxides of iron) to produce slag.

Question 8.
(i) Name the method of refining of metals such as germanium. (CBSE Delhi 2016)
(ii) In the extraction of Al, impure Al₂O₃ is dissolved in cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
(ii) What is the role of coke in the extraction of iron from its oxides?
Answer:
(i) Zone refining
(ii) Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]
This process is called leaching.
(iii) The coke serves as a fuel as well as a reducing agent. It burns to produce CO₂ and heat which raises the temperature to about 2200 K (in combustion zone).
C + O₂ → CO₂, ΔH = -393.4 kJ
It also reduces Fe₂O₃ to iron.
Fe₂O₃ + 3C → 2Fe + 3CO + Heat

Question 9.
(i) Indicate the principle behind the method used for the refining of zinc.
(ii) What is the role of silica in the extraction of copper?
(iii) Which form of the iron is the purest form of commercial iron? (CBSE Delhi 2015)
Answer:
(i) Zinc is refined by electrolytic refining:
In this method, the impure zinc is made anode, while a plate of pure zinc is made the cathode. These are placed in a suitable electrolytic bath containing suitable salt of the same metal (e.g. zinc sulphate) containing a small amount of dilute sulphuric acid. On passing current, zinc from the solution is deposited at the cathode, while an equivalent amount of zinc from anode goes into the electrolytic solution. Therefore, pure zinc is obtained at cathode.

(ii) During roasting, the copper pyrites are converted into a mixture of FeO and Cu₂O.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with ferrous oxide present to form fusible slag of iron silicate.

The slag being lighter floats and forms the upper layer and is removed through slag hole. Therefore, silica helps to remove FeO in the metallurgy of copper.

(iii) Wrought iron is the purest form of commercial iron.

Question 10.
(a) What is the difference between calamine and malachite?
(b) Why is zinc used instead of Cu for recovery of Ag from [Ag(CN)₂]?
(c) What is the role of cryolite in metallurgy of Al? (CBSE 2019C)
OR
(a) Give two points of differences between pig iron and cast iron.
(b) Outline the principle of zone refining.
Answer:
(a) Calamine is an ore of Zn while malachite is an ore of copper. Calamine is ZnCO₃ while malachite is CuCO₃.Cu(OH)₂

(b) Zinc is more electropositive than silver and therefore, zinc displaces silver from its solution.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag
On the other hand, copper is less electropositive than silver and therefore, cannot displace silver from its solution. Zn is more reactive than Cu, so reduction will be faster in case of Zn.

(c) Fused alumina is a bad conductor of electricity. Purified alumina is mixed with molten cryolite (Na₃AlF₆) and is electrolysed in an iron tank lined inside with carbon. The molten cryolite decreases the melting point and increases the electrical conductivity. It acts as a solvent.

OR

(a)

(b) Zone refining – It is based on the principle that impurities are more soluble in the melt than the solid state of the metal. Therefore, an impure metal on solidification will deposit crystals of pure metal and the impurities will remain behind in the molten part of the metal.

Question 11.
How will you convert the following:
(i) Impure Nickel to pure Nickel
(ii) Zinc blende to Zinc metal
(iii) [Ag(CN)₂]^– to Ag (CBSE Delhi 2019)
Answer:
(i) Impure nickel is heated in a stream of carbon monoxide forming volatile complex tetracarbonylnickel (0). The volatile complex is decomposed at high tempreature to give pure nickel.

This is called Mond’s process.

(ii) The zinc blende ore is concentrated by froth floatation process and then roasted in the presence of excess air at about 1200 K.
2ZnS + 3O₂ → 2ZnO + 2SO₂
Zinc oxide formed is reduced to zinc by heating with crushed coke at 1673 K in vertical fire clay retorts.

(iii) [Ag(CN)₂]^– complex is heated with zinc to get silver.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag

Question 12.
(a) Name the method of refining which is
(i) used to obtain semiconductor of high purity,
(ii) used to obtain low boiling metal.
(b) Write chemical reactions taking place in the extraction of copper from Cu₂S. (CBSE Delhi 2019)
Answer:
(a) (i) Zone refining
(ii) Distillation

(b) Cu₂S is first roasted and converted to oxide
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
The oxide is then easily reduced to metallic copper with coke.
Cu₂O + C → 2Cu + CO

Question 13.
Describe how the following steps can be carried out.
(i) Recovery of Gold from leached gold metal complex.
(ii) Conversion of Zirconium iodide to pure Zirconium.
(iii) Formation of slag in the extraction of copper.
(Write the chemical equations also for the reactions involved)
OR
Explain the use of the following:
(i) NaCN in Froth Floatation Method.
(ii) Carbon monoxide in Mond process.
(iii) Coke in the extraction of Zinc from Zinc Oxide.
Answer:
(i) Leached gold complex is treated with Zinc and gold is recovered by displacement method.
2Au[(CN)₂]r(aq) + Zn(s) → 2Au(s) + [Zn(CN)₄]^2-(aq)

(ii) Zirconium iodide is decomposed on a tungsten filament, electrically heated to 1800 K. Pure Zr metal is deposited on the filament.
ZrI₄ → Zr + I₂

(iii) Silica is added to the ore and heated. It helps to slag off iron oxide as iron silicate.
FeO + SiO₂ → FeSiO₃ (slag)

OR

(i) NaCN is used as depressants to separate two sulphide ores (ZnS and PbS) in Froth Floatation Method.
(ii) Carbon monoxide forms a volatile complex of nickel, nickel tetracarbonyl.
(iii) Coke is used as a reducing agent to reduce zinc oxide to zinc.

