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https://docs.google.com/document/d/1L1eVhE1pEE6NDOMWcLXD1fjjSWvn9aPa/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true VARIOUS DEFINITIONS & FEATURES RELATED TO THEM BAR MAGNET A system composed of two poles, equal in magnitude but opposite in polarity, placed at a small displacement apart is known as a bar magnet. It is also known as a magnetic dipole. POLE STRENGTH (MP) (Ia) The strength of a magnetic pole to attract magnetic materials towards it, is known as pole strength. (b) Greater the number of unit poles in a magnetic pole, greater will be its strength. (c) (d) Its unit is ampere-meter. Illustration 1: The magnetic flux density of magnetic field is 1.5 Wb/m2. A proton enters this field with a velocity of 2 × 107 m/s in a direction making an angle of 30° with the field. The force acting on the proton will be : (A) 2.4 × 10–12 Newton (B) 0.24 × 10–12 Newton (C) 24 × 10–12 Newton (D) 0.024 × 10–12 Newton Solution : (A) MAGNETIC LINES OF FORCE (a) The imaginary lines which represent the direction of magnetic field, are known as magnetic lines of force. (b) The imaginary path traced by an isolated (imaginary) unit north pole is defined as a line of force. (c) Magnetic lines of force are closed curves. Outside the magnet their direction is from north pole to south pole and inside the magnet these are from south to north pole. (d) They neither have origin nor end. (e) These lines do not intersect, because if they do so then it would mean two value of magnetic field at a single point, which is not possible. (f) The tangent drawn at any point to the line of force indicates the direction of magnetic field at that point. (g) At the poles of the magnet the magnetic field is stronger because the lines of force there are crowded together and away from the poles the magnetic field is week. i.e. magnetic field intensity α number of lines of force. (h) The number of magnetic lines of force passing through unit normal area is defined as magnetic induction (B) whereas the number of lines of force passing through any area is known as magnetic flux. (i) The lines of force can emerge out of the north pole of magnet at any angle and these can merge into the south pole at any angle. MAGNETIC FIELD (a) The space around a magnet in which a torque acts on a magnetic needle is known as magnetic field. (b) The space around a magnet in which a net force acts on a magnetic test pole is known as magnetic field. (c) The space around a magnet in which its effect is experienced is known as magnetic field. (d) There are four types of magnetic field : (i) Uniform magnetic field: (a) The magnetic field, in which the intensity of magnetic field is same at all points, is known as uniform magnetic field. (b) In such a magnetic field the magnetic lines of force are parallel and equidistant. e.g. the magnetic dines of force of earth's magnetic field. (ii) Non-uniform magnetic field: (a) The magnetic field, in which the intensity of magnetic field at different points is different, is known as non-uniform magnetic field. (b) It is represented by non-parallel lines of force. (iii) Varying magnetic field: (a) The magnetic field, which keeps on changing with respect to time is known as a variable magnetic field. (b) Example :– B = B0 sin ω t or B = B0 cos ω t (iv) Non-varying magnetic field: (a) The magnetic field which does not change with time is known as a constant magnetic field. (b) The direction of magnetic field is that in which a force acts on a unit test pole. (c) It can be produced by moving charges, current carrying loops, and variations in electric currents. MAGNETIC DIPOLE (a) The structures, which tend to align along the direction of magnetic field, are known as magnetic dipoles. (b) Bar magnet, current carrying solenoid, current carrying coil, current carrying coil, current loop, magnetic needle etc. are the examples of magnetic dipole. (c) It is not