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Magnetism and MatterMagnetism and Matter
PREPARED BY
Sri. Satishkumar. Umadi
Lecturer
N V Pre university collegeN V Pre university college
Gulbarga
MAGNETIC DIPOLEMAGNETIC DIPOLE
Two unlike magnetic poles of equal pole g p q pstrength separated by a small dist is called a magnetic dipole.a magnetic dipole.Ex: A bar magnet, a solenoid carrying current a circular wire carrying currentcurrent, a circular wire carrying current etc.
Magnetic dipole moment = pole strength xMagnetic dipole moment = pole strength x
magnetic length
The direction of magnetic moment ( ) isThe direction of magnetic moment ( ) is from South to North Pole. SI unit of is Am2
and unit of pole strength is Amand unit of pole strength is Am.
MAGNETIC FIELD AT A POINT ON AXIAL LINE:
The magnetic field at a point on the axial
MAGNETIC FIELD AT A POINT ON AXIAL LINE:
The magnetic field at a point on the axial line at a distance ‘r’ from the centre of a magnet of dipole moment ismagnet of dipole moment is
H i l h 2l i d bHere magnetic length 2l is assumed to be much smaller than distance of point(r) from the centre of magnet (r >>>2 l)the centre of magnet.(r >>>2 l) .
MAGNETIC FIELD AT A POINT ON
Th ti fi ld t i t t i l
MAGNETIC FIELD AT A POINT ON EQUATORIAL LINE
The magnetic field at a point on equatorial line at a distance ‘r’ from the centre of a
t f di l t imagnet of dipole moment is
Here magnetic length 2l is assumed to beHere magnetic length 2l is assumed to be much smaller than distance of point(r) (r >>>2 l) .2 l) .
TORQUE ON A MAGNETIC DIPOLE IN ATORQUE ON A MAGNETIC DIPOLE IN A UNIFORM MAGNETIC FIELD:
θτ = m. B. sinθIn vector form
POTENTIAL ENERGY OF A MAGNETICPOTENTIAL ENERGY OF A MAGNETIC DIPOLE IN A MAGNETIC FIELD:
The amount of work done in rotating dipole through a small angle ‘dθ’ isthrough a small angle dθ is
dw= τ.dθThe potential energy of dipole isThe potential energy of dipole is
U = ‐m B cos θU =U =
1. A magnetic needle kept in a non . ag et c eed e ept a ouniform magnetic field , it experiences 1 Torque but no force1. Torque but no force2. neither force nor torque3 f d t3. a force and a torque4. a force but not torque
• Solution : In a non uniform magnetic field it i b h d ff i fexperiences both torque and an effective force.
2. The magnetic moment of a diamagnetic atom is g1. Equal to zero2 equal to one2. equal to one 3. between zero and one4 much greater than one4. much greater than one
• Solution : magnetic moment of a diamagnetic atom is equal to zero.
3 A charged particle (charge q) is moving3. A charged particle (charge q) is moving in a circle of radius R with uniform
d ‘ ’ h i d ispeed ‘v’. The associated magnetic moment is
1. q v R /2 2. q v R3. q vR2/2 4. q vR2q / q
• Solution : The current associated with motion• Solution : The current associated with motion
• I = Q/t = q / (2 π R/v) = q v / 2 π R
• The magnetic moment =I A
• = q v x π R R / 2 π R = q v R/ 2q / q /
4. A bar magnet having a magnetic moment is free to rotate in a horizontal plane . A horizontal magnetic field of B= exists in the space Thefield of B= exists in the space. The work done in taking the magnet slowly f di ti ll l t th fi ld tfrom a direction parallel to the field to a direction 60 degree from the field is 1. 0.6 J 2. 2 J3. 6 J 4. 12 J
• Solution : Work done in rotating a dipole isSolution : Work done in rotating a dipole is W =
5. The dimensions of magnetic dipole moment will be1) LA 2) ML‐1 A3) ML‐1 A‐2 4) L2A3) ML A 4) L A
• Magnetic moment = m = q X 2l
= Ampere X meter X meter Ampere X meter X meter
= A m2
6 The work done in turning a magnet of6. The work done in turning a magnet of magnetic moment M by an angle of 90o
f h i idi i ifrom the magnetic meridian is n times the corresponding work done in turning it through an angle of 60o. The value of n is1) 1 2) 23) 1/2 4) 1/43) 1/2 4) 1/4
W k d iWork done is
7. A bar magnet has a magnetic moment of 2.5 JT‐1 and is placed in a magnetic field2.5 JT and is placed in a magnetic field of 0.2 T . Work done in turning the magnet from parallel to anti parallelmagnet from parallel to anti parallel position relative to the field direction is) 2) 01) zero 2) 0.5J3) 2.0 J 4) 1 J