Question 14.
Write the role of
(i) NaCN in the extraction of gold from its ore.
(ii) Cryolite in the extraction of aluminium from pure alumina.
(iii) CO in the purification of Nickel. (CBSE 2018)
Answer:
(i) Gold is leached out in the form of a complex with dil. solution of NaCN in the presence of air. Here NaCN acts as leaching agent.
(ii) It lowers the melting point of alumina and makes it a good conductor of electricity.
(iii) CO forms a volatile complex with nickel which is further decomposed to give pure Ni metal.

Question 15.
(i) Write the role of ‘CO’ in the purification of nickel.
(ii) What is the role of silica in the extraction of copper?
(iii) What type of metals are generally extracted by electrolytic method? (CBSE Delhi 2019)
Answer:
(i) Carbon monoxide forms a volatile complex, Ni(CO)₄, on heating nickel in a stream of CO while the impurities remain unaffected. The compound on further heating decomposes to give pure nickel.

Thus, CO helps to purify nickel.

(ii) During roasting, the copper pyrites are converted into a mixture of Cu₂O and FeO.

To remove FeO (basic), the roasted ore is mixed with silica and heated. Silica acts as a flux and combines with FeO to form fusible slag of iron silicate.
FeO + SiO₂ → FeSiO₃
The slag being lighter floats and forms the upper layer and is removed. Thus, silica helps to remove FeO.

(iii) Very reactive metals such as sodium, potassium, calcium, aluminium, etc. are extracted by electrolytic methods. Certain less reactive metals such as copper are also purified by using electro-refining method.

Question 16.
Write down the reactions taking place in blast furnace related to the metallurgy of iron in the temperature range 500 K to 800 K. What is the role of limestone in the metallurgy of iron?
OR
What happens when
(a) Silver is leached with NaCN in the presence of air?
(b) Copper matte is charged into silica-lined converter and hot air blast is blown?
(c) NaCN is added in an ore containing PbS and ZnS during concentration by froth floatation method? (CBSE Delhi Al 2019)
Answer:
Reactions taking place in blast furnace at 500K to 800K: The oxides of iron are reduced by CO.
FeO + CO → Fe + CO₂
3Fe₂O₃ + CO → 2Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3Fe + 4CO₂
Fe₂O₃ + CO → 2FeO + CO₂
Role of Limestone: It acts as a flux which decomposes to calcium oxide.
CaCO₃ → CaO + CO₂
Limestone
CaO combines with impurity (e.g. Si02) to form slag which is then removed.
CaO + SiO₂ → CaSiO₃

OR

(a) Silver is leached with dilute solution (0.5%) of NaCN in the presence of atmospheric oxygen. The metal dissolves forming the complex.
4Ag + 8CN^– + 2H₂O + O₂ → 4[Ag(CN)₂]^– + 4OH^–
The metal is extracted from water-soluble complex by zinc metal.
2[Ag(CN)₂]^– + Zn → [Zn(CN)₄]^2- + 2Ag

(b) The copper matte containing Cu₂S and FeS is put in silica-lined convertor and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → FeSiO₃
Cu2S or CuO gets converted to copper
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

(c) NaCN acts as a depressant in preventing ZnS from forming the froth. If forms a layer of zinc complex Na₂[Zn(CN)₄] on the surface of ZnS and thereby prevents it from forming the froth.
4NaCN + ZnS → Na₂[Zn(CN)₄] + Na₂S

Long Answer Type Questions (LA-II)

Question 1.
Explain the following:
(i) Although thermodynamically feasible, in practice magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why?
(ii) Why is zinc and not copper used for the recovery of silver from the complex [Ag(CN)₂]^–?
(iii) The extraction of Au by leaching with NaCN involves both oxidation and reduction. Justify giving equations. (CBSE Sample Paper 2011)
(iv) Lime-stone is used in the manufacture of pig iron from haematite. Why?
Answer:
(i) Inspection of Ellingham diagram shows that ∆G vs T curves for Al₂O₃ and MgO intersect at a point corresponding to very high temperature of the order of 2000 K. This means above this temperature, ∆G for the reaction:
Al₂O₃ + 3Mg → 2Al + 3Mgo
would become negative and hence reduction will be feasible. However, this temperature is very high so that the process is uneconomical and technologically difficult.

(ii) Zinc is a stronger reducing agent (E° = – 0.76 V) and more electropositive than copper (E° = + 0.34 V). Therefore, zinc is used for recovery of Ag from [Ag(CN₂)]^– complex.

(iii) During leaching process, gold (Au) is first oxidised by O₂ of the air to Au⁺ which then combines with CN^– ions to form the soluble complex, dicyanoaurate (I).

