by NCERT Solved Exercise Questions – Class 12 Physics Chapter 5 Magnetism And Matter
5.1 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 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, 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 8 × 1022 J T–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?
Ans – (a) The three independent measurements that make up the earth’s magnetic field are:
(i) Magnetic declination,
(ii) Dip angle, and
(iii) The earth’s magnetic field’s horizontal component
(b) The distance between a point and the North or South Pole affects the angle of dip at that point. Due of Britain’s proximity to the magnetic North Pole on the globe, the angle of dip there would be greater (it is roughly 70°) than in southern India.
(c)A massive bar magnet with its north pole close to the geographic South Pole and its south pole close to the geographic North Pole is hypothetically thought to be buried inside the earth.
From a magnetic north pole, magnetic field lines extend, and they come to an end at a magnetic south pole. As a result, Melbourne, Australia’s field lines on a map showing the earth’s magnetic field lines would appear to emerge from the ground.
(d)A compass will be free to travel in the horizontal plane while the earth’s field is exactly vertical to the magnetic poles if it is placed on the geomagnetic North Pole or South Pole. The compass can then point in any direction in this situation.
(e) Using the formula for magnetic field on the equatorial line of a magnetic dipole i.e. he magnetic field at point P equatorial position on earth can be calculated as
(f) Localized magnetic dipoles may form as a result of charged ion migration in the environment or magnetised mineral formations.
5.2 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) The magnetic field of the Earth varies over time. It takes a few hundred years for things to alter significantly. It is impossible to ignore how the magnetic field of the earth changes throughout time.
(b) Iron is molten in the Earth’s core. This particular type of iron is not magnetic. As a result, this is not regarded as the earth’s magnetism’s As a result, this is not regarded as the earth’s magnetism’s
(c) The energy that powers the currents in the outer conducting portions of the earth’s core is derived from the radioactivity of the planet’s interior. The earth’s magnetism is thought to be caused by these charged currents.
(d) Over the course of its 4 to 5 billion year history, the Earth’s magnetic field has been reversed numerous times. When rocks solidified, these magnetic fields were weakly recorded in them. The investigation of this rock’s magnetism can provide hints about the geomagnetic past.
(e)Due to the ionosphere, the Earth’s field significantly deviates from its dipole structure at considerable distances (more than 30,000 km). Because of the field of single ions in this area, the earth’s field is altered. These ions generate the magnetic field that surrounds them as they move.
(f)Charged particles travelling in a circle can be bent by a very weak magnetic field. This might not be apparent for a path with a huge radius. The deflection in relation to the enormous interstellar space.
5.3 A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet?
Ans –Angle between axis of bar magnet and external magnetic field, θ = 30°
Strength of external magnetic field, B = 0.25 T
Torque experienced by bar magnet, Τ = 4.5 × 10-2J
5.4 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?
Ans – (a) When a magnet returns to its initial state after being disturbed, an equilibrium is said to be stable. Bar magnet is at 0 = 0° Potential energy and is in a stable balance.
(b) In contrast, the moment in an unstable equilibrium is at a 180° angle to the magnetic field. The magnet will never return to its original position if it is rotated..
θ = 180°
Now, by putting values in equation (1), we get,
U = -0.32 × 0.15 × Cos 180°
U = 0.048 J
5.5 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?
Ans – The solenoid produces a magnetic field and an axis when current is supplied through it. Similar to a bar magnet, the magnetic field lines come from one end and enter the other. The solenoid’s two ends function as a bar magnet’s two poles.
In this case, the solenoid has 800 revolutions.
I = 3A
A = 2.5 x 10-4m2
The magnetic moment of the solenoid,
M = (IA) x number of turns
= 3 x 2.5 x 10-4 x 800
= 0.6 Am2
5.6 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 torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Ans –Magnetic field strength, B = 0.5 T
Magnetic moment, M = 3.12 T−1
Using x = MB sin θ, we get
x = 0.6 x 0.25 x sin 30
= 0.6 x 0.25 x 12
= 0.3 x 0.25 = 0.075 Nm
= 7.5 x 10-2 Nm.
5.7 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)?
Ans – (a) Magnetic moment, M = 1.5JT−1
Magnetic field strength, B = 0.22 T
(i) Initial angle of the axis with respect to the magnetic field is 1=0 °.
Final angle of 90 degrees between the magnetic field’s axis and its field
The following is a list of the labour needed to make the magnetic moment normal to the direction of the magnetic field:
W = −MB(cos θ2 – cos θ1)
= −1.5 × 0.22 (cos90 ° – cos0 °
= – 0.33 (0 – 1)
= 0.33 J
(ii) Initial angle of the axis with respect to the magnetic field is 1=0 °.
Final angle of 180 degrees between the magnetic field’s axis and its field
The following is a list of the labour needed to make the magnetic moment normal to the direction of the magnetic field:
W = −MB (cosθ2 – cos θ1)
= −1.5 × 0.22 (cos180 ° – cos0 °)
= – 0.33 (– 1 – 1)
= 0.66 J
(b) For case (i): θ= θ2=90° ∴ Torque, τ=MBsinθ
= MBsin90°
= 0.33 J
(ii): θ= θ2=180 ° ∴ Torque, τ=MBsinθ
= MBsin180°
= 0 J
5.8 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10 –4 m2 , 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 × 10–2 T is set up at an angle of 30° with the axis of the solenoid?
