Magnetic Properties of Matter
Questions for Short Answer
1. When a dielectric is placed in an electric field, it gets polarized. The electric field in a polarized material is less than the applied field. When a paramagnetic substance is kept in a magnetic field, the field in the substance is more than the applied field. Explain the reason for this opposite behavior.
ANSWER: In the case of a dielectric placed in an electric field, the dielectric gets polarized. Due to the polarization, there is a field inside it that is opposite to the applied field. Thus the net electric field in the polarised material is less than the applied field.
When a paramagnetic substance is kept in a magnetic field, the randomly oriented magnetic moments of the atoms get aligned along the magnetic field. In this case, unlike the polarized dielectric in the electric field, the magnetic moments of the aligned magnetic fields of the substance are in the same direction as the applied field. Thus the magnetized substance produces an extra magnetic field in the direction of the applied field. The resultant magnetic field in the material is greater than the applied field.
2. The property of diamagnetism is said to be present in all materials. Then, why are some materials paramagnetic or ferromagnetic?
ANSWER: When a substance is placed in an applied magnetic field, dipole moments are induced in the atoms. From Lenz's law, the magnetic field due to the induced magnetic moments opposes the original field. The resultant field is smaller than the applied field. This property is called diamagnetism and it is present in all materials. But some materials have a permanent atomic magnetic moment that gets aligned according to the applied field and produces an extra magnetic moment along the field. This paramagnetism or ferromagnetism property is much stronger than diamagnetism. Such types of materials are paramagnetic or ferromagnetic.
3. Do permeability and relative permeability have the same dimensions?
ANSWER: No. The unit of permeability is T-m/A. Hence its dimensions are,
[µ] =[MT⁻²I⁻¹][L][I⁻¹]
=[MLT⁻²I⁻²]
The relative permeability is a factor by which the magnetic field B is increased when a material is brought into the magnetic field. It is also the comparison factor by which the permeability of a material is greater than the permeability of a vacuum. Thus it is unitless and dimensionless.
4. A rod when suspended in a magnetic field stays in the east-west direction. Can we be sure that the field is in the east-west direction? Can it be in a north-south direction?
ANSWER: We cannot be sure that the field is in the east-west direction because it depends upon the substance of the rod. If the substance of the rod is paramagnetic or ferromagnetic then the field will be in the east-west direction but if it is diamagnetic then the field will be in the north-south direction.
5. Why cannot we make permanent magnets from paramagnetic materials?
ANSWER: Paramagnetic materials show a small magnetization in the direction of the field. When the applied field is removed, that small magnetization too disappears. So the retentivity is nil, thus permanent magnets cannot be made from paramagnetic materials.
6. Can we have magnetic hysteresis in paramagnetic or diamagnetic substances?
ANSWER: No, magnetic hysteresis is only shown by ferromagnetic substances.
7. When a ferromagnetic material goes through a hysteresis loop, its thermal energy is increased. Where does this thermal energy come from?
ANSWER: The aligning process of the atomic magnets does not occur simultaneously with the magnetizing field but lags behind in ferromagnetic materials. Even when the applied magnetic field is withdrawn, some magnetism is retained by it. To make it zero, an opposite magnetic field is applied which is called coercive force. So we have to spend energy to make the residual magnetization to zero. This energy is converted to thermal energy in a similar manner to that of the friction between two surfaces.
8. What are the advantages of using soft iron as a core, instead of steel, in the coils of galvanometers?
ANSWER: Soft iron has a small area of the hysteresis loop because the retentivity and the coercive force are smaller than steel. So it is easily magnetized and only a small magnetization is retained when the field is removed. Also, the loss of energy is small during periodic variations in the magnetizing fields. The area of the hysteresis loop of steel is larger, so retentivity and coercive force are higher. Energy loss is also higher in periodic variations of the magnetic fields in the case of steel. Hence soft iron is more suitable as a core in the coils of the galvanometer.
9. To keep valuable instruments away from the earth's magnetic field, they are enclosed in iron boxes. Explain.
ANSWER: Iron being ferromagnetic, has a relative permeability of the order of thousands. So the earth's magnetic field lines near the iron box pass through the material of the box and do not go inside the box. Thus the valuable instruments inside the iron box remain unaffected from the earth's field. 
Diagram for Q-9
OBJECTIVE-I
1. A paramagnetic material is placed in a magnetic field. Consider the following statements:
(A) If the magnetic field is increased, the magnetization is increased.
(B) If the temperature is increased the magnetization is increased.
(a) Both A and B are true.
(b) A is true but B is false.
(c) B is true but A is false.
(d) Both A and B are false.
ANSWER: (b)
EXPLANATION: In a paramagnetic material, the atomic dipoles align along the applied magnetic field. Thus the resulting field is more than the applied field. So the magnetization will increase if the magnetic field is increased. Statement A is correct.
