Magnetism and Matter PREPARED BY Sri. Satishkumar. Umadi Lecturer N V Pre university college N V Pre university college Gulbarga MAGNETIC DIPOLE Two unlike magnetic g p poles of equal q p pole strength separated by a small dist is called a magnetic dipole. Ex: A bar magnet, a solenoid carrying current a circular wire carrying current current, etc. Magnetic dipole moment = pole strength x Magnetic dipole moment = pole strength x magnetic length The direction of magnetic moment ( ) is The direction of magnetic moment ( ) is from South to North Pole. SI unit of is Am2 and unit of pole strength is Am and unit of pole strength is Am. MAGNETIC FIELD AT A POINT ON AXIAL LINE: MAGNETIC FIELD AT A POINT ON AXIAL LINE: The magnetic field at a point on the axial The magnetic field at a point on the axial line at a distance ‘r’ from the centre of a magnet of dipole moment is magnet of dipole moment is Here magnetic length 2l is assumed to be H i l h 2l i d b 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 MAGNETIC FIELD AT A POINT ON EQUATORIAL LINE The magnetic field at a point on equatorial Th ti fi ld t i t t i l line at a distance ‘r’ from the centre of a magnet of dipole moment is t f di l ti Here magnetic length 2l is assumed to be Here magnetic length 2l is assumed to be much smaller than distance of point(r) (r 2 l) . l) . >>>2 TORQUE ON A MAGNETIC DIPOLE IN A TORQUE ON A MAGNETIC DIPOLE IN A UNIFORM MAGNETIC FIELD: τ = m. B. sinθ θ In vector form POTENTIAL ENERGY OF A MAGNETIC POTENTIAL ENERGY OF A MAGNETIC DIPOLE IN A MAGNETIC FIELD: The amount of work done in rotating dipole through a small angle ‘dθ’ through a small angle dθ is is dw= τ.dθ The potential energy of dipole is The potential energy of dipole is U = ‐ m B cos θ U= U = 1.. A magnetic needle kept in a non ag et c eed e ept a o uniform magnetic field , it experiences 1 Torque but no force 1. Torque but no force 2. neither force nor torque 3 f 3. a force and a torque d t 4. a force but not torque • Solution : In a non uniform magnetic field it experiences both torque and an effective force. i b h d ff i f 2. The magnetic moment of a diamagnetic atom is g 1. Equal to zero 2 equal to one 2. equal to one 3. between zero and one 4 much greater than one 4. much greater than one • Solution : magnetic moment of a diamagnetic atom is equal to zero. 3 A 3. A charged particle (charge q) is moving charged particle (charge q) is moving in a circle of radius R with uniform speed ‘v’. The associated magnetic d‘ ’ h i d i moment is 1. q v R /2 2. q v R 3. q vR q 2//2 4. q vR q 2 • Solution Solution : The current associated with motion : 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 q R R // 2 π R = q v R/ 2 q / 4. A bar magnet having a magnetic moment is free to rotate in a horizontal plane . A horizontal magnetic field of B= field of B= exists in the space. The exists in the space The work done in taking the magnet slowly f from a direction parallel to the field to a di ti ll l t th fi ld t direction 60 degree from the field is 1. 0.6 J 2. 2 J 3. 6 J 4. 12 J • Solution Solution : Work done in rotating a dipole is : Work done in rotating a dipole is W = 5. The dimensions of magnetic dipole moment will be 1) LA 2) ML‐1 A 3) ML‐1 A‐2 3) ML 4) L2A 4) L • 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 of 6 The work done in turning a magnet of magnetic moment M by an angle of 90o f from the magnetic meridian is n times h i idi i i the corresponding work done in turning it through an angle of 60o. The value of n is 1) 1 2) 2 3) 1/2 3) 1/2 4) 1/4 4) 1/4 W kd Work done is i 7. A bar magnet has a magnetic moment of 2.5 JT‐1 and is placed in a magnetic field 2.5 JT and is placed in a magnetic field of 0.2 T . Work done in turning the magnet from parallel to anti parallel magnet from parallel to anti parallel position relative to the field direction is 1) zero ) 2) 0 2) 0.5J 3) 2.0 J 4) 1 J • Solution: work done = 8. Rate of change of torque with deflection θ 8 R t f h ft ith d fl ti θ is maximum for a magnet suspended freely in a uniform magnetic field of induction B when 1) θ = 45o 2) θ = 60 2) θ = 60o 3) θ = 0o 4) θ =90o 4) θ • Solution : • Rate Rate of change of torque with respect to is 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 30o 2 ampere meter is deflected through 30 from the magnetic meridian. The h i t l horizontal component of earth’s magnetic t f th’ ti field is 0.32 x 10‐4 T. The value of deflecting couple is ‐7 Nm 2) 2.5 x 10 ‐7 Nm 1) 64 x 10 ) ) 3) 32 x 10‐7 Nm 4) 16 x 10‐7 Nm Solution :Magnetic moment =m = q X 2l = 2 X 0.1 2 X 0.1 = 0.2 0.2 Torque on the couple