Ans -(a) The solenoid’s magnetic moment along its axis is computed as follows:
Magnetic moment associated with solenoid
M = NIA = 2000 × 4 × 1.6 × 10-4 = 1.28 A m2
(b) Force on The solenoid’s magnetic moment along its axis is computed as follows:
Torque x = MB sinθ = 1.28 × 7.5 × 10-2 × sin 30°
or
× = 4.8 × 10-2 N m
The torque tends to align the solenoid in the direction of magnetic field.
5.9 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 × 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?
Ans – Radius of the coil, r = 10 cm = 0.1 m
Cross-section of the coil, A = πr2 = π × (0.1)2 m2
Current in the coil, I = 0.75 A
5.10 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.
Ans – The location is in the northern hemisphere because the north tip is pointed downward at a 22° angle to the horizontal.
5.11 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.
Ans – Declination angle 0 is 12° west because the compass needle points 12° west of geographic north. The southern hemisphere is where the location is, and the angle of dip is 60°, as the north tip of the magnetic needle is 60° above horizontal. Calculations for the magnitude of the net magnetic field include
5.12 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.
Ans – Magnetic moment of the bar magnet, M = 0.48 J T−1
5.13 A short bar magnet placed in a horizontal plane has its axis aligned along 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.)
Ans –The earth’s magnetic field is uniform with a magnitude of 0.36 G
in the direction of geographical south to geographical north since the angle of dip is zero at that location.
Magnetic field at the equatorial line of the magnet is given
5.14 If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?
Ans – When a magnet is flipped 180 degrees, its south pole is located in the geographic south. Therefore, the null point will be located at a distance of x from the magnet’s centre, on the equatorial line.
5.15 A short bar magnet of magnetic moment 5.25 × 10–2 J T –1 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 45° 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.
Ans –
- Assume that the magnetic field produced by the magnet at point P forms an angle of = 45° with the field of the earth.
Therefore,
5.16 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 magnetisation of a ferromagnet?
Ans – (a) When paramagnetic materials are cooled, the tendency of the thermal agitation to interfere with the alignment of magnetic dipoles reduces. They hence exhibit stronger magnetization.
(b) A diamagnetic does not have intrinsic magnetic dipole moments in its atoms. A diamagnetic sample’s magnetic moment is always perpendicular to the direction of the field when it is placed in a magnetic field. The dipoles’ thermal motion has no impact on it.
(c) Because bismuth is diamagnetic, the core coil’s magnetic field will be slightly less than when the core is empty.
(d) The applied magnetic field affects a ferromagnetic material’s permeability. Lower magnetic fields are more permeable.
(e) The fact that the field lines virtually always intersect the material when a material has r > > 1 is one of the causes of this finding.
(f) The magnetization of a paramagnetic sample with saturated magnetization will be in the same.order as that of a ferromagnetic substance. However, to achieve the saturation magnetization, a magnetising field that is too powerful will be needed. Additionally, the atomic dipole strengths of paramagnetic and ferromagnetic materials may differ slightly.
5.17 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.
Ans – (a)Atomic dipoles are arranged in domains in a specimen of ferromagnet. The alignment and net magnetic moment of each dipole inside a domain are the same.
These domains are randomly distributed in an unmagnetized substance, resulting in zero magnetization. These domains line up in the field’s direction when the substance is exposed to an external magnetic field.
When the external field is removed, some energy is used in the alignment process, so the domains don’t totally return to their random places. The compound still has some magnetic properties. The energy used in the magnetization process is f’ not entirely returned. The remaining energy is released as heat.
(b) The area of a hysteresis loop is directly proportional to the heat energy dissipated. The area of the hysteresis curve is larger in carbon steel. As a result, it dissipates more heat energy.
(c) The memory or record of a magnetization’s hysteresis loop cycles is what gives a magnetization its value. These nuggets of knowledge line up with the magnetization cycle. Information can be stored in hysteresis loops.
(d)Ceramic is utilised to create memory storage in contemporary computers as well as to coat magnetic tapes in cassette players.
(e) If a region of space is encircled by soft iron rings, it can be protected from magnetic fields. These configurations draw the magnetic lines outside of the area.
5.18 A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian.
The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Ans – Let the neutral point be x feet away from the cable.
At this point, the cable’s magnetic field is equal in strength to the earth’s magnetic field and is pointing in the opposite direction.
5.19 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 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Ans – earth’s magnetic field at the place (H) = 0.39G = 0.39 × 10-4T
The angle of dip (δ) = 35º
Magnetic declination (θ) ∼ 0o
5.20 A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(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.
Ans – (a)
(b) The needle will change its orientation when the coil’s current is reversed and it is turned 90 degrees around its vertical axis. In this situation, the needle will point westward.
5.21 A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10–2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?
Ans -If the torques caused by B1 and B2 are equal and in opposition to each other, the dipole will be in equilibrium.
That is, MB1 sin = MB2 sin θ2
5.22 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 = 9.11 × 10 –31 kg). [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 – Mass of electron (me) = 9.11 × 10–31 kg
Horizontal magnetic field (B) = 0.04 G = 0.04 × 10-4T
5.23 A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10–23 J T –1. 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 – Dipole moment of each atomic dipole (M) = 1.5 × 10-23JT-1
Magnetic field (B1) = 0.64 T and Temperature (θ1) = 4.2K
5.24 A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Ans – Number of turns in the ring (n) = 3500
Relative permeability of the ferromagnetic core (μr) = 800
5.25 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.
Ans –