When the temperature is increased, the randomization of the individual atomic magnetic moments increases, thus decreasing the magnetization I for a given magnetic intensity. So the statement B is false.
Hence the option (b) is correct.
2. A paramagnetic material is kept in a magnetic field. The field is increased till the magnetization becomes constant. If the temperature is now decreased, the magnetization
(a) will increase
(b) decrease
(c) remain constant
(d) may increase or decrease.
ANSWER: (c)
EXPLANATION: When the temperature is decreased after the magnetization is constant, the alignment of individual magnetic dipoles does not vary. So the magnetization remains constant. Option (c) is correct.
3. A ferromagnetic material is placed in an external magnetic field. The magnetic domains
(a) increase in size
(b) decrease in size
(c) may increase and decrease in size
(d) have no relation with the field.
ANSWER: (c)
EXPLANATION: The magnetic domains inside a ferromagnetic material have the directions of magnetic moment randomly organized. When the ferromagnetic material is placed in an external magnetic field, the domains having the magnetic field direction along the external magnetic field direction, grow in size while those opposing it get reduced. Hence the option (c) is correct.
4. A long straight wire carries a current 'i'. The magnetizing field intensity H is measured at a point P close to the wire. A long cylindrical iron rod is brought close to the wire so that point P is at the center of the rod. The value of H at P will
(a) increase many times
(b) decrease many times
(c) remain almost constant
(d) become zero.
ANSWER: (c)
EXPLANATION: Since the cylindrical rod is long, end effects at the center of the rod may be neglected. So the magnetic intensity due to magnetization is zero. Thus the magnetizing field intensity is determined by the external sources only that is the current-carrying wire here. So H will remain almost constant at point P. Option (c) is correct.
5. The magnetic susceptibility is negative for
(a) paramagnetic materials only
(b) diamagnetic materials only
(c) ferromagnetic materials only
(d) paramagnetic and ferromagnetic materials.
ANSWER: (b)
EXPLANATION: The intensity of magnetization I is proportional to magnetic intensity H. Thus,
I = χH
The proportionality constant χ is called the magnetic susceptibility of the material. It is positive for paramagnetic and ferromagnetic materials but negative for diamagnetic materials. Option (b) is correct.
6. The desirable properties for making permanent magnets are
(a) High retentivity and high coercive force
(b) high retentivity and low coercive force
(c) low retentivity and high coercive force
(d) low retentivity and low coercive force.
ANSWER: (a)
EXPLANATION: Permanent magnets have high magnetization even without the external magnetic field. To make a permanent magnet the material should have high retentivity. i.e. even when the external magnetizing field is brought back to zero, the material still remains magnetized.
Also, this residual magnetization should not be easily lost when the opposite magnetizing field is encountered. i.e. the value of H needed to dis-align the domains forcibly (coercive force) should also be high. Option (a) is correct.
7. Electromagnets are made of iron because soft iron has
(a) high retentivity and high coercive force
(b) high retentivity and low coercive force
(c) low retentivity and high coercive force
(d) low retentivity and low coercive force.
ANSWER: (d)
EXPLANATION: Electromagnets need instant magnetization and instant demagnetization. For this, there should be very little magnetization left in the iron when the current is made zero. Also whatever magnetization is remained should get removed with little coercive force. So low retentivity and low coercive force, that is available in soft iron. Option (d) is correct.
OBJECTIVE-II
1. Pick the correct options.
(a) All electrons have magnetic moments.
(b) All protons have magnetic moments.
(c) All nuclei have a magnetic moment.
(d) All atoms have a magnetic moment.
ANSWER: (a), (b).
EXPLANATION: Electrons in an atom move about the nucleus in closed paths and hence constitute electric current loops. As current loops have a magnetic dipole moment, each electron has a magnetic moment due to its orbital motion. Also due to spin and resulting angular momentum each electron has a permanent magnetic moment. Option (a) is correct.
All protons have also a magnetic moment. Option (b) is correct.
Some nuclei may have a magnetic moment but not all. Similarly, in a number of atoms, the resultant magnetic moment is zero. So all atoms do not have a magnetic moment. Options (c) and (d) are incorrect.
2. Permanent magnetic moment of the atoms of a material is not zero. The material
(a) must be paramagnetic
(b) must be diamagnetic.
(c) must be ferromagnetic.
(d) may be paramagnetic.
ANSWER: (d)
EXPLANATION: Such material may be paramagnetic or ferromagnetic. It is not the condition for a sure paramagnetic material. Option (d) is correct.
3. The permanent magnetic moment of the atoms of material is zero. The material
(a) must be paramagnetic
(b) must be diamagnetic.
(c) must be ferromagnetic.