possible to separate out the two poles of a magnet. (d) Magnetic dipole is not a system composed of two poles because the existence of monopoles is not possible. A loop of single turn is also a magnetic dipole. One face of the loop behaves as north pole and the other face behaves as south pole. The face of the coil, in which current is anticlockwise, behaves as north pole and the face in which current is clockwise, behaves as south pole. MAGNETIC FLUX (a) The number of lines of force passing through a given area is defined as magnetic flux. (b) The magnetic flux passing through unit normal area is defined as magnetic induction (B). (c) When the magnetic field is normal to the plane then φ = BA, when A = 1m2 then φ = B. (d) When magnetic field makes an angle θ with the normal to the plane : (i) Magnetic flux linked with the plane = Area of the plane (A) × Component of magnetic field normal to the plane (B cos θ) i.e. φ = AB cos θ If the number of turns in the coil is N. then φ = NAB cos θ (ii) φ = magnetic field normal to the plane (B) × component of A in the direction of magnetic field (A cos θ) i.e. φ = BA cos θ In both cases q is the angle between . (iii) When are mutually parallel then θ = 0o and φ = BA. (iv) When are mutually perpendicular, θ = 90o and φ = 0. (v) When are antiparallel, then θ = 180o and φ = BA. (e) When the angle between B and the plane of coil is θ then φ = BA sin θ. If the number of turns in the coil is N then φ = NBA sin θ. (i) When the plane of coil is parallel to then θ = 0o and φ = 0. (ii) When and the plane of coil are mutually perpendicular i.e. θ = 90, then φ = BA. (iii) When and the plane of coil are mutually antiparallel i.e. θ = 180o, then φ = 0. (f) Magnetic flux linked with a small surface element dA where d = area of small element. = unit normal vector. (g) The flux linked with total area of the surface A (h) Positive magnetic flux: When the magnetic induction and the unit normal vector are in the same direction then φ is called the positive magnetic flux. φ = BA (i) Negative magnetic flux: When the magnetic induction and unit normal vector are mutually in opposite directions then φ is called negative magnetic flux φ = – BA (j) The flux emanating out of a surface is positive and the flux entering the surface is negative. (k) φ is a scalar quantity. (l) The net magnetic flux coming out of a closed surface is always zero, i.e. or MAGNETIC FLUX DENSITY OR MAGNETIC INDUCTION (a) The magnetic lines of force passing through unit normal area in a magnetic field is defined as magnetic induction. (b) (c) Direction of magnetic induction: The direction in the magnetic field in which if a current carrying conductor is placed then no force acts on it, is known as the direction of magnetic induction. (d) Magnetic induction is a vector quantity. (e) The magnetic induction due to a bar magnet (i) In axial position (ii) In equatorial position Here MAGNETISING FIELD OR INTENSITY OF MAGNETIC FIELD H (a) The ratio of magnetic induction produced in vacum (Bo) and magnetic permeability of vacuum is defined as magnetising field (H), i.e. (b) The intensity of magnetic field due to a pole of strength mp at a distance r from it is (c) Due to a small magnet H = Illustration 2: 1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising force generated in the material will be : (A) 1.2 × 10–3 A/m (B) 2.6 × 10–3 A/m (C) 2.6 × 10–4 A/m (D) 2 × 103 A/m Solution: (D) Illustration 3: In the above problem, the value of intensity of magnetisation in A/m will be : (A) 7.96 × 106 (B) 7.96 × 10–6 (C) 3.98 × 103 (D) zero Solution: (A) Illustration 4: In the above question 115 the relative permeability of the material will be : (A) 4.98 × 103 (B) 4.98 × 10–3 (C) 2.98 × 10–3 (D) 3.98 × 103 Solution: (D) MAGNETIC MOMENT (M) (a) If a magnet of length l and magnetic moment M is bent in the form of a semicircular are then its new magnetic moment will be M' = (b) The magnetic moment of an electron due to