PERIOD OF OSCILLATION OF COMPASS NEEDLE PERIOD OF OSCILLATION OF COMPASS NEEDLE IN A UNIFORM MAGNETIC FIELD The period of oscillation of a magnet in g uniform magnetic field ‘B’ is ‘‘m’ is magnetic moment and ‘I’ is moment ’ i ti t d ‘I’ i t of inertia of magnet 10. Two magnets are held together in a 10 Two magnets are held together in a vibration galvanometer and are allowed to oscillate in the earth’s magnetic field ill i h h’ i fi ld with like poles together and 12 oscillations per minute are made but for unlike poles together, only 4 oscillations p g , y per minute are executed. The ratio of their magnetic moments is their magnetic moments is 1) 5 : 4 2) 3 : 1 3) 3 5 3) 3 : 5 4) 1 3 4) 1 : 3 • Solution: Period of rotation when like poles are together T1= 60/12 = 5 sec are together = T 60/12 5 sec • Period of rotation when unlike poles are together = T h / 2= 60/ 4 = 15 sec 11 For 11. For a vibration magnetometer, the time a vibration magnetometer the time period of the suspended bar magnet can b be reduced by d db 1) moving it towards south pole of earth 2) moving it towards north pole of ) g p earth 3) moving it towards equator 3) moving it towards equator 4) 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 towards time period can be reduced by moving towards equator. 12. A magnet oscillates in earth’s magnetic field with time period T. If the mass is field with time period T. If the mass is quadrupled, then new time will be 1) 2T, motion remaining SHM 2) 2) 4T, motion remaining SHM i i i S 3) T/2, motion remaining SHM 4) Unaffected but motion not SHM S l ti Solution : period of oscillation i d f ill ti Moment of inertia is directly proportional to its mass. mass 13. Two bar magnets of the same mass, same 13 T b t f th length and breadth having magnetic moments M and 2M are joined together p pole for pole and suspended by a string. p p y g The 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 again is reversed and the combination is again made to oscillate in the same field, the time of oscillations is the 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 For polarity reversed m2 = 2m – = 2m – m = m m=m . 14 The 14. The time period of a freely time period of a freely suspended magnet does not depend upon 1) length of the magnet 2) the pole strength of the magnet g 3) the horizontal component of earth’ss magnetic field earth magnetic field 4) the length of suspension • Time Time period depends on Inertia and magnetic period depends on Inertia and magnetic moment (length) of the magnet . There fore time period does not depend on length of the time period does not depend on length of the suspension. 15. The time period of a freely suspended 15 The time period of a freely suspended thin magnet is 4 seconds. If it is broken i l in length in two equal parts and one hi l d part is suspended in the same way, then its time period (in seconds) will be 1) 2 3) 0 5 3) 0.5 2) 4 4) 0 25 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 per minute in H 0 3 x 10‐4 T. By T By what amount the field should be increased so that the number of increased 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 T 3) 0.9 x 10‐4 T 4) 1.2 x 10‐4 T • Solution: S l ti 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 S 3. 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 y oscillations is 1) infinity 1) infinity 2) zero 2) zero 3) 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 and A line passing through North and South pole of earths magnet is called magnetic axis. A line passing through North l h h h and South of earth is called geographic axis. The magnetic axis is inclined at an a ge angle INCLINATION OR DIP AT A PLACE: INCLINATION OR DIP AT A PLACE The dip at a place is the angle made by total magnetic field BE of the earth with the g horizontal in the magnetic meridian HE = B = BE cos I ; Z cos I ; ZE = B = BE sin I sin I HE2 + ZE2 = BE2 ZE → vertical comp of earths magnetic field vertical comp of earths magnetic field HE → Horizontal comp of earths magnetic field. 19 19. The angle of dip is 90 The angle of dip is 90o at 1) earth’s magnetic poles 2) equator ) 3) both (a) and (b) 4) none of these • Solution Solution : At magnetic poles , the earths : 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 north 20 A bar magnet is placed with its north pole pointing earth’s north. The points of zero magnetic field will be in which ti fi ld ill b i hi h direction from the centre of the magnet? 