• Solution: work done =
8 R t f h f t ith d fl ti θ8. Rate of change of torque with deflection θis maximum for a magnet suspended freely in a uniform magnetic field of induction B when
1) θ = 45o2) θ = 60o2) θ = 603) θ = 0o4) θ =90o4) θ =90o
• Solution :
• Rate of change of torque with respect to is• Rate of change of torque with respect to is maximum if
9. A magnet 10 cm having a pole strength of 2 ampere meter is deflected through 30o2 ampere meter is deflected through 30from the magnetic meridian. The h i t l t f th’ tihorizontal component of earth’s magnetic field is 0.32 x 10‐4 T. The value of deflecting couple is1) 64 x 10‐7 Nm2) 2.5 x 10‐7 Nm) )3) 32 x 10‐7 Nm 4) 16 x 10‐7 Nm
Solution :Magnetic moment =m = q X 2l
= 2 X 0.1 = 0.2 2 X 0.1 0.2
Torque on the couple
PERIOD OF OSCILLATION OF COMPASS NEEDLEPERIOD OF OSCILLATION OF COMPASS NEEDLE IN A UNIFORM MAGNETIC FIELD
The period of oscillation of a magnet in uniform magnetic field ‘B’ is g
‘ ’ i ti t d ‘I’ i t‘m’ is magnetic moment and ‘I’ is moment of inertia of magnet
10 Two magnets are held together in a10. Two magnets are held together in a vibration galvanometer and are allowed
ill i h h’ i fi ldto oscillate in the earth’s magnetic field with like poles together and 12 oscillations per minute are made but for unlike poles together, only 4 oscillations p g , yper minute are executed. The ratio of their magnetic moments istheir magnetic moments is1) 5 : 4 2) 3 : 13) 3 5 4) 1 33) 3 : 5 4) 1 : 3
• Solution: Period of rotation when like poles are together T 60/12 5 secare together = T1= 60/12 = 5 sec
• Period of rotation when unlike poles are h /together = T2= 60/ 4 = 15 sec
11 For a vibration magnetometer the time11. For a vibration magnetometer, the time period of the suspended bar magnet can b d d bbe reduced by1) moving it towards south pole of earth2) moving it towards north pole of ) g pearth3) moving it towards equator3) moving it towards equator4) any one of them
• Time period of vibration magnetometer is• Time period of vibration magnetometer is
• For same moment of inertia(I) and magnetic moment (m)
• T can be reduced if horizontal component of earth’s magnetic field H is increased . But
• H is high if dip is zero. This occurs at equator. Hence time period can be reduced by moving towardstime period can be reduced by moving towards equator.
12. A magnet oscillates in earth’s magnetic field with time period T. If the mass isfield with time period T. If the mass is quadrupled, then new time will be
1) 2T, motion remaining SHM2) i i i S2) 4T, motion remaining SHM3) T/2, motion remaining SHM4) Unaffected but motion not SHM
S l ti i d f ill tiSolution : period of oscillation
Moment of inertia is directly proportional to its massmass.
13 T b t f th13. Two bar magnets of the same mass, same length and breadth having magnetic moments M and 2M are joined together pole for pole and suspended by a string. p p p y gThe time period of the assembly in a magnetic field of strength H is 3 seconds.magnetic field of strength H is 3 seconds. If now the polarity of one of the magnets is reversed and the combination is againis reversed and the combination is again made to oscillate in the same field,
the time of oscillations isthe time of oscillations is 1) 3 sec 2) 3 sec) )3) 9 sec 4) 6 sec
Solution : For pole to pole combination
m1 = 2m + m = 3m 1
For polarity reversed
m = 2m – m = mm2 = 2m – m = m
.
14 The time period of a freely14. The time period of a freely suspended magnet does not depend upon1) length of the magnet2) the pole strength of the
magnetg3) the horizontal component of
earth’s magnetic fieldearth s magnetic field4) the length of suspension
• Time period depends on Inertia and magnetic• Time period depends on Inertia and magnetic moment (length) of the magnet . There fore time period does not depend on length of thetime period does not depend on length of the suspension.