Gold is then extracted from this complex by displacement method by using a more electropositive zinc metal. In this method, zinc acts as a reducing agent and it reduces Au⁺ to Au. Zinc itself gets oxidised to Zn²⁺ ions which combine with CN^– ions to form soluble complex, sodium tetracyanozincate (II).

Thus, extraction of gold by leaching with NaCN involves both oxidation and reduction.

(iv) Haematite is an ore of iron and contains silica (SiO₂) as the main impurity. The purpose of limestone is to remove SiO₂ as calcium silicate (CaSiO₃) slag.

Question 2.
Name the principal ore of aluminium. Explain the significance of leaching in the extraction of aluminium. (CBSE AI 2013)
Answer:
The principal ore of aluminium is bauxite, Al₂O₃.2H₂O. Leaching process is used to concentrate the ore of aluminium, bauxite, which is contaminated with impurities of silica (SiO₂), iron oxide (Fe₂O₃), titanium oxide (TiO₂), etc. Leaching is done by treating the powdered ore with hot cone. (45%) solution of NaOH at about 473-523 K and 35-36 bar pressure.
Al₂O₃.2H₂O(s) + 2NaOH(aq) + H₂O(l) → 2Na [Al(OH)₄](aq)
The impurities of ferric oxide and silica are insoluble and are removed by filtration.
The solution containing sodium aluminate is neutralised by passing C02 gas and hydrated alumina is precipitated.
2Na [Al(OH)₄] (aq) + CO₂ (g) → Al₂O₃. x H₂O (s) + 2Na HCO₃(s)
Hydrated alumina is heated to 1473 K to get pure alumina.

Question 3.
Describe the principle controlling each of the following processes:
(i) Vapour phase refining of titanium metal.
(ii) Froth floatation method of concentration of sulphide ore.
(iii) Recovery of silver after silver ore was leached with NaCN. (CBSE AI 2011, CBSE Delhi 2011)
Answer:
(i) The principle of vapour phase refining is that certain metals are converted to their volatile compounds while the impurities are not affected during compound formation. The compound formed decomposes on heating to give pure metal. During refining of titanium metal, the titanium metal is heated with l₂ to form a volatile compound Til₄, which on further heating at higher temperature decomposes to give pure titanium metal.

(ii) This is based upon the different wetting properties of the ore and the gangue particles with oil and water. The mineral particles become wet by oil and the gangue particles by water. When the ore is mixed with water containing small quantities of pine oil and then by agitating the water by blowing air violently, the froth is formed. The froth carries the lighter ore particles to the surface and the heavier impurities sink to the bottom.

(iii) This is a chemical method and is useful for the ores which are soluble in certain suitable solvents but the impurities are not soluble. The impurities left undissolved are removed by filtration. The ore of silver, Ag₂S (argentite), reacts with NaCN as

Sodium sulphide thus formed is oxidised to sodium sulphate by blowing air into the solution. This helps the reaction to occur in the forward direction.
4Na₂S + 5O₂ + 2H₂O → 2Na₂SO₄ + 4NaOH + 2S

The above solution after filtration and removing insoluble impurities is heated with zinc to get silver.

Question 4.
Describe the role of the following:
(i) NaCN in the extraction of silver from a silver ore
(ii) Iodine in the refining of titanium
(iii) Cryolite in the metallurgy of aluminium (CBSE 2010)
Answer:
(i) The role of NaCN in the extraction of silver is to do the leaching of silver ore in the presence of atmospheric air from which silver is obtained later by replacement with zinc as:

(ii) Iodine is used to form a volatile unstable compound of the metal, which is decomposed to get the pure metal. For example, titanium is heated with iodine to 523 K forming unstable titanium iodide which is decomposed to Ti as

(iii) Cryolite is added to bauxite ore before electrolysis because of the following reasons:
(a) It acts as a solvent.
(b) It lowers the melting point of alumina to about 1173 K.
(c) Addition of cryolite to alumina increases the electrical conductivity.

Question 5.
Describe the role of the following:
(i) SiO₂ in the extraction of copper from copper matte
(ii) NaCN in froth floatation process
Answer:
(i) The copper matte containing Cu₂S and FeS is put in silica lined converter. Some silica is also added and hot air blast is blown to convert remaining FeS to FeO, which is removed as slag with silica.
2FeS + O₂ → 2FeO + 2SO₂
FeO + Si0₂ → FeSiO₃
Cu₂S or CuO gets converted to copper.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2Cu₂O + Cu₂S → 6Cu + SO₂

(ii) NaCN in froth floatation process is used to separate one sulphide ore from another and is known as depressant.
For example, sodium cyanide can be used as a depressant in the separation of zinc sulphide ore (ZnS) and lead sulphide ore (PbS). Sodium cyanide forms a layer of zinc complex, Na₂[Zn(CN)₄], on the surface of ZnS and therefore, prevents it from forming the froth. Therefore, it acts as a depressant.

However, NaCN does not prevent PbS from forming the froth. Thus, it selectively prevents ZnS from coming to the froth but allows PbS to come with the froth. Thus, the two ores can be separated by the use of a depressant.