1) north and south 2) east and west 3) north‐east 3) north east and south and south‐ west 4) north‐east and south‐east • E Neutral point BH N(earth) S(earth) ( ) ( ) N S Neutral point W BH 21 Earth’s 21. Earth s magnetic field has a horizontal magnetic field has a horizontal component except at 1) equator ) 2) magnetic poles of earth 3) a latitude of 60o o 4) an inclination of 60 ) Solution: Horizontal component of magnetic field is HE E = BE cos I This is zero if Cos I = 0 i.e . I = 90 (at poles) 22. Vertical component of earth 22 Vertical component of earth’ss magnetic field is zero at a place where angle of dip is where angle of dip is 1) 45 ) o 3) 0o 2) 90 ) o 4) 60o Solution: Vertical component of earth magnetic field is p g ZE = BE sin I . This is zero if Sin I = 0 Hence I Sin I 0 Hence I = 0 0 This occurs at magnetic equator 23. A magnet suspended at 30o with the magnetic meridian makes an angle of 45o with the horizontal. What is the actual ith the hori ontal What is the act al value of dip? 1 1. 2. 2 3. 4. Horizontal component H Horizontal component H H=Hcos30 Hcos30 I = 45 let the dip at place be I let the dip at place be I R V R V Then R cos I =Hcos30 R sin I= V Taking 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 are 1) 30o, 60o 2) 90o, 0o 3) 30o 90o 4) 0o, 0o • Solution Solution : At poles only vertical component of : 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 where 25 The angle of dip at a certain place where the horizontal and vertical components of the earth’s magnetic field are equal is f h h’ i fi ld li 1) 30o 3) 60 ) o 2) 45o 4) 75 ) o H = V Hence R cos I = R sin I tanI = 1 Hence I = 450 MAGNETIC PROPERTIES OF MATERIALS MAGNETIC PROPERTIES OF MATERIALS MAGNETIC PERMEABILITY µ: RELATIVE PERMEABILITY MAGNETIC INDUCTION OR MAGNETIC MAGNETIC 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. M/H. Ratio of magnetization attained (M) to the magnetic intensity (H) of magnetizing field i i i ( ) f i i fi ld the magnetic field in the interior of solenoid is h i fi ld i h i i f l id i B = Bo + Bm B = µoH + µoM = µo [H + M] ∴ µr = 1 + χ Where µr is relative permeability of medium HYSTERESIS 1) The area of hysteresis loop gives the The area of hysteresis loop gives the energy loss per cycle. Hence the materials which have small area of hysteresis loop are which have small area of hysteresis loop are used for making transformer core. 2) The materials which have high retentivity 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 material form graph. More is the coercivity of material, the materials cannot loose magnetism easily. 26 M 26. Most suitable material for t it bl t i lf making transformer core is 1) steel 2) nickel ) pp 4) soft iron ) 3) copper solution: The area of hysteresis loop represents y p p the 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 27. The B – B H curves (i) and (ii) show in fig. H (i) d (ii) h i fi are respectively associated with 1) diamagnetic and paramagnetic ) g g 2) Paramagnetic and diamagnetic 3) sift iron and steel 4) steel and sift iron sift iron • Explanation : 1. Curve (1) has high retentivity and low ( ) g y coercivity. It represents hysteresis for soft iron. 2. Curve (2) has low retentivity and high coercivity It represents hysteresis loop for coercivity . It represents hysteresis loop for steel. 28 Magnetic 28. Magnetic susceptibility for susceptibility for diamagnetic materials is 1) 1) small and negative ll d ti 2) small and positive 3) large and positive 4) 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 29. Susceptibility is positive and large for is positive and large for 1) paramagnetic substance 2) diamagnetic substance ) di i b 3) non‐ magnetic substance 4) ferromagnetic substance • Feromagnetic Feromagnetic substances strongly magnetise substances strongly magnetise in the direction of applied external magnetic field Hence they have high magnetic field. Hence they have high magnetic susceptibility ( soft iron has 8000) 30 To 30. To shield an instrument from shield an instrument from external magnetic field, it is placed in a cabin made of 1) wood 2) iron 3) ebonite 4) diamagnetic substance Solution :Iron has very large magnetic :Iron has very large magnetic permeability . Hence it conducts magnetic flux largely There fore magnetic lines pass through largely. There fore magnetic lines pass through iron cabin. Thus the instrument is shielded from magnetic field from magnetic field. 