(d) may be paramagnetic.
ANSWER: (b)
EXPLANATION: In this case, there will be an induced dipole magnetic moment in the atoms when kept in a magnetic field. From Lenz's law, this induced magnetic moment will oppose the original field. The resultant field in such materials is, therefore, smaller than the applied field. This phenomenon is diamagnetism. Option (b) is correct.
4. Which of the following pairs has quantities of the same dimensions?
(a) magnetic field B and magnetizing field intensity H
(b) magnetic field B and intensity of magnetization.
(c) magnetizing field intensity H and intensity of magnetization I.
(d) longitudinal strain and magnetic susceptibility.
ANSWER: (c), (d).
EXPLANATION: H and I are intensities. Their dimensions are the same. Option (c) is correct.
Longitudinal strain is a dimensionless quantity. Susceptibility is a proportionality constant between H and I, both having the same dimensions. Thus susceptibility is also dimensionless. Option (d) is correct.
5. When a ferromagnetic material goes through a hysteresis loop, the magnetic susceptibility
(a) has a fixed value
(b) maybe zero
(c) maybe infinity
(d) maybe negative.
ANSWER: (b), (c), (d).
EXPLANATION: Since the magnetic susceptibility =I/H, this ratio may sometimes be zero and sometimes infinity when the material goes through a hysteresis loop. It is because there comes a stage when I = zero but H is non-zero and also sometimes it is the opposite. Hence options (b) and (c) are correct.
It may also be negative sometimes because H and I may have opposite directions. Option (d) is correct.
6. Mark out the correct options.
(a) Diamagnetism occurs in all materials
(b) Diamagnetism results from the partial alignment of the permanent magnetic moment.
(c) The magnetizing field intensity H is always zero in free space.
(d) The magnetic field of the induced magnetic moment is opposite to the applied field.
ANSWER: (a), (d).
EXPLANATION: Diamagnetism occurs due to induced magnetic moments that are opposite to the original field according to Lenz's law. And it happens in all materials. So options (a) and (d) are correct, option (b) is incorrect.
In a vacuum, I = zero. The magnetizing field intensity
H =B/µₒ -I
=B/µₒ, may not be zero. Option (c) is incorrect.
EXERCISES
1. The magnetic intensity H at the center of a long solenoid carrying a current of 2.0 A, is found to be 1500 A/m. Find the number of turns per centimeter of the solenoid.
ANSWER: The magnetic intensity at the center of a long solenoid,
H =B/µₒ =µₒni/µₒ
→H =ni.
Given i =2 A, H =1500 A/m. So,
1500 =n*2
→n =750 turns/m
=750/100 turns/cm
=7.5 turns/cm.
2. A rod is inserted as the core in the current-carrying solenoid of the previous problem, (a) What is the magnetic intensity H at the center? (b) If the magnetization I of the core is found to be 0.12 A/m, find the susceptibility of the material of the rod. (c) Is the material paramagnetic, diamagntic or ferromagnetic?
ANSWER: (a) Since the rod is inserted as a core, it will be long as the solenoid. So at the center, the end effects may be neglected. Thus there will be no effect of the rod on the magnetic intensity at the center. The magnetic intensity H will remain the same i.e. equal to 1500 A/m.
(b) Given, magnetization of the core
I = 0.12 A/m.
Since I =χH
(Where χ is susceptibility)
→χ =I/H
= 0.12/1500
= 8.0x10⁻⁵.
(c) Since the susceptibility of the material χ, is positive and very small it is paramagnetic.
3. The magnetic field inside a long solenoid having 50 turns/cm is increased from 2.5x10⁻³ T to 2.5 T when an iron core of cross-sectional area 4 cm² is inserted into it. Find (a) the current in the solenoid, (b) the magnetization I of the core, and (c) the pole strength developed in the core.
ANSWER: (a) The magnetic field inside the solenoid, B =2.5x10⁻³ T.
Number of turns in the solenoid,
n =50 turns/cm =5000 turns/m.
If i is the current in the solenoid, then
B =µₒni
→i =B/µₒn
=2.5x10⁻³/(4π*10⁻⁷*5000) A
=0.4 A.
(b) Number of turns in the solenoid,
n =50 turns/cm =5000 turns/m.
After the core is inserted, the magnetic field inside the solenoid becomes, B =2.5 T. Now,
B =µₒ(H+I)
→H+I =B/µₒ
→I =B/µₒ -H
=B/µₒ -ni
=2.5/(4π*10⁻⁷) -5000*0·4 A/m
=2.0x10⁶ -2000 A/m
≈2.0x10⁶ A/m.
(c) Suppose there gets a pole strength m developed in the core. Area of the cross-section of the core A =4 cm² =4x10⁻⁴ m². The magnetization I of the core is given as,
I = m/A
→m = IA
= 2.0x10⁶*4x10⁻⁴ A-m
= 800 A-m.