its orbital motion is 1μB whereas that due to its spin motion it is . i.e. Morbital = and Mspin = s Here μB = Bohr magneton (i) The value of Bohr magneton μB = (ii) μB = 0.93 × 10–23 Amp-m2 (c) Other formulae of M: (i) M = niπr2 (ii) (iii) (iv) M = nμB (d) Resultant magnetic moment : (i) When two bar magnets are lying mutually perpendicular to each other, then (ii) When two coils, each of radius r and carrying current i, are lying concentrically with their planes at right angles to each other, then if M1 = M2 Illustration 5: A square loop OABCO of side carries a current i. It is placed as shown in figure. Find the magnetic moment of the loop. Solution: Magnetic moment of the loop can be written as, Here, or Ans. Illustration 6: Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i = 2.0 A. Solution : By assuming two equal and opposite currents in BE, two current carrying loop (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence M Where M = ia = (2.0)(0.1)(0.1) = 0.02 A-m2 ∴ Mnet = () (0.02) A – m2 = 0.028 A-m2 Illustration 7: Two identical bar magnets each of length L and pole–strength m are placed at right angles to each other with the north pole of one touching the south pole of the other. Evaluate the magnetic moment of the system. Solution: As magnetic moment is a vector MR = with and as here M1 = M2 = mL and θ = 90°, so MR = (M2 + M2 + 2MM cos 90)1/2 = And, i.e., φ = tan–1(1) = 45° INTENSITY OF MAGNETISATION (I) (a) The magnetic moment per unit volume of a material is defined as intensity of magnetisation (I). i.e. I = , when V = 1m3 then I = M. (b) The unit of intensity of magnetisation is ampere/meter and its dimensions are M0L–1T0A1 (c) The pole strength per unit area of cross-section is defined as intensity of magnetisation. i.e. when A = 1m2 then A=1m2 then I = mp (d) It is a vector quantity whose direction is along the magnetic field. (e) In para- and ferro-magnetic materials its direction is in the direction of H and in dia-magnetic materials it is opposite to that of H. (f) I is produced in materials due to spin motion of electrons. (g) The value of I and its direction in a material depend on the nature of that material. (h) I - M curve (i) For para magnetic materials (ii) For diamagnetic materials (iii) For ferromagnetic materials (i) I - H curve (j) Its value depends on temperature. (k) It is produced on account of induction in a material. (l) For low magnetising field I ∝ H. i.e. I ∝ H or I = χHt (m) χ is a dimensionless constant i.e. it carries no unit. (n) Other types of intensity of magnetisation: (i) Mass intensity of magnetisation (Imass)- (ii) Molar intensity of magnetisation- Imolar = IMW = Mo. Wt.× mass intensity of magnetisation = MW × Im (iii) Molecular intensity of magnetisation- MAGNETIC SUSCEPTIBILITY OR (χ OR K) (a) The ratio intensity of magnetisation (I) in a material and the magnetising field (H) is defined as magnetic susceptibility (χ). i.e. When H = 1 Oersted then χ = I. (b) The intensity of magnetisation induced in a material by unit magnetising field is defined as magnetic susceptibility. (c) It has no unit and no dimensions. (d) It is a measure of ease with which a material can be magnetised by a magnetised by a magnetising field (H). (e) Magnetic susceptibility of various materials- (i) For diamagnetic materials – χ = low and negative (ii) For paramagnetic materials – χ = low but positive (iii) For ferromagnetic materials – χ = high and positive (f) For paramagnetic substances it is inversely proportional to temperature i.e. (g) For low magnetising field the value of χ is constant. (h) Different types of magnetic susceptibility- (i) Volume susceptibility (ii) Mass or specific susceptibility χm (iii) Molar susceptibility χMW χMW = χm × MW χ = Specific susceptibility × Mol. Wt. (iv) Molecular susceptibility χm χm = Atomic wt. × specific susceptibility = A (i) χ–T curve (j) χ– : For paramagnetic substances For paramagnetic substances Illustration 