15 The time period of a freely suspended15. The time period of a freely suspended thin magnet is 4 seconds. If it is broken i l h i l din length in two equal parts and one part is suspended in the same way, then its time period (in seconds) will be
1) 2 2) 43) 0 5 4) 0 253) 0.5 4) 0.25
••
When a magnet is broken into two parts along length its magnetic moment
•
16. A magnet makes 5 oscillations per minute in H 0 3 x 10‐4 T Byper minute in H = 0.3 x 10‐4 T. By what amount the field should be increased so that the number ofincreased so that the number of oscillations is 10 in the same time?
1) 0.3 x 10‐4 T 2) 0.6 x 10‐4 T3) 0.9 x 10‐4 T 4) 1.2 x 10‐4 T
S l ti• Solution:
17. A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth’s horizontal magnetic field of 24µT. When a horizontal field of 18µT is produced opposite to the earth’s field by placing a current carrying wire, the new time period will be 1. 1 S 2. 2 S3. 3 S 4. 4 S
Solution : Time period of rotation of a magnet is
18. When a magnet is suspended horizontally at the neutral point, its frequency of p , q yoscillations is1) infinity 2) zero1) infinity 2) zero3) neither 4) data incomplete
• period of oscillation
• If H = zero then frequency of oscillation = 0
THE EARTHS MAGNETISM:THE EARTHS MAGNETISM:
A line passing through North andA line passing through North and South pole of earths magnet is called
l h h hmagnetic axis. A line passing through North and South of earth is called geographic axis. The magnetic axis is inclined at an angle a g e
INCLINATION OR DIP AT A PLACEINCLINATION OR DIP AT A PLACE:The dip at a place is the angle made by
total magnetic field BE of the earth with the horizontal in the magnetic meridiang
H = B cos I ; Z = B sin IHE = BE cos I ; ZE = BE sin I
2 2 2HE2 + ZE2 = BE
2
Z → vertical comp of earths magnetic fieldZE → vertical comp of earths magnetic field HE → Horizontal comp of earths magnetic field.
19 The angle of dip is 90o at19. The angle of dip is 90o at1) earth’s magnetic poles)2) equator3) both (a) and (b)4) none of these
• Solution : At magnetic poles , the earthsSolution : At magnetic poles , the earths magnetic field is vertical . The angle between magnetic field and horizontal is 90 degree.magnetic field and horizontal is 90 degree. Hence the dip at poles is 90 degree.
20 A bar magnet is placed with its north20. A bar magnet is placed with its north pole pointing earth’s north. The points of
ti fi ld ill b i hi hzero magnetic field will be in which direction from the centre of the magnet?
1) north and south2) east and west3) north‐east and south‐ west3) north east and south west4) north‐east and south‐east
• EE
Neutral point
BBH
( ) ( )N(earth) S(earth)
N S
Neutral point BH
W
21 Earth’s magnetic field has a horizontal21. Earth s magnetic field has a horizontal component except at )1) equator2) magnetic poles of earth3) a latitude of 60o
4) an inclination of 60o)Solution: Horizontal component of magnetic field is HE = BE cos IE E
This is zero if Cos I = 0 i.e . I = 90 (at poles)
22 Vertical component of earth’s22. Vertical component of earth s magnetic field is zero at a place where angle of dip iswhere angle of dip is
1) 45o 2) 90o) )3) 0o 4) 60o
Solution: Vertical component of earth magnetic field isp g
ZE = BE sin I . This is zero if Sin I = 0 Hence I = 0Sin I 0 Hence I 0
This occurs at magnetic equator
23. A magnet suspended at 30o with the magnetic meridian makes an angle of 45oith the hori ontal What is the act alwith the horizontal. What is the actual
value of dip?1 21. 2.
3. 4.
Horizontal component HHorizontal component H H=Hcos30 Hcos30 I = 45
let the dip at place be I R Vlet the dip at place be I R V Then R cos I =Hcos30
R sin I= VTaking the ratio
Tan I = V / H cos 30
But tan(I)= V/H thus
24. The angles of dip at the earth’s magnetic poles and equator respectively are1) 30o, 60o 2) 90o, 0o3) 30o 90o 4) 0o, 0o
• Solution : At poles only vertical component of• Solution : At poles only vertical component of magnetic field exists. Hence dip is 90 degree
• At equator only horizontal component of• At equator only horizontal component of earth magnetic field exists. Hence dip is zero.
