Ncert Solution Class 12th Physics Chapter – 5 Magnetism and Matter
| Textbook | NCERT |
| class | Class – 12th |
| Subject | Physics |
| Chapter | Chapter – 5 |
| Chapter Name | Magnetism and Matter |
| Category | Class 12th Physics Question & Answer |
| Medium | English |
| Source | last doubt |
Ncert Solution Class 12th Physics Chapter – 5 Magnetism and Matter
?Chapter – 5?
✍Magnetism and Matter✍
?Question & Answer?
Q .1) Answer the following:
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?
(c)If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
(d) In which direction would a compass free to move in the vertical plane point to, is located right on the geomagnetic north or south pole?
(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of the magnetic moment 8 × 1022 JT-1 located at its centre. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
Answer:
(a) The three independent quantities conventionally used for specifying earth’s magnetic field are:
(i) Magnetic declination,
(ii) Angle of dip, and
(iii) Horizontal component of earth’s magnetic field
(b) The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.
(c) It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole. Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.
(d) If a compass is located on the geomagnetic North Pole orSouth Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction. This quantity is of the order of magnitude of the observed field on earth.
(f) Yes, there are several local poles on earth’s surface oriented in different directions. Amagnetised mineral deposit is an example of a local N-S pole.
Q .2) Answer the following:
(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such a distant past?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
(f ) Interstellar space has an extremely weak magnetic field of the order of 10–12 T. Can such a weak field be of any significant consequence? Explain.
Answer:
(a) Earth’s magnetic field varies with time and it takes a couple of hundred years to change by an obvious sum. The variation in the Earth’s magnetic field with respect to time can’t be ignored.
(b) The Iron core at the Earth’s centre cannot be considered as a source of Earth’s magnetism because it is in its molten form and is non-ferromagnetic.
(c) The radioactivity in the earth’s interior is the source of energy that sustains the currents in the outer conducting regions of the earth’s core. These charged currents are considered to be responsible for the earth’s magnetism.
(d) The Earth’s magnetic field reversal has been recorded several times in the past about 4 to 5 billion years ago. These changing magnetic fields were weakly recorded in rocks during their solidification. One can obtain clues about the geomagnetic history from the analysis of this rock magnetism.
(e) Due to the presence of ionosphere, the Earth’s field deviates from its dipole shape substantially at large distances. The Earth’s field is slightly modified in this region because of the field of single ions. The magnetic field associated with them is produced while in motion.
(f) A remarkably weak magnetic field can deflect charged particles moving in a circle. This may not be detectable for a large radius path. With reference to the gigantic interstellar space, the deflection can alter the passage of charged particles.
Q .3) A short bar magnet placed with its axis at 30º with a uniform externalmagnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 x 10-2J. What is the magnitude of magnetic moment of the magnet?
Answer:
Magnetic field strength, B= 0.25 T
Torque on the bar magnet, T= 4.5 ×10-2J
Angle between the bar magnet and the external magnetic field,θ= 30°
Torque is related to magnetic moment (M) as:
T= MB sin θ
Hence, the magnetic moment of the magnet is 0.36 J T-1.
Q .4) A short bar magnet of magnetic moment m = 0.32 JT–1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Answer:
Given,
Magnetic moment of magnet, m = 0.32
JT-1
Magnetic field strength, B = 0.15 T
(a) Stable equilibrium: When the magnetic moment is along the magnetic field i.e.
θ = 00
(b) Unstable equilibrium: When the magnetic moment is at 180° with the magnetic field i.e.
θ = 1800
(c) We know that,
U = – m.B = -mBcos
By putting the given values:
U = (-0.32)(0.15)(cos) = -0.048 J
00
Therefore, Potential energy of the system in stable equilibrium is -0.048 J
Similarly,
U = (-0.32)(0.15)(cos) = 0.048 J
1800
Therefore, Potential energy of the system in unstable equilibrium is 0.048J.
Q .5) A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Number of turns in the solenoid, n = 800
Area of cross-section, A = 2.5 × 10−4 m2
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M = n I A
= 800 × 3 × 2.5 × 10−4
= 0.6 J T−1
Q .6) If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Answer:
Magnetic field strength, B = 0.25 T
Magnetic moment, M= 0.6 T-1
The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
T = MB sinθ
= 0.6 x 0.25 sin30°
= 7.5 x 102 J
Q .7) A bar magnet of magnetic moment 1.5 J T–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?