31 The 31. The relative permeability of a relative permeability of a substance is 1.001. The substance is 1) paramagnetic 2) diamagnetic 3) ferromagnetic 4) none of the above 4) none of the above A paramagnetic material has a small and positive value of magnetic susceptibility. ii l f i ibili Hence the material is a paramagnetic 32 If 32. If the relative permeability iron is the relative permeability iron is 2000, its absolute permeability in SI units is units is 1) 8π x 10‐4 H/m 2) 8π x 10‐3 H/m 3) 4) • Absolute permeability = = 4πx10 4πx10‐7x2000 = 8πx10‐4 H/m 33 Th 33. The relative permeability of a l ti bilit f substance is 0.9999. the substance is 1) f 1) ferromagnetic ti 2) paramagnetic 3) di 3) diamagnetic ti 4) none of the above • Solution: S l ti Hence the substance is diamagnetic 34 Which 34. Which of the following is most of the following is most suitable for making cores of electromagnets? 1) soft iron ) 3) steel 2) air ) 4) copper Solution: Since soft iron can be magnetized and demagnetized easily, it is most suitable for core of electromagnets 35 A 35. A magnetizing force of 360 Am ti i f f 360 A ‐11 produces a magnetic flux density of 0 6T in a ferromagnetic material The 0.6T in a ferromagnetic material. The susceptibility of the material is 1) 1625 1) 1625 2) 1329 2) 1329 3) 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 is angle of dip at that place is 1) 30O 2) 60o 3) 90o 3) 90 4) 45o 4) 45 • solution : 37. In which type of materials the 37 In hich t pe of materials the magnetic susceptibility does not d depend on the temperature? d th t t ? 1. Diamagnetic 2 F 2. Ferromagnetic ti 3. Ferrite 4 4. paramagnetic ti • Solution : Diamagnetic material 38. Nickel shows ferromagnetic property at 38 Nickel shows ferromagnetic property at room temperature . If the temperature is increased beyond curie temperature increased beyond curie temperature then it will show 1 paramagnetism 1. paramagnetism 2. anti‐ferromagnetism 3 diamagnetism 3. diamagnetism 4. no magnetic property • Solution Solution ; Above curie temperature a F.M ; Above curie temperature a F M substance becomes a p.m. substance Hence (1) 39. Which of the following is not a magnetic substance? 1) brass 2) iron 3) cobalt 4)nickel Explanation: Brass is not a magnetic substance 40 M 40. Most of the substances show the f h b h h property of 1) Diamagnetism 1) Diamagnetism 2) Para magnetism 3) ferromagnetism 3) ferromagnetism 4) non of the above Solution : Most of the substances are paramagnetic 41. Magnetic cores should have 1) small permeability ) p y 2) zero permeability 3) large permeability 3) large permeability 4) none of the above SSolution :If the material has large permiability l i If h i lh l i bili then it conducts magnetic lines largely. Hence magnetic core should have large permeability. i h ld h l bili 42. When all the domains of a 42 When all the domains of a ferromagnetic material are in the direction of the applied magnetic direction of the applied magnetic field, the related term is 1) permeability ) bili 2) reluctance ) l 3) retentivity 4) saturation Solution: When all domains are arranged along g , the direction of magnetic field , then it is called magnetic saturation. 43. The value of magnetic permeability is g p y maximum for a 1) diamagnetic substance 1) diamagnetic substance 2) paramagnetic substance 3) ferromagnetic substance 3) ferromagnetic substance 4) none of the above Solution: For ferromagnetic materials susceptibility is positive and high(of order 8000). Magnetic permeability permeability is also maximum is also maximum for F.M. materials 44. For making permanent magnets, we use 1) hard steel 1) hard steel 2) sift iron 2) sift iron 3) copper 4) brass Solution: The hysteresis loop of hard steel shows th t it h hi h l that it has high value of retentivity and f t ti it d coercivity. Therefore it is used for making permanent magnets. t t 45. The relative permeability of iron is of the order of is of the order of 1) zero 2) 10‐4 3) 1 ) 4) 10 ) 4 SSolution; Since Iron is good conductor of l i Si I i d d f magnetic flux it will have magnetic permeability high of order 10 bili hi h f d 104 46 The 46. The relation between relative relation between relative permeability and magnetic susceptibility is ibili i 1) µr = 1 ‐ 3) µr = 3 (1 + ) 3) µr = 1+ 4) µr = 1/ Solution : 47 The 47. The relation between B (flux relation between B (flux density), H (magnetizing force) and I (i t it f I (intensity of magnetization) ti ti ) 1) B = µo (H + I) 2) B = µo (H – I) 3) B = µ0(2H+I) 4) B = µ ) µ0 ((H+2I)) • Solution: 48. The 48 The properties like properties like retentivity and retentivity and coercivity of a permanent magnet should be should be 1. high –high 2. low‐low 3. low‐high 4. high‐low • Solution : retentivity should be high and coercivity should be high for a permanent coercivity should be high for a permanent magnet.

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