25 The angle of dip at a certain place where25. The angle of dip at a certain place where the horizontal and vertical components f h h’ i fi ld l iof the earth’s magnetic field are equal is
1) 30o 2) 45o
3) 60o 4) 75o) )
H = V Hence R cos I = R sin ItanI = 1 Hence I = 450
MAGNETIC PROPERTIES OF MATERIALSMAGNETIC PROPERTIES OF MATERIALS
MAGNETIC PERMEABILITY µ:
RELATIVE PERMEABILITY
MAGNETIC INDUCTION OR MAGNETICMAGNETIC INDUCTION OR MAGNETIC FLUX DENSITY OR MAGNETIC FIELD:
MAGNETIC INTENSITY (H) OR MAGNETIZING FORCE:MAGNETIZING FORCE:
Magnetic intensity does not depend on medium inside solenoid.
MAGNETIC SUSCEPTIBILITY (χ):MAGNETIC SUSCEPTIBILITY (χ):
χ = M/H.χ M/H.
Ratio of magnetization attained (M) to the i i i ( ) f i i fi ldmagnetic intensity (H) of magnetizing field
h i fi ld i h i i f l id ithe magnetic field in the interior of solenoid is B = Bo + Bm
B = µoH + µoM = µo [H + M]
∴ µr = 1 + χWhere µr is relative permeability of medium
HYSTERESIS
1) The area of hysteresis loop gives the1) The area of hysteresis loop gives the energy loss per cycle. Hence the materials which have small area of hysteresis loop arewhich have small area of hysteresis loop are used for making transformer core.
2) The materials which have high retentivity2) The materials which have high retentivity are used for making strong artificial magnets.
3) The coercivity of materials can be known form graph More is the coercivity of materialform graph. More is the coercivity of material, the materials cannot loose magnetism easily.
26 M t it bl t i l f26. Most suitable material for making transformer core is
1) steel 2) nickel3) copper 4) soft iron) pp )
solution: The area of hysteresis loop represents y p pthe loss of energy in every cycle. The hysteresis area for soft iron is very small . Hence soft iron is used for making the core of transformer.
27 Th B H (i) d (ii) h i fi27. The B – H curves (i) and (ii) show in fig. are respectively associated with1) diamagnetic and paramagnetic2) Paramagnetic and diamagnetic) g g3) sift iron and
steelsteel4) steel and
sift ironsift iron
• Explanation :
1. Curve (1) has high retentivity and low ( ) g ycoercivity. It represents hysteresis for soft iron.
2. Curve (2) has low retentivity and high coercivity It represents hysteresis loop forcoercivity . It represents hysteresis loop for steel.
28 Magnetic susceptibility for28. Magnetic susceptibility for diamagnetic materials is
1) ll d ti1) small and negative2) small and positive3) large and positive4) large and negative) g g
Solution :Diamagnetic substances magnetize in opposite direction when they are placed in external magnetic field. Hence they posses small and negative magnetic susceptibility.
29 Susceptibility is positive and large for29. Susceptibility is positive and large for 1) paramagnetic substance) di i b2) diamagnetic substance3) non‐magnetic substance4) ferromagnetic substance
• Feromagnetic substances strongly magnetise• Feromagnetic substances strongly magnetise in the direction of applied external magnetic field Hence they have high magneticfield. Hence they have high magnetic susceptibility ( soft iron has 8000)
30 To shield an instrument from30. To shield an instrument from external magnetic field, it is placed in a cabin made of1) wood 2) iron3) ebonite 4) diamagnetic substance
Solution :Iron has very large magneticSolution :Iron has very large magnetic permeability . Hence it conducts magnetic flux largely There fore magnetic lines pass throughlargely. There fore magnetic lines pass through iron cabin. Thus the instrument is shielded from magnetic fieldfrom magnetic field.
31 The relative permeability of a31. The relative permeability of a substance is 1.001. The substance is1) paramagnetic 2) diamagnetic3) ferromagnetic4) none of the above4) none of the above
A paramagnetic material has a small and i i l f i ibilipositive value of magnetic susceptibility.
Hence the material is a paramagnetic
32 If the relative permeability iron is32. If the relative permeability iron is 2000, its absolute permeability in SI units isunits is
1) 8π x 10‐4 H/m 2) 8π x 10‐3 H/m
3) 4)
• Absolute permeability =
= 4πx10‐7x2000 4πx10 x2000
= 8πx10‐4 H/m
33 Th l ti bilit f33. The relative permeability of a substance is 0.9999. the substance is1) f ti1) ferromagnetic2) paramagnetic3) di ti3) diamagnetic4) none of the above
S l ti• Solution:
Hence the substance is diamagnetic
34 Which of the following is most34. Which of the following is most suitable for making cores of electromagnets?electromagnets?
