Chapter 5 Magnetism And Matter Solutions
Question - 11 : - At a certain location inAfrica, a compass points 12º west of the geographic north. The north tip of themagnetic needle of a dip circle placed in the plane of magnetic meridian points60º above the horizontal. The horizontal component of the earth’s field ismeasured to be 0.16 G. Specify the direction and magnitude of the earth’s fieldat the location.
Answer - 11 : -
Angle of declination,θ =12°
Angle of dip, 
Horizontal component ofearth’s magnetic field, BH = 0.16 G
Earth’smagnetic field at the given location = B
We can relate B and BHas:Earth’s magnetic fieldlies in the vertical plane, 12° West of the geographic meridian, making anangle of 60° (upward) with the horizontal direction. Its magnitude is 0.32G.
A short bar magnet has amagnetic moment of 0.48 J T−1. Give the direction and magnitude ofthe magnetic field produced by the magnet at a distance of 10 cm from thecentre of the magnet on (a) the axis, (b) the equatorial lines (normalbisector) of the magnet.
Answer - 12 : -
Magnetic moment ofthe bar magnet, M = 0.48 J T−1
(a) Distance, d =10 cm = 0.1 m
The magnetic field atdistance d, from the centre of the magnet on the axis is given bythe relation:
Where,
= Permeability of free space =
The magnetic fieldis along the S − N direction.
(b) The magnetic field at adistance of 10 cm (i.e., d = 0.1 m) on the equatorial line ofthe magnet is given as:
The magnetic fieldis along the N − S direction.
Question - 13 : - A short bar magnetplaced in a horizontal plane has its axis aligned along the magneticnorth-south direction. Null points are found on the axis of the magnet at 14 cmfrom the centre of the magnet. The earth’s magnetic field at the place is 0.36G and the angle of dip is zero. What is the total magnetic field on the normalbisector 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 amagnet is equal and opposite to the horizontal component of earth’s magneticfield.)
Answer - 13 : -
Earth’s magnetic fieldat the given place, H = 0.36 G
The magnetic field at adistance d, on the axis of the magnet is given as:Where,
= Permeability of freespace
M = Magnetic moment
The magnetic field atthe same distance d, on the equatorial line of the magnet is givenas:
Total magnetic field,
Hence, the magneticfield is 0.54 G in the direction of earth’s magnetic field.
If the bar magnet inexercise 5.13 is turned around by 180º, where will the new null points belocated?
Answer - 14 : -
The magnetic field onthe axis of the magnet at a distance d1 = 14 cm,can be written as:
Where,
M = Magnetic moment
= Permeability of freespace
H = Horizontal component of the magneticfield at d1
If the bar magnet isturned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magneticfield at a distance d2, on the equatorial line of themagnet can be written as:
Equating equations(1) and (2), we get:
The new null points willbe located 11.1 cm on the normal bisector.
Question - 15 : - A short bar magnet ofmagnetic moment 5.25 × 10−2 J T−1 is placedwith its axis perpendicular to the earth’s field direction. At what distancefrom the centre of the magnet, the resultant field is inclined at 45º withearth’s field on
(a) its normal bisectorand (b) its axis. Magnitude of the earth’s field at the place is given to be0.42 G. Ignore the length of the magnet in comparison to the distancesinvolved.
Answer - 15 : -
Magnetic moment of thebar magnet, M = 5.25 × 10−2 J T−1
Magnitude of earth’s magneticfield at a place, H = 0.42 G = 0.42 × 10−4 T
(a) The magnetic field at adistance R from the centre of the magnet on the normalbisector is given by the relation:
Where,
= Permeability of freespace = 4π × 10−7 Tm A−1
When the resultant fieldis inclined at 45° with earth’s field, B = H
(b) The magnetic field at adistanced 
from the centre of themagnet on its axis is given as:
The resultant fieldis inclined at 45° with earth’s field.
Question - 16 : - A short bar magnet ofmagnetic moment 5.25 × 10−2 J T−1 is placedwith its axis perpendicular to the earth’s field direction. At what distancefrom the centre of the magnet, the resultant field is inclined at 45º withearth’s field on
(a) its normal bisectorand (b) its axis. Magnitude of the earth’s field at the place is given to be0.42 G. Ignore the length of the magnet in comparison to the distancesinvolved.
Answer - 16 : -
Magnetic moment of thebar magnet, M = 5.25 × 10−2 J T−1
Magnitude of earth’s magneticfield at a place, H = 0.42 G = 0.42 × 10−4 T
(a) The magnetic field at adistance R from the centre of the magnet on the normalbisector is given by the relation:
Where,
= Permeability of freespace = 4π × 10−7 Tm A−1
When the resultant fieldis inclined at 45° with earth’s field, B = H
(b) The magnetic field at adistanced from the centre of themagnet on its axis is given as:
The resultant fieldis inclined at 45° with earth’s field.
Question - 17 : - Answer the followingquestions:
(a) Why does a paramagneticsample display greater magnetisation (for the same magnetising field) whencooled?
(b) Why is diamagnetism, incontrast, almost independent of temperature?
(c) If a toroid uses bismuthfor 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 aferromagnetic material independent of the magnetic field? If not, is it morefor lower or higher fields?
(e) Magnetic field lines arealways nearly normal to the surface of a ferromagnet at every point. (This factis analogous to the static electric field lines being normal to the surface ofa conductor at every point.) Why?