8: The magnetic induction along the axis of an air solenoid is 0.03 Tesla. On placing an iron core inside the solenoid the magnetic induction becomes 15 Tesla. The relative permeability of iron core will be : (A) 300 (B) 500 (C) 700 (D) 900 Solution: (B) ABSOLUTE MAGNETIC PERMEABILITY (Μ) (a) The ratio of magnetic induction (B) to magnetising field (H) is defined as magnetic permeability (μ). (b) The extent to which magnetic permeability of that medium. (c) It is the characteristic property of a magnetic material because it represents the amplification of magnetising field in that material. (d) Its value is always positive and is different for different materials. (e) For materials its value can be greater or less than μ0. (f) Its value depends on H and T. (h) μ = μ0 [1 + χ] (i) μ = μ0μr (j) (i) For feeromagnetic materials μ = high (ii) For paramagnetic materials μ = low (iii) For diamagnetic materials μ = very low (k) The unit of magnetic permeability is Weber per ammere-meter or Henry per meter and its dimensions are M1L0T–2A–2. RELATIVE PERMEABILITY (ΜR) (a) The ratio of magnetic permeability of medium (μ) to the magnetic permeabiliy of free space (μ0) is defined as relative permeability (μr). i.e. (b) (c) (d) The limit unto which a magnetic field penetrates matter, is known as relative permeability of that material. (e) It has no unit and no dimensions. (f) μr = 1 + χ (g) Relative permeability of various substances- (i) For diamagnetic substances the value of μr is slightly less than one i.e. μr < 1. (ii) For paramagnetic substances the value of μr is slightly greater than one i.e. μr > 1. (iii) For ferromagnetic substance the value of μr is much greater than one i.e. μr >> 1. MAGNETIC DIPOLE IN A UNIFORM MAGNETIC FIELD (i) When a dipole is placed (or suspended) in a uniform magnetic field, then no net force acts on it i.e. the resultant force acting on it is zero. (ii) When a dipole is placed in a uniform magnetic field then a torque acts on it which is given by Newton-meter (iii) The magnitude of torque is given by Where m is the pole strength and 2l is the length of dipole. (iv) The units of M are ampere-meter2 or Joule/Tesla and its dimensions are M0L2A1 (v) When a magnetic dipole is placed in a non-uniform magnetic field, then a torque as well as a force both act on it. The force acting on the dipole is given by (vi) (a) The work done in rotating a dipole in a magnetic field from an angle θ1 to angle θ2 with the field direction is given by- W = MB (cos θ1 – cos θ2) Joule (b) If initially the dipole is lying along the field direction, then θ1 = 0 ∴ W = MB (1 – cos θ) (vii) Potential energy (U): (a) The potential energy of a dipole in a magnetic field is given by U = (b) Special cases: When θ = 0, then U = – MB When θ = 90o, U = 0 When θ = 180o, U = – MB (–1) = MB Illustration 9: A Magnet is suspended in the magnetic meridian with a untwisted wire. The upper end of the wire is rotated through 180° to deflect the magnet by 30° from magnetic meridian. Now this magnet is replaced by another magnet. Now the upper end of the wire is rotated through 270° to deflect the magnet by 30° from magnetic meridian. Compare the magnetic moments of magnets. Solution: If φ be the twist of the wire, then , where C being restoring couple per unit twist of wire. Here φ1 = 180° – 30° = 150° = (150 × π/150) radian φ2 = (270° – 30° )= 240° = (240 × π/180) radian If M be the magnetic moment and H, the horizontal component of earth’s field, then = MH sinθ = MH sinθ If M1 and M2 be the magnetic moments of the two magnets respectively, then 1 = M1H sinθ for first magnet 2 = M2H sinθ for second magnet ∴ M1 : M2 = 5 : 8 CLASSIFICATION OF MATERIALS (i) The root cause of magnetism in matter is the motion of electric charges. (ii) The motion of electrons and protons in atoms is responsible for their magnetic properties. (iii) The variation in the number of fundamental charged particles and variation in their arrangement in different materials are responsible for differences in their magnetic properties. (iv) On the basis of mutual interactions or behaviour of various materials in an external magnetic field, the materials are divided in three main categories. (1) Diamagnetic substances (2) Paramagnetic substances (3) Ferromagnetic substances Comparative study of these materials: Property Diamagnetic substances Paramagnetic substances Ferromagnetic substances Cause of magnetism Orbital motion substances Spin motion of electrons Ferromagnetic substances Explanation of magnetism On the basis of orbital motion of electrons On the basis of spin and orbital motion of electrons On the basis of domains formed. Principle Electron principle Electron principle Domain principle Behaviour in a non-uniform magnetic field These are repelled in an external magnetic field i.e. have a tendency to move from high to low field region. These are feebly attracted in an external magnetic field i.e. have a tendency to move from low to high field region. These are strongly attracted in an external magnetic field. i.e. have an easy tendency to move from low to high field region. State of magnetisation These are weakly magnetised in a direction opposite to that of applied magnetic field These get weekly magnetised in the direction of applied magnetic field These get strongly magnetised in the direction of applied magnetic field. When a rod of the material is suspended between the pole pieces of a magnet. The materials align themselves at right angles to the direction of magnetic field. The materials align themselves along the direction of magnetic field. The materials easily align themselves in the direction of magnetic field. Liquid or power in a watch glass when placed between the pole pieces (a) when poles are far apart. (b) When poles are close to each other. (a) The liquid gets bulged in the middle (b) The liquid gets depressed in the middle. (a) The liquid gets depressed in the middle (b) The liquid gets bulged in the middle. (a) The liquid is very much depressed in the middle. (b) The liquid gets very much bulged in the middle. When the material in the form of liquid is filled in the U-tube and placed between pole pieces. Liquid level in that limb gets depressed Liquid level in that limb rises up. Liquid level in that limb rises up very much. On placing the gaseous materials between pole pieces. The gas expands at right angles to the magnetic field. The gas expands in the direction of magnetic field. The gas rapidly expands in the direction of magnetic field. The value of magnetic induction B B < B0 B > B0 Here B0 = magnetic induction in vacuum B > > B0 Magnetic susceptibility χ Low and negative χ ≈ 1 Low but positive χ ≈ 1 Positive and high χ ≈ 102. Dependence of χ on temperature Does not depend on temperature (except Bi at low temperature) Inversely proportional to temperature . This is called Curie law where C = Curie constant. This is called Cure-Weiss law. TC = Curie temperature. Dependence of χ on H does not depend does not depend does not depend Relative permeability (μr) μr < 1 μr > 1 μr > > 1 μr ≈ 102 Intensity of magnetisation (Ι) Ι is a direction opposite to that of H and its value is very low. Ι is in the direction of H but value is low. Ι is in the direction of H and value is very high. Ι-H Curves Magnetic moment (M) The value of M is very low (≈ 0 and is in opposite direction to H.) The value of M is very low and is in the direction of H. The value of M is very high and is in the direction of H. Transition of materials (at Curie temperature) These do not change On cooling, these get converted to ferro- magnetic materials at Curie temperature. These get converted into paramagnetic materials above Curie temperature. χ–T Curve The property of magnetism Diamagnetism is found in those materials in the atoms of which the number electrons is even. Paramagnetism is found in those materials in the atoms of which the