4. A bar magnet of length 1 cm and cross-sectional area 1.0 cm² produces a magnetic field of 1.5x10⁻⁴ T at a point in end-on position at a distance of 15 cm away from the center. (a) Find the magnetic moment M of the magnet. (b) Find the magnetization I of the magnet. (c) Find the magnetic field B at the center of the magnet.
ANSWER: (a) Let the magnetic moment of the bar magnet = M. Distance of the point from the magnet in end-on position, r = 15.0 cm =0.15 m. The field is given as,
B =(µₒ/4π)*2M/r³
→M =½(4π/µₒ)Br³
=½(10⁷)*1.5x10⁻⁴*(0.15)³ A-m²
=2.50 A-m².
(b) Volume of the magnet,
V =(1.0 cm²)*(1.0 cm)
=1.0 cm³
=1.0x10⁻⁶ m³
Magnetization of the magnet,
I =M/V
=2.50/(1.0x10⁻⁶) A/m
=2.5x10⁶ A/m.
(c) Magnetic moment M =m*2l
→m =M/2l =2.5/0.01 A-m
→m =250 A-m.
Magnetic intensity H due to a magnetic pole of pole strength m,
H =m/(4πr²)
The intensity at the center will be equal and in the same direction, hence total magnetic intensity at the center of the magnet,
H =2m/(4πr²).
Its direction will be from north to south pole of the magnet. The direction of magnetization I will be along the direction of the magnetic moment of the magnet which is from south pole to north pole of the magnet. So the directions of H and I are opposite to each other.
The net magnetic field at the center will be
B =µₒ(H+I),
{with appropriate directions}
→B = -µₒ*2m/(4πr²) +µₒI
= -(µₒ/4π)*2m/r² +µₒI
r =0.5 cm =0.005 m, hence
B = -10⁻⁷*2*250/(0.005)²+4π*10⁻⁷*2.5x10⁶ T
= -2+3.14 T
=1.14 T
5. The susceptibility of annealed iron at saturation is 5500. Find the permeability of annealed iron at saturation.
ANSWER: Susceptibility of annealed iron at saturation, χ =5500.
Permeability at saturation is,
µ =µₒ(1+χ)
=4πx10⁻⁷(1+5500)
=6.9x10⁻³
6. The magnetic field B and the magnetic intensity H in a material are found to be 1.6 T and 1000 A/m respectively. Calculate the relative permeability µᵣ and the susceptibility χ of the material.
ANSWER: B =µH, where µ is the permeability of the material. Hence,
µ =B/H
=1.6/1000
=0.0016
Relative permeability µᵣ =µ/µₒ
→µᵣ =0.0016/(4πx10⁻⁷)
=1.3x10³
Now the relation between susceptibility of the material and permeability is
µ =µₒ(1+χ)
Where χ is the susceptibility.
0.0016 =4πx10⁻⁷(1+χ)
→1+χ =1.3x10³
→χ ≈ 1.3x10³.
7. The susceptibility of magnesium at 300 K is 1.2x10⁻⁵. At what temperature will the susceptibility increase to 1.8x10⁻⁵?
ANSWER: The susceptibility of a paramagnetic material is inversely proportional to the absolute temperature. Hence,
χ/χ' =T'/T
→T' =χT/χ'
=1.2x10⁻⁵*300/1.8x10⁻⁵
=200 K.
8. Assume that each iron atom has a permanent magnetic moment equal to 2 Bohr magnetons (1 Bohr magneton equals 9.27x10⁻²⁴ A-m²). The density of atoms in iron is 8.52x10²⁸ atoms/m³. (a) Find the maximum magnetization I in a long cylinder of iron. (b) Find the maximum magnetic field B on the axis inside the cylinder.
ANSWER: (a) Magnetization I =M/V
Hence the maximum magnetization, taking one m³ of iron,
I =2*9.27x10⁻²⁴*8.52x10²⁸ A/m
=1.58x10⁶ A/m.
(b) At this magnetization, the magnetic field,
B =µₒ(H+I)
Since it is a long cylinder, end effects may be taken as zero.
So, B =µₒI
=4πx10⁻⁷*1.58x10⁶ T
=1.98 T =2.0 T.
9. The coercive force for a certain permanent magnet is 4.0x10⁴ A/m. The magnet is placed inside a long solenoid of 40 turns/cm and a current is passed in the solenoid to demagnetize it completely. Find the current.
ANSWER: Given, n =40 turns/cm
→n =4000 turns/m
Coercive force,
H =4.0x10⁴ A/m
H =ni
→i =H/n
= 4.0x10⁴/4000 A
= 10 A.
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