(f ) Would the maximumpossible magnetisation of a paramagnetic sample be of the same order ofmagnitude as the magnetization of a ferromagnet?
Answer - 17 : -
(a) Owing to therandom thermal motion of the molecules, the alignments of dipoles get disruptedat high temperatures. On cooling, this disruption is reduced. Hence, aparamagnetic sample displays greater magnetisation when cooled.
(b) Each molecule ofthe diamagnetic material is not a magnetic dipole in itself. Hence, the randomthermal motion of the molecules of the diamagnetic material (which is relatedto the temperature) does not affect the diamagnetism of the material.
(c) Bismuth is adiamagnetic substance. Hence, a toroid with a bismuth core has a magnetic fieldslightly less than a toroid whose core is empty.
(d) The permeabilityof ferromagnetic materials is not independent of the applied magnetic field. Itis greater for a lower field.
(e) The permeabilityof a ferromagnetic material is always greater than one. Hence, magnetic fieldlines are always nearly normal to the surface of such materials at everypoint. The proof of this fact is based on the boundary conditions of themagnetic fields at the interface of two media.
(f) Yes, the maximumpossible magnetisation of a paramagnetic sample will be of the same order ofmagnitude as the magnetisation of a ferromagnet. This requires high magnetisingfields for saturation.
Question - 18 : - Answer the followingquestions:
(a) Explain qualitatively onthe basis of domain picture the irreversibility in the magnetisation curve of aferromagnet.
(b) The hysteresis loop of asoft iron piece has a much smaller area than that of a carbon steel piece. Ifthe material is to go through repeated cycles of magnetisation, which piecewill dissipate greater heat energy?
(c) ‘A system displaying ahysteresis loop such as a ferromagnet, is a device for storing memory?’ Explainthe meaning of this statement.
(d) What kind offerromagnetic material is used for coating magnetic tapes in a cassette player,or for building ‘memory stores’ in a modern computer?
(e) A certain region ofspace is to be shielded from magnetic fields.
Suggest a method.
Answer - 18 : -
The hysteresis curve (B-H curve)of a ferromagnetic material is shown in the following figure.
(a) It can be observed fromthe given curve that magnetisation persists even when the external field isremoved. This reflects the irreversibility of a ferromagnet.
(b)The dissipated heatenergy is directly proportional to the area of a hysteresis loop. A carbonsteel piece has a greater hysteresis curve area. Hence, it dissipates greaterheat energy.
(c)The value ofmagnetisation is memory or record of hysteresis loop cycles of magnetisation.These bits of information correspond to the cycle of magnetisation. Hysteresisloops can be used for storing information.
(d)Ceramic is used forcoating magnetic tapes in cassette players and for building memory stores inmodern computers.
(e)A certain region ofspace can be shielded from magnetic fields if it is surrounded by soft ironrings. In such arrangements, the magnetic lines are drawn out of the region.
Question - 19 : - A long straighthorizontal cable carries a current of 2.5 A in the direction 10º south of westto 10° north of east. The magnetic meridian of the place happens to be 10ºwest of the geographic meridian. The earth’s magnetic field at the location is0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignorethe thickness of the cable). (At neutral points, magnetic field dueto a current-carrying cable is equal and opposite to the horizontal componentof earth’s magnetic field.)
Answer - 19 : -
Current in thewire, I = 2.5 A
Angle of dip at the givenlocation on earth, 
= 0°Earth’s magneticfield, H = 0.33 G = 0.33 × 10−4 T
The horizontal componentof earth’s magnetic field is given as:
HH = H cos The magnetic field atthe neutral point at a distance R from the cable is given bythe relation:
Where,
= Permeability of free space = 
Hence, a set ofneutral points parallel to and above the cable are located at a normal distanceof 1.51 cm.
A telephone cable at aplace has four long straight horizontal wires carrying a current of 1.0 A inthe same direction east to west. The earth’s magnetic field at the place is0.39 G, and the angle of dip is 35º. The magnetic declination is nearly zero.What are the resultant magnetic fields at points 4.0 cm below the cable?
Answer - 20 : -
Number of horizontalwires in the telephone cable, n = 4
Current in eachwire, I = 1.0 A
Earth’s magnetic fieldat a location, H = 0.39 G = 0.39 × 10−4 T
Angle of dip at thelocation, δ = 35°
Angle ofdeclination, θ ∼ 0°
For a point 4 cm below the cable:
Distance, r =4 cm = 0.04 m
The horizontal componentof earth’s magnetic field can be written as:
Hh = Hcosδ − B
Where,
B = Magnetic field at 4 cm due tocurrent I in the four wires
= Permeability offree space = 4π × 10−7 Tm A−1
= 0.2 × 10−4 T= 0.2 G
∴ Hh =0.39 cos 35° − 0.2
= 0.39 × 0.819 −0.2 ≈ 0.12 G
The vertical componentof earth’s magnetic field is given as:
Hv = Hsinδ
= 0.39 sin 35° = 0.22 G
The angle made by thefield with its horizontal component is given as:
The resultant field atthe point is given as:
For a point 4 cm above the cable:
Horizontal component ofearth’s magnetic field:
Hh = Hcosδ + B
= 0.39 cos 35° + 0.2 =0.52 G
Vertical component ofearth’s magnetic field:
Hv = Hsinδ
= 0.39 sin 35° = 0.22 G
Angle, θ 
= 22.9°
And resultant field:
