{"id":35859,"date":"2022-11-25T22:21:37","date_gmt":"2022-11-25T16:51:37","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=35859"},"modified":"2022-11-26T10:13:41","modified_gmt":"2022-11-26T04:43:41","slug":"ts-inter-2nd-year-physics-study-material-chapter-8","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-8\/","title":{"rendered":"TS Inter 2nd Year Physics Study Material Chapter 8 Magnetism and Matter"},"content":{"rendered":"
Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material<\/a> 8th Lesson Magnetism and Matter Textbook Questions and Answers.<\/p>\n Very Short Answer Type Questions<\/span><\/p>\n Question 1. This will happen only when magnetic dipole is in non-uniform magnetic field.<\/p>\n Question 2. Similarly when it is taken to south pole then north pole of compass is attracted by south pole and it will align itself along magnetic meridian line.<\/p>\n Question 4. I = (\\(\\frac{M}{V}\\)) where M = the magnetic moments and V = volume of the given material.<\/p>\n Magnetic intensity is a vector, dimensions L-1<\/sup> A. Question 5. <\/p>\n Question 6. Question 7. Hence magnetic field lines are always closed loops.<\/p>\n Question 8. The angle between true geographic north to the north shown by magnetic compass is called “magnetic declination or simply declinations (D).” Question 9. Question 10. Question 11. Question 12. <\/p>\n Question 13. Short Answer Questions<\/span><\/p>\n Question 1. Effect of temperature on Ferromagnetic substances : Question 2. Let a solenoid of radius ‘a’ and length 2l contains n turns and a current ‘I’ is passed through it.<\/p>\n Magnetic moment of solenoid (M) = nlA.<\/p>\n Consider a circular element of thickness dx of solenoid at a distance x from its centre. Choose any point ‘P’ on the axis of solenoid at a distance ‘r’ from centre of the axis. Magnetic field at point P Question 3. Question 4. <\/p>\n Question 5. Magnetic declination (D) : Angle of dip or inclination (I): At a given place horizontal component of earth’s magnetic field HE<\/sub> = BE<\/sub> cos I. Vertical component of earth’s magnetic field ZE<\/sub> = BE<\/sub> sin I.<\/p>\n Tangent of dip tan I = \\(\\frac{Z_E}{H_E}\\)<\/p>\n Question 6. i) When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point a’. It indicates that all atomic magnets of the sample are parallel to applied field.<\/p>\n ii) When applied magnetic field is gradually decreased to zero still then some magnetic intensity will remain in the material.<\/p>\n Retentivity or Remanence : iii) When applied magnetic field (B) is reversed then magnetic intensity of the sample gradually decreases and finally it is magnetised in opposite direction upto saturation say point’d’.<\/p>\n Coercivity : In hysteresis diagram the-value of B on -ve X-axis gives coercivity.<\/p>\n iv) When direction of magnetic field is reversed and gradually increased again we can reach the point of saturation a’.<\/p>\n Area of hysteresis loop is large for ferromagnetic substances with high permeability value.<\/p>\n Question 7. For 1st<\/sup> Case: For 2nd<\/sup> Case : Question 8. Magnetic field along the axis of a solenoid B = \u00b50<\/sub>nI.<\/p>\n Where n is number of turns per unit length. When number of turns of a solenoid is doubled then magnetic field at the given point on the axis of solenoid will also double.<\/p>\n Long Answer Questions<\/span><\/p>\n Question 1. The magnetic field at ‘P’ makes some angle ‘\u03b8’ with X – axis. So resolve \\(\\mathrm{d} \\bar{B}\\) into components \\(\\mathrm{d} \\bar{B}_x\\) and \\(\\mathrm{d} \\bar{B}\\)\u22a5. Sum of \\(\\mathrm{d} \\bar{B}\\)\u22a5 is zero. Because \\(\\mathrm{d} \\bar{B}\\)\u22a5 component by an element d\/ is cancelled by another diametrically opposite component. From fig \\(\\mathrm{d} \\bar{B}_x\\) = dB. cos \u03b8. Where Question 2. When a bar magnet is cut into two parts it will behave like two weak bar magnets. Similarly when a solenoid is cut into two parts and current is circulated through them they will also act as two solenoids of weak magnetic properties. Analogy between solenoid and bar magnet.<\/p>\n Let a solenoid of radius ‘a’ and length 2l contains n turns and a current ‘I’ is passed through it.<\/p>\n Magnetic moment of solenoid M = nlA. \u2234 Total magnetic field B is obtained by integrating dB with in the limits -1 to 1. This value is similar to magnetic field at any point on the axial line of magnetic dipole. Thus, a bar magnet and a solenoid produce ! similar magnetic fields.<\/p>\n <\/p>\n Question 3. From principles of angular. \u2234 Time period of oscillation of a magnetic needle placed in a magnetic field T = 2\u03c0\\(\\sqrt{\\frac{I}{MB}}\\) Question 4. Where HE<\/sub> is horizontal component of earth’s magnetic field.