CBSE Class 12 Physics Chapter-5 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 5 – Magnetism And Matter 5 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-5 Important Questions
CBSE Class 12 Physics Important Questions Chapter 5 – Magnetism And Matter
5 Marks Questions
1. A particle of mass m and charge q moving with a uniform speed 
normal to a uniform magnetic field B describes a circular path of radius & Derive expressions for (1) Radius of the circular path (2) time period of revolution (3) Kinetic energy of the particle?
Ans. A particle of mass (m) and change (q) moving with velocity normal to 
describes a circular path if
Since Time period of Revolution
During circular path = 
=> 
(
)
=> T = 
Kinetic energy K.E = 
=> KE = 
2. Write an expression for the force experienced by the charged particle moving in a uniform magnetic field B With the help of labeled diagram explain the working of cyclotron? Show that cyclotron frequency does not depend upon the speed of the particle?
Ans. Force experienced by the charged particle moving at right angles to uniform magnetic field with velocity 
is given by 
= q (
) Initially Dee 
is negatively charged and Dee 
is positively charged so, the positive ion will get accelerated towards Dee since the magnetic field is uniform and acting at right angles to the plane of the Dees so the ion completes a circular path in when ions comes out into the gap, polarity of the Dee’s gets reversed used the ion is further accelerated towards Dee with greater speed and cover a bigger semicircular path. This process is separated time and again and the speed of the ion becomes faster till it reaches the periphery of the dees where it is brought out by means of a deflecting plate and is made to bombard the target.
Since F = qVBsin900 provides the necessary centripetal force to the ion to cover a circular path so we can say 
=> r = 
Time period = 
V = 
frequency is independent of velocity
3. (a) Obtain an expression for the torque acting on a current carrying circular loop.
(b) What is the maximum torque on a galvanometer coil 5 cm 
12 cm of 600 turns when carrying a current of 10-5 A. in a field where flux density is
?
Ans. ABCD is a rectangular loop of length (L), breadth (b) and area (A). Let I be the Current flowing in the anti clockwise direction. Let 
be the angle between the normal to the loop and magnetic field
Force acting on arm AB of the loop
Force on arm CD
Force on arm BC
Force on arm DA
Since 
are equal and opposite and also acts along the same line, hence they cancel each other.
are also equal and opposite but their line of action is different, so they form a couple and makes the rectangular loop rotate anti clockwise.
Thus 
= either force 
distance
For loop of N turns
Where M is magnetic moment of the loop.
Torque will be maximum when = 90o
4. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor of two. Calculate by what factor, the voltage sensitivity changes?
Ans. Current sensitivity 
Voltage sensitivity 
Resistance of a galvanometer increases when n and A are changed
Given 
= 2R
Then n = 
and A = 
New current sensitivity
New voltage sensitivity
From (i) and (iii)
n’A’=
Using equation (iv)
Thus voltage sensitivity decreases by a factor of 
.
5. (a) Show how a moving coil galvanometer can be converted into an ammeter?
(b) A galvanometer has a resistance 30
and gives a full scale deflection for a current of 2mA. How much resistance in what way must be connected to convert into?
(1) An ammeter of range 0.3A
(2) A voltammeter of range 0.2V.
Ans. (a) A galvanometer can be converted into an ammeter by connecting a low resistance called shunt parallel to the galvanometer.
Since G and RS are in parallel voltage across then is same 
(b) (1) I = 0.3A G = 30 Ig = 2mA = 
Sheent (S) = 
S = 0.2
(2) G = 30
, Ig = 2mA =, V = 0.2V
Shunt Resistance (R) 
R = 70
6. A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me=
). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Ans. Energy of an electron beam, E= 18 keV =
Charge on an electron, e= 
E=
Magnetic field, B = 0.04 G
Mass of an electron, me= 
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 
Where,
θ= Angle of declination
Therefore, the up and down deflection of the beam is 3.9 mm.
7. A sample of paramagnetic salt contains 
atomic dipoles each of dipole moment
. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)
Ans. Number of atomic dipoles, n=
Dipole moment of each atomic dipole, M= 
When the magnetic field, 
= 0.64 T
The sample is cooled to a temperature, 
= 4.2°K
Total dipole moment of the atomic dipole, 
= 
= 
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, 
When the magnetic field, 
= 0.98 T
Temperature, 
= 2.8°K
Its total dipole moment =
According to Curie’s law, we have the ratio of two magnetic dipoles as:
Therefore, 
is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.
8. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is
. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Ans. Number of horizontal wires in the telephone cable, n= 4
Current in each wire, 
= 1.0 A
Earth’s magnetic field at a location, H= 0.39 G =
Angle of dip at the location, 
Angle of declination, 
For a point 4 cm below the cable:
Distance, r = 4 cm = 0.04 m
The horizontal component of earth’s magnetic field can be written as:
Where,
B= Magnetic field at 4 cm due to current I in the four wires
= Permeability of free space =
= 0
= 0.2 G
∴ 
= 
The vertical component of earth’s magnetic field is given as:
Hv= Hsin
The angle made by the field with its horizontal component is given as:
The resultant field at the point is given as:
s
For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field:
= 0.39 cos 35° + 0.2 = 0.52 G
Vertical component of earth’s magnetic field:
= 0.39
Angle, 
And resultant field:
9. 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.
Ans. The hysteresis curve (B–H curve) 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.
10. 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 to roid 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 ferromagnetic 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?
Ans. (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.
11. A short bar magnet of magnetic moment 
is 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 s
(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.
Ans. Magnetic moment of the bar magnet, M= 
Magnitude of earth’s magnetic field at a place, H= 0.42 G =
(a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
Where,
= Permeability of free space =
When the resultant field is inclined at 
with earth’s field, B= H
(b) The magnetic field at a distanced
from the centre of the magnet on its axis is given as:
The resultant field is inclined at with earth’s field.
12. If the bar magnet in exercise 5.13 is turned around by 
where will the new null points be located?
Ans. The magnetic field on the axis of the magnet at a distance d1= 14 cm, can be written as:
…(i)
Where,
M = Magnetic moment
= Permeability of free space
H= Horizontal component of the magnetic field at 
If the bar magnet is turned through 
then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance 
, on the equatorial line of the magnet can be written as:
……(2)
Equating equations (1) and (2), we get:
s
The new null points will be located 11.1 cm on the normal bisector.
13. A bar magnet of magnetic moment 
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)?
Ans. (a) Magnetic moment, M= 
Magnetic field strength, B= 0.22 T
(i) Initial angle between the axis and the magnetic field, 
= 0° Final angle between the axis and the magnetic field, 
The work required to make the magnetic moment normal to the direction of magnetic field is given as:
(ii) Initial angle between the axis and the magnetic field,
Final angle between the axis and the magnetic field, 
The work required to make the magnetic moment opposite to the direction of magnetic field is given as:
(b) For case (i): 
∴Torque, 
For case (ii): 
∴ Torque, ss
14. Answer the following questions regarding earth’s magnetism:
(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 
.
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, if 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 magnetic moment 
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?
Ans. (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 or South 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.
(e) Magnetic moment, M=
Radius of earth, r= 
Magnetic field strength,
Where,
= Permeability of free space = 
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. A magnetised mineral deposit is an example of a local N-S pole.
15. Answer the following questions:
(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 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.
[Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]
Ans. (a) Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected.
(b) Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth’s magnetism.
(c) The radioactivity in earth’s interior is the source of energy that sustains the currents in the outer conducting regions of earth’s core. These charged currents are considered to be responsible for earth’s magnetism.
(d) Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.
(e) Earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.
(f) An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.