1) soft iron 2) air) )3) steel 4) copper
Solution: Since soft iron can be magnetized and demagnetized easily, it is most suitable for core of electromagnets
35 A ti i f f 360 A 135. A magnetizing force of 360 Am‐1
produces a magnetic flux density of 0 6T in a ferromagnetic material The0.6T in a ferromagnetic material. The susceptibility of the material is1) 1625 2) 13291) 1625 2) 13293) 2105 4) 1914
Solution:Solution:
36. At a certain place, the horizontal component of earth’s magnetic field is
times the vertical component. The angle of dip at that place isangle of dip at that place is 1) 30O 2) 60o
3) 90o 4) 45o3) 90o 4) 45o
• solution :
37 In hich t pe of materials the37. In which type of materials the magnetic susceptibility does not d d th t t ?depend on the temperature?1. Diamagnetic 2 F ti2. Ferromagnetic 3. Ferrite4 ti4. paramagnetic
• Solution : Diamagnetic material
38 Nickel shows ferromagnetic property at38. Nickel shows ferromagnetic property at room temperature . If the temperature is increased beyond curie temperatureincreased beyond curie temperature then it will show 1 paramagnetism1. paramagnetism 2. anti‐ferromagnetism3 diamagnetism3. diamagnetism 4. no magnetic property
• Solution ; Above curie temperature a FM• Solution ; Above curie temperature a F.M substance becomes a p.m. substance Hence (1)
39. Which of the following is not amagnetic substance?magnetic substance?
1) brass 2) iron3) cobalt 4)nickel3) cobalt 4)nickel
Explanation:
Brass is not a magnetic substance
40 M f h b h h40. Most of the substances show the property of
1) Diamagnetism1) Diamagnetism2) Para magnetism3) ferromagnetism3) ferromagnetism4) non of the above
Solution :
Most of the substances are paramagnetic
41. Magnetic cores should have 1) small permeability) p y2) zero permeability3) large permeability3) large permeability4) none of the above
S l i If h i l h l i biliSolution :If the material has large permiability then it conducts magnetic lines largely. Hence
i h ld h l bilimagnetic core should have large permeability.
42 When all the domains of a42. When all the domains of a ferromagnetic material are in the direction of the applied magneticdirection of the applied magnetic field, the related term is) bili ) l1) permeability 2) reluctance3) retentivity 4) saturation
Solution: When all domains are arranged along the direction of magnetic field , then it is g ,called magnetic saturation.
43. The value of magnetic permeability is g p ymaximum for a1) diamagnetic substance1) diamagnetic substance 2) paramagnetic substance3) ferromagnetic substance3) ferromagnetic substance4) none of the above
Solution: For ferromagnetic materials susceptibility is positive and high(of order 8000). Magnetic permeability is also maximumpermeability is also maximum for F.M. materials
44. For making permanent magnets, we use1) hard steel 2) sift iron1) hard steel 2) sift iron3) copper 4) brass
Solution: The hysteresis loop of hard steel shows th t it h hi h l f t ti it dthat it has high value of retentivity and coercivity. Therefore it is used for making
t tpermanent magnets.
45. The relative permeability of iron is of the order ofis of the order of1) zero 2) 10‐43) 1 4) 104) )
S l i Si I i d d fSolution; Since Iron is good conductor of magnetic flux it will have magnetic
bili hi h f d 104permeability high of order 104
46 The relation between relative46. The relation between relative permeability and magnetic
ibili isusceptibility is1) µr = 1 ‐ 3) µr = 3 (1 + )3) µr = 1+ 4) µr = 1/
Solution :
47 The relation between B (flux47. The relation between B (flux density), H (magnetizing force) and I (i t it f ti ti )I (intensity of magnetization)1) B = µo (H + I) 2) B = µo (H – I)3) B = µ0(2H+I) 4) B = µ0 (H+2I)) µ0 ( )
• Solution:Solution:
48 The properties like retentivity and48. The properties like retentivity and coercivity of a permanent magnet should beshould be 1. high –high 2. low‐low3. low‐high 4. high‐low
• Solution : retentivity should be high and coercivity should be high for a permanentcoercivity should be high for a permanent magnet.