<\/p>\n The relation between resultant magnetic field BE<\/sub>, horizontal magnetic field HE<\/sub> and angle of dip or inclination ‘I’ of earth’s magnetic field is HE<\/sub> = BE<\/sub> cos I. \u2014 (2) Question 5. For ferromagnetic materials susceptibility is +ve and \u03c7 >> 1 and \u03c7 is nearly in the order of 1000. For diamagnetic substances \u03c7 is small and negative, magnetisation M and magnetic intensity H are in opposite direction. Question 6. The net magnetic flux is zero for both surfaces. This law is true for any surface of any shape. Consider an irregular Gaussian surface as shown in figure. Consider a smdl vector area element \u2206S of closed surface ‘S’ placed in a \u03a6 magnetic field B. Flux through \u2206s is say \u2206\u03a6.<\/p>\n When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point ‘a’. It indicates that all atomic magnets of the sample are parallel to applied field.<\/p>\n Now flux (i.e., Number of magnetic field lines through unit area) through the element \u2206S is given by \u2206\u03a6B<\/sub> = B . \u2206S.<\/p>\n Let us divided the total Gaussian surface ‘S’ into number of small areas \u2206S1<\/sub>, \u2206S2<\/sub>, \u2206S3<\/sub>, ………… \u2206Sn<\/sub>.<\/p>\n The total flux through the Gaussian surface Magnetic monopole is not existing so there is no source or sinks of B’ in Gaussian surface. The simplest possible source is magnetic dipole i.e., bar magnet. Magnetic field lines of a bar magnet are closed curves of loops. So all the lines entering the Gaussian surface must leave from it.<\/p>\n Hence net magnetic flux through a closed surface is zero.<\/p>\n Question 7. ii) When applied magnetic field is gradually decreased to zero still then some magnetic intensity will remain in the material.<\/p>\n Retentivity or Remanence : iii) When applied magnetic field (B) is reversed then magnetic intensity of the sample gradually decreases and finally it is magnetised in opposite direction upto saturation say point d\u2019.<\/p>\n Coercivity : In hysteresis diagram the value of B on -ve X-axis gives coercivity.<\/p>\n iv) When direction of magnetic field is reversed and gradually increased again we can reach the point of saturation a’. Area of hysteresis loop is large for ferromagnetic substances with high permeability value.<\/p>\n Application of hysteresis in electromagnets. Hysteresis curve allows us to select suitable materials for permanent magnets and for electromagnets.<\/p>\n For materials to use as permanent magnets they must have high retentivity and high coercivity.<\/p>\n For electromagnets ferromagnetic substances of high permeability and low retentivity are used because when current is switched off it must loose magnetic properties quickly.<\/p>\n In case of transformers and telephone diaphragms they are subjected to prolonged AC cycles. For these applications the hysteresis curve of ferromagnetic materials must be narrow.<\/p>\n In this way hysteresis loop helps us to select magnetic materials for various applications.<\/p>\n Problems<\/span><\/p>\n Question 1. Question 2. Question 3. <\/p>\n Question 4. Question 5. Question 6. Question 7. Question 8. Intext Questions and Answer<\/span><\/p>\n Question 1. Question 2. b) The bar magnet is oriented 180\u00b0 to the magnetic field. Hence, it is in unstable equilibrium. Question 3. Question 4. Question 5. i) Initial angle between the axis and the magnetic field, \u03b81<\/sub> = 0\u00b0. ii) Initial angle between the axis and the magnetic field, \u03b81<\/sub> = 0\u00b0. b) For case CD : \u03b8 = \u03b82<\/sub> = 90\u00b0 then For case (ii): \u03b8 = \u03b82<\/sub> = 180\u00b0 then <\/p>\n Question 6. a) The magnetic moment along the axis of the solenoid is calculated as : b) Magnetic field, B = 7.5 \u00d7 10-2<\/sup> T Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 \u00d7 10-2<\/sup> Nm.<\/p>\n Question 7. Question 8. Question 9. Question 10. b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is <\/p>\n Question 11. b) The magnetic field at a distance R’ from the centre of the magnet on its axis is given as: Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter Textbook Questions and Answers. TS Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter Very Short Answer Type Questions Question 1. A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter 2nd Year Physics Study Material 8th Lesson Magnetism and Matter<\/h2>\n
\nA magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field?
\nAnswer:
\nFor a magnetic dipole placed in a magnetic field, some net force is experienced It implies that force on the two poles of dipole is not equal.<\/p>\n
\nThere is no question in text book. [TS Mar. 19, 17, May 14]
\nQuestion 3.
\nWhat happens to compass needles at the Earth’s poles? [TS Mar. 19, 17, May 14]
\nAnswer:
\nWhen a compass is taken to earth poles say north pole then south pole of compass will adhere to north pole. It will align it self along magnetic meridian line.<\/p>\n
\nWhat do you understand by the ‘magnetisation’ of a sample?
\nAnswer:
\nMagnetisation (T) :
\nIt is the ratio of magnetic moment per unit volume.<\/p>\n
\nUnit : Ampere\/metre : Am-1<\/sup><\/p>\n
\nWhat is the magnetic moment associated with a solenoid?