Answer:
(a) Magnetic moment, M= 1.5 J T-1
Magnetic field strength, B= 0.22 T
(i) Initial angle between the axis and the magnetic field, θ1= 0°
Final angle between the axis and the magnetic field, θ2= 90°
The work required to make the magnetic moment normal to the direction of magnetic field is given as:
W = – MB ( cos θ2 – cos θ1)
= -1.5 x 0.22 ( cos 90° – cos 0°)
= -0.33 (0 – 1)
= 0.33 J
(ii) Initial angle between the axis and the magnetic field, θ1= 0°
Final angle between the axis and the magnetic field, θ2= 180°
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:
W = – MB (cos θ2 – cos θ1)
= -1.5 x 0.22 (cos 180° – cos 0°)
= -0.33 (-1 – 1)
= 0.66 J
(b) For case (i): θ = θ2 = 90°
∴Torque, T = MB sinθ
= 1.5 x 0.22 sin 90°
=0.33 J
Forcase (ii): θ = θ2 = 180°
∴Torque, T = MB sin θ
=MB sin 180° = 0
Q .8) A closely wound solenoid of 2000 turns and area of cross-section 1.6 x 10-4m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 x 10-2T is set up at an angle of 30º with the axis of the solenoid?
Answer:
Number of turns on the solenoid,n = 2000
Area of cross-section of the solenoid, A= 1.6 ×10-4 m2
Current in the solenoid, I= 4 A
(a) The magnetic moment along the axis of the solenoid is calculated as:
M= nAI
= 2000 ×1.6 ×10-4 × 4
= 1.28 Am2
(b) Magnetic field, B = 7.5 ×10-2T
Angle between the magnetic field and the axis of the solenoid, θ= 30°
Torque, T = MB sin θ
= 1.28 x 7.5 x 10-2 sin30°
= 4.8 x 10-2 Nm
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is
Q .9) A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 x 10-2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s-1. What is the moment of inertia of the coil about its axis of rotation?
Answer:
Number of turns in the circular coil, N= 16
Radius of the coil, r= 10 cm = 0.1 m
Cross-section of the coil, A= πr2= π× (0.1)2m2
Current in the coil, I= 0.75 A
Magnetic field strength, B= 5.0 ×10-2T
Frequency of oscillations of the coil, v= 2.0 s-1
∴Magnetic moment, M= NIA
= 16 × 0.75 × π× (0.1)2
= 0.377 J T-1
Where,
I= Moment of inertia of the coil
Hence,the moment of inertia of the coil about its axis of rotation is 1.19 x 10-4 kg m2
Q .10) A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Answer:
Horizontal component of earth’s magnetic field, BH= 0.35 G
Angle made by the needle with the horizontal plane = Angle of dip = ? = 220
Earth’s magnetic field strength = B
We can relate B and BH as:
Q .11) At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the
horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Answer:
Angle of declination, θ = 12°
Angle of dip, ? = 600
Horizontal component of earth’s magnetic field, BH = 0.16 G
Earth’s magnetic field at the given location = B
We can relate B and BH as:
Earth’s magnetic field lies in the vertical plane, 12°West of the geographic meridian, making an angle of 60°(upward) with the horizontal direction. Its magnitude is 0.32 G.
Q .12) A short bar magnet has a magnetic moment of 0.48 J T–1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis,
(b) the equatorial lines (normal bisector) of the magnet.
Answer:
Magnetic moment of the bar magnet, M= 0.48 J T -1
(a) Distance, d = 10 cm = 0.1 m
The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:
Where,
μ0= Permeability of free space = 4π x 10-7 Tm A-1
The magnetic field is along the S – N direction.
(b) The magnetic field at a distance of 10 cm (i.e., d= 0.1 m) on the equatorial line of the magnet is given as:
The magnetic field is along the N – S direction.
Q .13) A short bar magnet placed in a horizontal plane has its axis alignedalong the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
Earth’s magnetic field at the given place, H= 0.36 G
The magnetic field at a distance d, on the axis of the magnet is given as:
Where, μ0= Permeability of free space
M= Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:
Total magnetic field, B = B1 + B2
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
Q .14) If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?
Answer:
The magnetic field on the axis of the magnet at a distance d1= 14 cm, can be written as:
Where, M = Magnetic moment
μ0= Permeability of free space
H= Horizontal component of the magnetic field at d1
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance d2, on the equatorial line of the magnet can be written as:
Equating equations (1) and (2), we get:
The new null points will be located 11.1 cm on the normal bisector.
Q .15) A short bar magnet of magnetic moment 5.25 x 10-2J T-1is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º with earth’s field on
(a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer:
Magnetic moment of the bar magnet, M = 5.25 ×10 – 2J T – 1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 ×10 – 4T
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
Where, μ0= Permeability of free space = 4π ×10 – 7Tm A – 1
When the resultant field is inclined at 45°with earth’s field, B = H
(b) The magnetic field at a distanced R’ from the centre of the magnet on its axis is given as:
The resultant field is inclined at 45°with earth’s field.
Q .16) Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?
(f ) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?
Answer:
(a) Owing to therandom thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.
(b) The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.
(c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
(d) The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa.
(e) The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.
(f) The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.
Q .17) Answer the following questions:(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?
(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?
(e) A certain region of space is to be shielded from magnetic fields.
Suggest a method.
(e) A certain region of space is to be shielded from magnetic fields.
Suggest a method.
Answer: The hysteresis curve (B-Hcurve) of a ferromagnetic material is shown in the following figure.
(a) It can be observed from the given curve that magnetisation persists even when the external field is removed. This reflects the irreversibility of a ferromagnet.
(b) The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat energy.
(c) The value of magnetisation is memory or record of hysteresis loop cycles of magnetisation. These bits of information correspond to the cycle of magnetisation. Hysteresis loops can be used for storing information.
(d) Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.
(e) A certain region of space can be shielded from magnetic fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.
Answer:Current in the wire = 2.5 A
The earth’s magnetic field at a location, R= 0.33 G = 0.33 x 10-4 T
Angle of dip is zero, δ = 0
Horizontal component of earth’s magnetic field, BH= R cosδ = 0.33 x 10-4 Cos 0 = 0.33 x 10-4 T
Magnetic field due to a current carrying conductor, Bc = (μ0/2π) x (I/r)
Bc= (4π x 10-7/2π) x (2.5/r) = (5 x 10-7/r)
BH = Bc
0.33 x 10-4 = 5 x 10-7/r
r = 5 x 10-7/0.33 x 10-4
= 0.015 m = 1.5 cm
Hence neutral points lie on a straight line parallel to the cable at a perpendicular distance of 1.5cm.
Answer:Earth’s magnetic field, Be = 0.39 G
Angle of dip, δ = 35°
Number of wires, n = 4
declination = 0
distance , r = 4 cm = 4
Current carried by the cable, I = 1 A
∴ Horizontal component of earth’s magnetic field is given by,
BH = Be cos δ
BH = 0.39 cos 35°
= 0.3195 G
Vertical component of earth’s magnetic field, Bv = Be sin δ
Bv = 0.39 sin 35°
= 0.224 G
Magnetic field produced by telephone cable having 4 wires is,
Now,
We will find the resultant field below the cable.
As per the right hand thumb rule, the direction of B’ will be opposite to BH at a point below the cable.
Therefore at a point 4 cm below the cable, resultant horizontal component of earth’s field
RH = BH – B’
= 0.3195 – 0.2
= 0.1195 G
Resultant vertical component of earth’s field
Rv = Bv = 0.224 G (unchanged)
∴ Resultant of earth’s field
R = RH2+RV2 = (0.1195)2+(0.224)2 = 0.0143+0.0500 = 0.0643 = 0.254 G
Now,
Resultant field above the cable is given as per right hand thumb rule which says that, at a point above the cable, B’ will be in the same direction as BH.
Hence, at a point 4 cm above the cable,
RH = BH+B’ = 0.3195+0.2 = 0.5195 GRV = BV = 0.224 G
∴Resultant magnetic field is,
R=RH2+RV2
= (0.5195)2+(0.224)2
R = 0.566 G.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Answer:Answer:Answer:Energy of an electron beam, E= 18 keV = 18 × 103eV
Charge on an electron, e= 1.6 × 10 –19C
E= 18 × 103× 1.6 × 10 – 19J
Magnetic field, B = 0.04 G
Mass of an electron, me= 9.11 × 10 – 19 kg
Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:
The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.
Let the up and down deflection of the electron beam be x = r (1-cosθ)
Where,
θ= Angle of declination
Therefore, the up and down deflection of the beam is 3.9 mm.
Answer:Number of atomic dipoles, n= 2.0 × 1024
Dipole moment of each atomic dipole, M= 1.5 × 10 – 23J T – 1
When the magnetic field, B1= 0.64 T
The sample is cooled to a temperature, T1= 4.2°K
Total dipole moment of the atomic dipole, Mtot= n × M
= 2 × 1024× 1.5 × 10 – 23
= 30 J T – 1
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment,
When the magnetic field, B2= 0.98 T
Temperature, T2= 2.8°K
Its total dipole moment = M2
According to Curie’s law, we have the ratio of two magnetic dipoles as:
Therefore, 10.336 J T-1 is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.
Answer:Mean radius of a Rowland ring, r = 15 cm = 0.15 m
Number of turns on a ferromagnetic core, N = 3500
Relative permeability of the core material, μr= 800
Magnetising current, I = 1.2 A
The magnetic field is given by the relation:
B
Where,
Ã�¼0= Permeability of free space = 4π× 10 – 7Tm A – 1
Therefore, the magnetic field in the core is 4.48 T.
Q .25) The magnetic moment vectors μs and μl associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:μs= -(e/m) S,
μl= -(e/2m)l
Which of these relations is in accordance with the result expected classically?
Outline the derivation of the classical result.
Answer:The magnetic moment associated with the orbital angular momentum is valid with the classical mechanics.
The magnetic moment associated with the orbital angular momentum is given as
For current i and area of cross-section A, we have the relation:
Magnetic moment