\nAnswer:
\nMagnetic moment associated with a solenoid (M) = nIA. Where
\nn = Number of turns in solenoid;
\nI = Current through it;
\nA = Area vector<\/p>\n
\nWhat are the units of magnetic moment, magnetic induction and magnetic field? [AP Mar. 16. May 17, 16; TS Mar. 16]
\nAnswer:
\n1. Magnetic moment m is a vector. Unit A-m\u00b2, dimensions L-2<\/sup> A.
\n2. Magnetic induction (B) and magnetic field (B) are used with same meaning. Magnetic induction B is a vector.
\nUnit: Tesla (T), Dimension : MT-2<\/sup>A-1<\/sup>.<\/p>\n
\nMagnetic lines form continuous closed loops. Why? [AP Mar. ’19, ’16, May ’18; TS May ’18, Mar. ’17]
\nAnswer:
\nIn magnetism magnetic monopole (single pole) is not existing. The simple possible way is to take a magnetic dipole. So the path a free magnetic needle or compass starts from north pole and terminates at south pole forms a loop.<\/p>\n
\nDefine magnetic declination. [TS Mar. 18, May 18, 17, 16; AP Mar. 18, 14, May 17, 16]
\nAnswer:
\nMagnetic declination (D) :
\nThe magnetic meridian at a place makes some angle (D) with true geographic north and south direction.<\/p>\n
\n<\/p>\n
\nDefine magnetic inclination or angle of dip. [AP Mar. ’17, ’15; TS Mar. ’15]
\nAnswer:
\nMagnetic inclination or angle of dip (I) :
\nIt is the angle of total magnetic field BE at a given place with the surface of earth.
\n(OR)
\nThe angle between horizontal to earth’s surface and net magnetic field of earth BE at that point.<\/p>\n
\nClassify the following materials with regard to magnetism: Manganese, Cobalt, Nickel, Bismuth, Oxygen, Copper. [AP Mar. 19. 18. 17, 16, 15; TS Mar. 16. 15]
\nAnswer:
\nManganese : Paramagnetic substance
\nCobalt : Ferromagnetic substance
\nNickel : Ferromagnetic substance
\nBismuth : Diamagnetic substance
\nOxygen : Paramagnetic substance
\nCopper : Diamagnetic substance<\/p>\n
\nIn the magnetic meridian of a certain place, the horizontal component of the earth\u2019s magnetic field is 0.26 G and the dip angle is 60\u00b0. What Is the magnetic field of the earth at this location?
\nAnswer:
\nGiven HE = 0.26 G; Dip angle = 60
\nBut Dip angle = \\(\\frac{H_E}{B_E}\\) = cos \u03b8 \u21d2 BE<\/sub> = HE<\/sub> cos \u03b8
\n\u2234 Magnetic field of earth = 0.26 \u00d7 cos 60\u00b0 = 2 \u00d7 0.26 = 0.52 G<\/p>\n
\nDefine Magnetisation of a sample. What is its SI unit?
\nAnswer:
\nMagnetisation (I) : It is the ratio of net magnetic moment per unit volume.
\nI = \\(\\frac{m_{net}}{V}\\) where mnet<\/sub> = the vectorial sum of magnetic moments of atoms in bulk material and V is volume of the given material.
\nMagnetic intensity is a vector, dimensions L-1<\/sup> A.
\nUnit: Ampere\/metre : A m-1<\/sup>.<\/p>\n
\nDefine Magnetic susceptibility. Mention its unit. [AP Mar. ’15]
\nAnswer:
\nMagnetic susceptibility (\u03c7) :
\nIt is a measure for the response of magnetic materials to an external field.
\n
\nIt is a dimensionless quantity.<\/p>\n
\nWhat are Ferromagnetic materials? Give examples. What happens to a ferromagnetic material at Curie temperature?
\nAnswer:
\nFerromagnetism:<\/p>\n\n
\nEx: Manganese, Iron, Cobalt, Nickel etc.<\/li>\n<\/ol>\n
\nWhen ferromagnetic substances are heated upto Curie temperature, they will be converted into paramagnetic substances.<\/p>\n
\nDerive an expression for the axial field of a solenoid of radius “r”, containing “n” turns per unit length and carrying current “I”.
\nAnswer:
\nThe behaviour of a magnetic dipole and a current carrying solenoid are similar.<\/p>\n
\n<\/p>\n
\n
\nThis is similar to magnetic field at any point on the axial line of magnetic dipole.<\/p>\n
\nThe force between two magnet poles separated by a distance ‘d’ in air is ‘F. At what distance between them does the force become doubled?
\nAnswer:
\nForce between two magnetic poles F = \\(\\frac{\\mu_0}{4 \\pi} \\frac{\\mathrm{m_1m_2}}{\\mathrm{d^2}}\\)
\n
\n
\nWhen separation between the poles is reduced by \u221a2 times their force between them is doubled.<\/p>\n
\nCompare the properties of para, dia and ferromagnetic substances. [TS & AP June ’15]
\nAnswer:<\/p>\n\n\n
\n Paramagnetic substances<\/td>\n Diamagnetic substances<\/td>\n Ferromagnetic substances<\/td>\n<\/tr>\n \n 1. Feebly attracted by magnets.<\/td>\n 1. Repelled by magnets.<\/td>\n 1. Strongly attract by magnets.<\/td>\n<\/tr>\n \n 2. Susceptibility is +ve and nearly equals to one.
\n\u03c7 = i<\/td>\n2. Susceptibility is -ve and less than one.
\n\u03c7 < 1<\/td>\n2. Susceptibility is + ve and large.
\n\u03c7 > > 1<\/td>\n<\/tr>\n\n 3. In a magnetic filed they move from weak field to strong field.<\/td>\n 3. They move from strong field to weak field.<\/td>\n 3. They move from weak field to strong field.<\/td>\n<\/tr>\n \n 4. They have individual atomic magnetic moments but total magnetic moment is zero.
\nEx: Aluminium, sodium etc.<\/td>\n4. Individual atomic magnetic moment is zero.
\nEx: Bismuth, copper, lead.<\/td>\n4. They have individual atomic magnetic moments. These atoms will form domains. Magnetic moment of all atoms in adomain is in same direction.
\nEx: Iron, cobalt, nickel.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\nExplain the elements of the Earth’s magnetic field and draw a sketch showing the relationship between the vertical component, horizontal component and angle of dip.
\nAnswer:
\nEarth’s magnetism :
\nThe magnetic field of earth is believed to arise due to electrical currents produced by convective motion of metallic fluids in outer core of earth. This effect is also known as the “dynamo effect”.<\/p>\n\n
\nThe angle between true geographic north to the mag-netic north shown by magnetic compass is called “magnetic declination or simply declination (D).”
\n<\/p>\n
\nThe angle of dip is the angle of total magnetic field BE at a given place with the surface of earth.<\/p>\n
\n<\/p>\n
\nDefine retentivity and coercivity. Draw the hysteresis curve for soft iron and steel. What do you infer from these curves?
\nAnswer:
\nHysteresis loop :
\nMagnetic hysteresis loop a graph between magnetic field (B) and magnetic intensity (H) of a ferromagnetic substance. It is as shown in figure.
\n<\/p>\n
\nThe magnetic intensity (H) of a material at applied magnetic field B – 0 is called “retentivity”. In hysteresis loop value of H on +ve Y-axis i.e., at B = 0 gives retentivity.<\/p>\n
\nThe -ve value of magnetic field (B^applied (i.e., in opposite direction of mag netisation) at which the magnetic intensity (H) inside the sample is zero is called “coer-civity”.<\/p>\n
\nIf B is the magnetic field produced at the centre of adrcular coil of one turn of length L carrying current I then what is the manetic field at the centre of the same coil which is made into 10 turns?
\nAnswer:
\nOne turn coil means it is a circular loop.<\/p>\n
\nMagnetic field at the centre of a loop B1<\/sub> = \\(\\frac{\\mu_0}{2} \\frac{\\mathrm{I}}{\\mathrm{r_1}}\\)
\nGiven that length of the wire = L.
\n<\/p>\n
\nGiven wire is made into a coil of 10 turns \u21d2 n = 10
\n\u2234 2\u03c0r2<\/sub>10 = L \u21d2 r2<\/sub> = \\(\\frac{L}{2 \\pi}.\\frac{1}{10}\\)
\nFor a coil of n turns magnetic field at its centre B2<\/sub> = \\(\\frac{\\mu_0}{2} \\frac{\\mathrm{nI}}{\\mathrm{r_2}}\\)
\n
\n
\nWhen same length of wire is made into a coil of n turns then B2<\/sub> at centre = n\u00b2 times previous value B1<\/sub>.
\n\u21d2 B2<\/sub> = n\u00b2 B1<\/sub> for given current T.<\/p>\n
\nIf the number of turns of a solenoid is doubled, keeping the other factors constant, how does the magnetic held at the axis of the solenoid change?
\nAnswer:
\nA solenoid will produce almost uniform magnetic field (B) along its axis.<\/p>\n
\nIn our case number of turns of a solenoid is doubled keeping others as constant i.e., length of solenoid L is not changed and permeability p0, and current T not changed. So new number of turns n2<\/sub> = 2n1<\/sub>.
\n\u2234 New magnetic field at the same given point B2<\/sub> = \u00b50<\/sub>n2<\/sub>I.
\nBut n2<\/sub> = 2n1<\/sub>; \u2234 B2<\/sub> = \u00b50<\/sub>2n1<\/sub>I = 2B1<\/sub><\/p>\n
\nDerive an expression for the magnetic field at a point on the axis of a current carrying circular loop.
\nAnswer:
\nConsider a circular loop of radius R’ carrying a steady current i. Consider any point P’ on the axis of the coil (say X – axis).
\n
\nFrom Biot – Savart’s law magnetic field at
\n<\/p>\n
\n
\nTotal magnetic field due to all elements on the circular loop
\n<\/p>\n
\nProve that a bar magnet and a solenoid produce similar fields. (IMP)
\nAnswer:
\nMagnetic field lines suggest that the behaviour of a current-carrying solenoid and a bar magnet are similar.<\/p>\n
\nConsider a circular element of thickness dx of solenoid at a distance x from its centre. Choose any point ‘P’ on the axis of solenoid at a distance r from centre of the axis.
\n<\/p>\n
\n<\/p>\n
\nA small magnetic needle is set into oscillations in a magnetic field B. Obtain an expression for the time period of oscillation.
\nAnswer:
\nLet a small compass of magnetic moment m and moment of inertia ‘I’ is placed in a uniform magnetic field \u2019B\u2019.
\nLet the compass is set into oscillation in a horizontal plane.
\nTorque on the needle is \u03c4 = MB sin \u03b8.
\nWhere 0 angle between M and B.
\nAt equilibrium the magnitude of deflecting torque and restoring torque are equal.
\n\u2234 Restoring torque \u03c4 = I\\(\\frac{\\mathrm{d}^2 \\theta}{\\mathrm{dt}^2}\\) = – MB sin \u03b8.
\n– ve sign indicates that restoring torque is in opposite direction of deflecting torque.
\n\u03b8 is small; sin \u03b8 = \u03b8.
\n
\nMagnetic needle in a uniform field<\/p>\n
\nSimple harmonic motion \\(\\frac{MB}{I}\\) = \u03c9\u00b2
\n\u21d2 \u03c9 = \\(\\sqrt{\\frac{MB}{I}}\\)
\n<\/p>\n
\nand magnetic field at that point B = 4\u03c0\u00b2\\(\\frac{I}{MT^2}\\).<\/p>\n
\nA bar magnet, held horizontally, is set into angular oscillations in the Earth’s magnetic field. It has time periods T1<\/sub> and T2<\/sub> at two places, where the angles of dip are \u03b81<\/sub> and \u03b82<\/sub> respectively. Deduce an expression for the ratio of the resultant magnetic fields at the two places.
\nAnswer:
\nLet a bar magnet is held horizontally at a given place where earth’s magnetic field is BB<\/sub>. When it is set into vibration in a horizontal plane it will oscillate with a time period T = 2\u03c0\\(\\sqrt{\\frac{I}{MH_E}}\\) ……….. (1)<\/p>\n
\nGiven at place ‘1’ angle of dip = \u03b81<\/sub>
\nGiven at place ‘2’ angle of dip = \u03b82<\/sub>.
\n<\/p>\n
\nDefine magnetic susceptibility of a material. Name two elements one having positive susceptibility and other having negative susceptibility. [AP Mar. ’15]
\nAnswer:
\nSusceptibility \u03c7 :
\nThe ratio of magnetisation of a sample (I) to the magnetic intensity (H) is called “susceptibility”.
\nSusceptibility \u03c7 = \\(\\frac{I}{H}\\)
\nIt is a dimensionless quantity.
\nIt is a measure of how a magnetic material responds to external magnetic field.<\/p>\n
\nEx: Iron, Cobalt, Nickel.
\nFor paramagnetic substances \u03c7 is positive and nearly equals to one (\u03c7 \u2245 1).
\nEx : Calcium, Aluminium, Platinum.<\/p>\n
\nEx : Bismuth, Copper, Mercury, Gold.<\/p>\n
\nObtain Gauss’ Law for magnetism and explain it.
\nAnswer:
\nGauss Law in magnetism: Gauss law states that the net magnetic flux through any closed loop is zero.
\n\\(\\oint \\overrightarrow{\\mathrm{B}} \\cdot \\overrightarrow{\\mathrm{ds}}=0\\)
\nExplanation :
\nConsider Gaussian surface I & II as shown in figure. In both cases the number of magnetic field lines entering the surface is equal to number of magnetic field lines leaving the surface.
\n<\/p>\n
\n
\nWhere all’ term includes surface area of all surface elements \u2206S1<\/sub>, \u2206S2<\/sub>, ………… \u2206Sn<\/sub>.<\/p>\n
\nWhat do you understand by “hysteresis”? How does this property influence the choice of materials used in different appliances where electromagnets are used?
\nAnswer:
\nHysteresis loop :
\nMagnetic hysteresis loop is a graph between magnetic field (B) and magnetic intensity (H) of a ferromagnetic substance. It is as shown in figure.
\ni) When applied magnetic field B is gradually increased then magnetic intensity in the material will also gradually increases and reaches a saturation point ‘a’. It indicates that all atomic magnets of the sample are parallel to applied field.
\n<\/p>\n
\nThe magnetic intensity (H) of a material at applied magnetic field B = 0 is called “retentivity”. In hysteresis loop value of H on +ve Y-axis i.e., at B = 0 gives retentivity.<\/p>\n
\nThe -ve value of magnetic field (B) applied (i.e., in opposite direction of magnetisation) at which the magnetic inten-sity (H) inside the sample is zero is called “coercivity”.<\/p>\n
\nWhat is the torque acting on a plane coil of “n” turns carrying a current “i” and having an area A, when placed in a constant magnetic field B?
\nSolution:
\nNumber of turns = n ; Current = i;
\nArea of coil = A; Magnetic field = B.
\nTorque on a current carrying coil in a magnetic field \u03c4 = ni (\\(\\overline{\\mathrm{A}}\\times\\overline{\\mathrm{B}}\\)) = niAB sin \u03b8.<\/p>\n
\nAcoilof 20 turns has an area of 800 mm\u00b2 and carries a current of 0.5 A. If it is placed in a magnetic field of intensity 0.3 T with its plane parallel to the field, what is the torque that it experiences?
\nSolution:
\nNumber of turns n = 200; Current i = 0.5 A;
\nArea A = 800 mm\u00b2 = 800 \u00d7 10-6<\/sup> m\u00b2 (\u2235 1 mm\u00b2 = 10-6<\/sup> m\u00b2) ;
\nMagnetic field B = 0.3 T.
\nArea of coil parallel to the field \u21d2 Angle between area vector \\(\\overline{\\mathrm{A}}\\) and magnetic filed \\(\\overline{\\mathrm{B}}\\) =90\u00b0. Since area vector \\(\\overline{\\mathrm{A}}\\) is perpendicular to area of the coil.
\n\u2234 \u03c4 = niBA = 200 \u00d7 0.5 \u00d7 0.3 \u00d7 800 \u00d7 10-6<\/sup>
\n= 3 \u00d7 800 \u00d7 10-6<\/sup> = 2.4 \u00d7 10-3<\/sup> N-m.<\/p>\n
\nIn the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression for the magnetic moment (\u00b5) of the electron in a Hydrogen atom in terms of its angular momentum L.
\nSolution:
\nCharge of electron = e ;
\nAngular momentum = L;
\nMass of electron = me<\/sub>;
\nMagnetic moment of electron = \u00b5 = ?
\nWhen electron revolves in orbit current i = \\(\\frac{e}{T}\\) ; Where time period T = 2\u03c0 \\(\\frac{r}{\u03c5}\\)
\n\u2234 Current i = \\(\\frac{\\mathrm{ev}}{2 \\pi \\mathrm{r}}\\)
\nMagnetic moment of electron in orbit \u00b5 = iA
\n
\nBut me<\/sub> \u03c5r = L
\n\u2234 Magnetic moment of electron in orbit
\n\u00b5 = \\(\\frac{e}{2m_e}\\)L<\/p>\n
\nA solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid?
\nSolution:
\nLength of solenoid l = 22.5 cm = 22.5 \u00d7 10-2<\/sup>m
\nNumber of turns N = 900; Current I = 0.8 A.
\na) Magnetising field near the centre H = ?
\nThe behaviour of a solenoid is equal to that of a bar magnet.
\nMagnetic field due to a solenoid B = \u00b50<\/sub>nI.
\nMagnetising field H = \\(\\frac{B}{\\mu_0}\\) = nI.
\nWhere n = Nil.
\n\u2234 H = \\(\\frac{900}{22.5 \\times 10^-2}\\) \u00d7 0.8 = 3200 Am-1<\/sup>.
\nA solenoid will give a uniform magnetic field along its axis.
\nMagnetising field far away from ends = 3200 Am-1<\/sup>.<\/p>\n
\nA bar magnet of length 0.1 m and with a magnetic moment of 5 Am\u00b2 is placed in a uniform magnetic field of intensity 0.4 T, with its axis making an angle of 60\u00b0 with the field. What is the torque on the magnet? [Mar. ’14]
\nSolution:
\nLength of bar magnet l = 0.1 m.;
\nMagnetic moment m = 5 Am\u00b2.
\nMagnetic field B = 0.4 T ;
\nAngle with field \u03b8 = 60\u00b0.
\nTorque on the magnet \u03c4 = mB sin \u03b8.
\n= 5 \u00d7 0.4 \u00d7 sin 60\u00b0
\n= 5 \u00d7 0.4 \u00d7 0.8660 = 1.732 N-m.<\/p>\n
\nIf the Earth’s magnetic field at the equator is about 4 \u00d7 10-5<\/sup> T, what is its approximate magnetic dipole moment? (radius ofi Earth = 6.4 \u00d7 106<\/sup> m)
\nSolution:
\nRadius of earth r = 6.4 \u00d7 106<\/sup> m ; Magnetic field near equator B = 4 \u00d7 10-5<\/sup> T
\nMagnetic dipolemoment of earth = M.
\n<\/p>\n
\nThe horizontal component of the earth’s magnetic field at a certain place is 2.6 \u00d7 10-5<\/sup> T and the angle of dip is 60\u00b0. What is the magnetic field of the earth at this location?
\nSolution:
\nHorizontal component of earth’s magnetic field HE = 2.6 \u00d7 10-5<\/sup> T
\n
\nAngle of dip ‘I’ = 60\u00b0.
\nEarth’s magnetic field BE<\/sub> = ?
\nAngle between BE<\/sub> and HE<\/sub> is called dip angle ‘I’.
\nHE<\/sub> = BE<\/sub> = cos \u03b8 \u21d2 BE<\/sub> = HE<\/sub>\/cos \u03b8 = 2.6 \u00d7 10-5<\/sup>\/ cos 60\u00b0
\n\u2234 BE<\/sub> = 2.6 \u00d7 10-5<\/sup>\/ (0.5) = 5.2 \u00d7 10-5<\/sup>T.<\/p>\n
\nA solenoid, of insulated wire, is wound on a core with relative permeability 400. If the number of turns per metre is 1000 and the solenoid carries a current of 2A, calculate H, B and the magnetisation M.
\nSolution:
\nRelative permeability \u00b5r<\/sub> = 400 ;
\nCurrent I = 2A;
\nNumber of turns \/ metre = n = 1000.
\nMagnetising force H = nl = 1000 \u00d7 2 = 2000 Am-1<\/sup>.
\nMagnetic field along axis of solenoid B = ?
\nWhen solenoid is on a magnetic material B
\n= \u00b5r<\/sub> nl = 400 \u00d7 1000 \u00d7 2 = 8 \u00d7 105<\/sup> Am-1<\/sup><\/p>\n
\nA short bar magnet placed with its axis at 30\u00b0 with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 \u00d7 10-2<\/sup> J. What is the magnitude of magnetic moment of the magnet?
\nSolution:
\nMagnetic field strength, B = 0.25 T; Torque on the bar magnet, \u03c4 = 4.5 \u00d7 10-2<\/sup> J
\nAngle between the bar magnet and the external magnetic field, \u03b8 = 30\u00b0
\nTorque is related to magnetic moment (M) as : \u03c4 = MB sin \u03b8
\n
\nHence, the magnetic moment of the magnet is 0.36 J T-1<\/sup>.<\/p>\n
\nA short bar magnet of magnetic moment m = 0.32 J T-1<\/sup> 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?
\nSolution:
\nMoment of the bar magnet, M = 0.32 J T-1 <\/sup>;
\nExternal magnetic field, B = 0.15 T
\na) The bar magnet is aligned along the magnetic field. This system is considered as being instable equilibrium. Hence, the angle 0, between the bar magnet and the magnetic field is 0\u00b0.
\nPotential energy of the system
\n= -MB cos \u03b8
\n= – 0.32 \u00d7 0.15 cos 0\u00b0 = – 4.8 \u00d7 10-2<\/sup> J<\/p>\n
\nPotential energy = – MB cos \u03b8 ;
\nwhere \u03b8 = 180\u00b0
\n= -0.32 \u00d7 0.15 cos 180\u00b0 = 4.8 \u00d7 10-2<\/sup> J.<\/p>\n
\nA closely wound solenoid of800 turns and area of cross-section 2.5 \u00d7 10-4<\/sup> m\u00b2 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?
\nSolution:
\nNumber of turns in the solenoid, n = 800 ;
\nArea of cross-section, A = 2.5 \u00d7 10-4<\/sup> m\u00b2
\nCurrent in the solenoid, I = 3.0 A
\nA current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length. The magnetic moment associated with the given current-carrying solenoid is calculated as :
\nM = nIA = 800 \u00d7 3 \u00d7 2.5 \u00d7 10-4<\/sup> = 0.6 J T-1<\/sup>.<\/p>\n
\nIf the solenoid in exercise 8.5 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\u00b0 with the direction of applied field?
\nSolution:
\nMagnetic field strength, B = 0.25 T ;
\nMagnetic moment, M = 0.6 T-1<\/sup>
\nThe angle \u03b8, between the axis of the solenoid and the direction of the applied field is 30\u00b0.
\nTherefore, the torque acting on the solenoid is given as : \u03c4 = MB sin \u03b8
\n\u03c4 = 0.6 \u00d7 0.25 sin 30\u00b0 = 7.5 \u00d7 10-2<\/sup> J.<\/p>\n
\nA bar magnet of magnetic moment 1.5 JT-1<\/sup> lies aligned with the direction of a uniform magnetic field of 0.22 T.
\na) 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 die field direction, (ii) opposite to the field direction?
\nb) What is the torque on the magnet in cases (i) and (ii)?
\nSolution:
\na) Magnetic moment, M = 1.5 J T-1<\/sup> ;
\nMagnetic field strength, B = 0.22 T<\/p>\n
\nFinal angle between the axis and the magnetic field, \u03b82<\/sub> = 90\u00b0.
\nThe work required to make the magnetic moment normal to the direction of magnetic field is given as :
\nW = -MB (cos \u03b82<\/sub> – cos \u03b81<\/sub>)
\n\u2234 B = -1.5 \u00d7 0.22 (cos 90\u00b0 – cos 0\u00b0)
\n= -0.33 (0 – 1) = 0.33 J.<\/p>\n
\nFinal angle between the axis and the magnetic field, \u03b82<\/sub> = 180\u00b0.
\nThe work required to make the magnetic moment opposite to the direction of magnetic field is given as :
\nW = – MB (cos \u03b82<\/sub> – cos \u03b81<\/sub>)
\n\u2234 W = – 1.5 \u00d7 0.22 (cos 180\u00b0 – cos 0\u00b0)
\n= -0.33 (-1 – 1) = 0.66 J.<\/p>\n
\nTorque, \u03c4 = MB sin \u03b8 \u03c4 = 1.5 \u00d7 0.22 sin 90\u00b0
\n= 0.33 J<\/p>\n
\nTorque, \u03c4 = MB sin \u03b8 \u03c4 = MB sin 180\u00b0 = 0 J<\/p>\n
\nA closely wound solenoid of 2000 turns and area of cross-section 1.6 \u00d7 10-4<\/sup> m\u00b2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
\na) What is the magnetic moment associated with the solenoid?
\nb) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 \u00d7 10-2<\/sup> T is set up at an angle of 30\u00b0 with the axis of the solenoid?
\nSolution:
\nNumber of turns on the solenoid, n = 2000,
\nArea of cross-section of the solenoid, A = 1.6 \u00d7 10-4<\/sup> m\u00b2;
\nCurrent in the solenoid, I = 4 A<\/p>\n
\nM = nAI = 2000 \u00d7 1.6 \u00d7 10-4<\/sup> \u00d7 4 = 1.28 Am\u00b2<\/p>\n
\nAngle between the magnetic field and the axis of the solenoid, 0 = 30\u00b0
\nTorque, \u03c4 = MB sin \u03b8
\n\u2234 \u03c4 = 1.28 \u00d7 7.5 \u00d7 10-2<\/sup> sin 30\u00b0
\n= 4.8 \u00d7 10-2<\/sup> Nm.<\/p>\n
\nA 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 8.0 \u00d7 10-2<\/sup>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<\/sup>. What is the moment of inertia of the coil about its axis of rotation?
\nSolution:
\nNumber of turns in the circular coil, N = 16;
\nRadius of the coil, r = 10 cm = 0.1 m
\nCross-section of the coil, A = \u03c0r\u00b2 = \u03c0 \u00d7 (0.1)\u00b2 m\u00b2 ;
\nCurrent in the coil, I = 0.75 A
\nMagnetic field strength, B = 5.0 \u00d7 10-2<\/sup> T ;
\nFrequency of oscillations of the coil, v = 2.0 s-1<\/sup>
\n\u2234 Magnetic moment, M = NIA = Nl\u03c0r\u00b2
\n= 16 \u00d7 0.75 \u00d7 \u03c0 \u00d7 (0.1)\u00b2
\n\u2234 M = 0.377 J T-1<\/sup>
\n
\nHence, the moment of inertia of the coil about its axis of rotation is 1.19 \u00d7 10-4<\/sup> kg m\u00b2.<\/p>\n
\nA magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22\u00b0 with the horizontal. The horizontal component of the earth\u2019s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth\u2019s magnetic field at the place.
\nSolution:
\nHorizontal component of earth\u2019s magnetic field, BH<\/sub> = 0.35 G
\nAngle made by the needle with the horizontal plane = Angle of dip = \u03b4 = 22\u00b0
\nRelation between B and BH<\/sub> is BH<\/sub> = B cos \u03b4
\n
\nHence, the strength of earth\u2019s magnetic field at the given location is 0.377 G.<\/p>\n
\nAt a certain location in Africa, a compass points 12\u00b0 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\u00b0 above the horizontal. The horizontal component of the earth\u2019s field is measured to be 0.16 G. Specify the direction and magnitude of the earth\u2019s field at the location.
\nSolution:
\nAngle of declination, \u03b8 = 12\u00b0;
\nAngle of dip, \u03b4 = 60\u00b0
\nHorizontal component of earth\u2019s magnetic field, BH<\/sub> = 0.16 G
\nEarth\u2019s magnetic field at the given location = B
\nRelation between B and BH is BH<\/sub> = B cos \u03b8
\n
\nEarth\u2019s magnetic field lies in the vertical plane, 12\u00b0 West of the geographic meridian, making an angle of 60\u00b0 (upward) with the horizontal direction. Its magnitude is 0.32 G.<\/p>\n
\nA short bar magnet has a magnetic moment of 0.48 J T-1<\/sup>. Give the direction and mag-nitude 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.
\nSolution:
\nMagnetic moment of the bar magnet,
\nM = 0.48 J T-1<\/sup>.
\na) Distance, d = 10 cm = 0.1 m
\nThe magnetic field at distance d, from the centre of the magnet on the axis is
\n
\nThe magnetic field is along the S – N direction.<\/p>\n
\n
\nThe magnetic field is along the N – S direction.<\/p>\n
\nA short bar magnet of magnetic moment 5.25 \u00d7 10-2<\/sup> J T-1<\/sup> is placed with its axis perpendicular to the earth\u2019s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45\u00b0 with earth\u2019s field on
\na) its normal bisector and
\nb) its axis. Magnitude of the earth\u2019s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
\nSolution:
\nMagnetic moment of the bar magnet,
\nM = 5.25 \u00d7 10-2<\/sup> J T-1<\/sup>
\nMagnitude of earth\u2019s magnetic field at a place, H = 0.42 G = 0.42 \u00d7 10-4<\/sup> T
\na) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation :
\nB = \\(\\frac{\\mu_0M}{4\\pi R^3}\\) ; When the resultant field is inclined at 45\u00b0 with earth\u2019s field, B = H
\n
\n(OR) R = 0.05 m = 5 cm<\/p>\n
\n\\(\\frac{\\mu_02M}{4\\pi R^3}\\)
\nB = The resultant field is inclined at 45\u00b0 with earth\u2019s field. \u2234 B’ = H.
\n<\/p>\n","protected":false},"excerpt":{"rendered":"