majority of electron spins are in the same direction. Ferro-magnetism is found in those materials which when placed in an external magnetic field are strongly magnetised. Examples Cu, Ag, Au, Zn, Bi, Sb, NaCl, H2O air and diamond etc. Al, Mn, Pt, Na, CuCl2, O2 and crown glass. Fe, Co, Ni, Gd, Fe3O4 etc. Nature of effect Distortion effect Orientation effect Hysteresis effect CURIE LAW AND CURIE TEMPERATURE (a) Curie law: (i) The magnetic susceptibility of paramagnetic substances in inversely proportional to its absolute temperature i.e. (ii) Where C = Curie constant, T = absolute temperature (iii) On increasing temperature, the magnetic susceptibility of paramagnetic materials decreases and vice versa. (iv) The magnetic susceptibility of ferromagnetic substances does not change according to curie law. (b) Curie temperature (TC): (i) The temperature above which a ferromagnetic material behaves like a paramagnetic material is defined as curie temperature (TC). (ii) The minimum temperature at which a ferromagnetic substance is converted into paramagnetic substance is defined as curie temperature. (iii) For various ferromagnetic materials its values are different. e.g. for Ni for Fe for CO (iv) At this temperature the ferromagnetism of the substances suddenly vanishes. CURIE-WEISS LAW χ–T Cure OTHER IMPORTANT FEATURES AND FORMULAE (A) Angle of declination and Geographical meridian: (i) The horizontal component of earth's magnetic field is from S to N. (ii) If at any place the angle of dip is θ and magnetic latitude is λ then, tan θ = 2 tan λ The total intensity of earth's magnetic field here Here M and R are the magnetic moment of bar magnet of earth and radius of earth respectively. (iii) At magnetic equator of earth λ = 0° and at poles λ = 90°. (iv) At the poles and equator of earth, the values of total intensity are 0.66 and 0.33 oersted respectively. (B) (i) In vacuum :– B0 = μ0H (ii) In medium :– B = μH (iii) Resultant magnetic field due to Ιand H :– (a) B = B1 + BH = μ0Ι + μ0H B = μ0 (Ι + H) B = μ0H (1 + Ι/H) or B = μ0 (1 + χ)H + = μH (iv) (a) μ = μ0 [1 + χ] (b) μ = μ0μr (c) (v) (a) (b) (vi) (a) (b) (vii) The magnetic potential due to a small magnet at a point distant r is given by :– (viii) The mutual interaction force between two small magnets of moments M1 and M2 is given by (C) Magnetic torque τ (a) τ = MB sin θ (b) τ = BiNA sin θ (c) (D) Magnetic potential energy (UB) (a) (b) ASSIGNMENT 1. The magnetism of atomic magnet is due to - (A) only spin motion of electrons (B) only orbital motion of electrons (C) both spin and orbital motion of electrons (D) the motion of protons. Solution: (C) The magnetism of atomic magnet is due to both spin an orbital motions of electrons. 2. The earth's magnetic field inside an iron box as compared to that outside the box, is- (A) less (B) more (C) zero (D) same. Solution: (A) The earths magnetic field inside an iron box is less than that outside the box. 3. The magnetic susceptibility of a paramagnetic materials varies with absolute temperature T as - (A) (B) (C) (D) . Solution: (C) The magnetic susceptibility of a paramagnetic substance is inversely proportional to the absolute temperature i.e. 4. There are two points A and B on the extended axis of a 2 cm long bar magnet. Their distances from the centre of the magnet are x and 2x respectively. The ratio of magnetic fields at points A and B will be- (A) 8 : 1 approximately (B) 4 : 1 (approximately) (C) 4 : 1 (D) 8 : 1. Solution: (A) The intensity of magnetic field on the axis 5. The resultant magnetic moment of neon atom will be- (A) infinity (B) zero (C) mB (D) . Solution: (B) Neon atom is diamagnetic, hence its net magnetic moment is zero. 6. A magnetic material of volume 30 cm3 is placed in a magnetic field of intensity 5 oersted. The magnetic moment produced due to it is 6 amp-m2. The value of magnetic induction will be- (A) 0.2517 Tesla (B) 0.025 Tesla (C) 0.0025 Tesla (D) 25 Tesla. Solution: (A) B = m0 (I + H) I = H = 5 oersted = m0 = 4 × 10–7 Wb/amp-m = 0.2517 Tesla 7. The mass of an iron rod is 80 gm and its magnetic moment is 10 A–m2. If the density of iron is 8gm/C.C. hen the value of intensity of magnetisation will be- (A) 106 A/m (B) 104 A/m (C) 102 A/m (D) 10 A/m Solution: (A) = 106 A/m 8. The main difference between electric lines of force and magnetic lines of force is- (A) Electric lines of force are closed curves whereas magnetic lines of force are open curves. (B) Electric lines of force are open curves whereas magnetic lines of force are closed curves. (C) Magnetic lines of force cut each other whereas electric lines of force do not cut. (D) Electric lines of force cut each other whereas magnetic lines of force do not cut. Solution: (B) The magnetic lines of force are in the form of closed curves whereas electric lines of force are open curves. 9. The mass of a specimen of a ferromagnetic material is 0.6 kg. and its density is 7.8×103 kg/m3. If the area of hysteresis loop of alternating magnetising field of frequency 50 Hz is 0.722 MKS units then the hysteresis loss per second will be- (A) 277.7×10–5 Joule (B) 277.7×10–6 Joule (C) 277.7×10–4 Joule (D) 277.7×10–4 Joule. Solution: (A) WH = VAft or = 277.7 × 10–5 Joule 10. The horizontal component of flux density of earth's magnetic field is 1.7×10–5 tesla. The value of horizontal component of intensity of earth's magnetic field will be? (A) 24.5 A/m (B) 13/5A/m (C) 0.135 A/m (D) 1.35 A/m. Solution: (B) = 13.5 A/m 11. A magnetising field of 2×103 amp/m produces a magnetic flux density of 8 Tesla in an iron rod. The relative permeability of the rod will be- (A) 102 (B) 10o (C) 104 (D) 101 Solution: (C) or 12. In a hydrogen atom the electron is revolving in a circular path of radius 5.1×10–11m with a frequency 6.8×1015 Hz. The equivalent magnetic moment will be- (A) 8 × 10–24 A–m2 (B) 8 × 10–22 A–m2 (C) 8.9 × 10–20 A–m2 (D) 8 × 10–18 A–m2 Solution: (A) M = iA = efr2 or M = 1.6×10–19×6.8×1015×3.14×(5.1×10–11)2 = 8.9 ×10–24 A–m2 13. A cube of side l is placed in a magnetic field of intensity B. The magnetic flux emanating out of it will be- (A) zero (B) Bl2 (C) 2Bl2 (D) 6Bl2 Solution: (A) The net flux coming out of a closed surface is zero. Hence the flux coming out of the cube will be zero. 14. A circular disc of area is placed in a uniform magnetic field of intensity Tesla. The flux crossing the disc will be- (A) 23 Weber (B) 23×10–2 Weber (C) 23×10–3 Weber (D) 23×10–4 Weber Solution: (C) = 0.0008 + 0.0015 = 0.0023 Weber = 23×10–4 Weber 15. The total magnetic flux in a material, which produces a pole of strength mp when a magnetic material of cross-sectional area A is placed in a magnetic field of strength H, will be- (A) μ0 (AH + mp) (B) μ0 AH (C) μ0 mp (D) μ0 [mp AH + A] Solution: (A) φ = BA ....(A) ....(B) From eqs. (A) and (B) .....(C) From eqs. (B) and (C) 16. The relation between μ and H for a specimen of iron is as follows- The value of H which produces flux density of 1 Tesla will be- (A) 250 A/m (B) 500 A/m (C) 750 A/m (D) 103 A/m Solution: (B) B = μH or or 1 = 0.4 + 12×10–4 H H = 500 A/m 17. The correct curve between X and for paramagnetic materials is- Solution: (A) 18. The SI unit of magnetic flux is- (A) Weber (B) Maxwel (C) Tesla (D) Gauss Solution: (A) 19. The intensity of magnetic field due to an isolated pole of strength mp at a point distant r from it will be- (A) (B) mpr2 (C) (D) Solution: (A) The magnetic intensity due to an isolated pole of strength mp at a distance r = 20. A uniform magnetic field is directed from left towards right in the plane of paper. When a piece of soft iron is placed parallel to the field. The magnetic lines of force passing through it will be- Solution: (C) The magnetic lines of force in a ferrmagnetic material are crowded together. 21. A bar magnet of magnetic moment M is cut into two equal parts. The magnetic moment of either of the parts will be- (A) (B) 2M (C) 2M (D) Solution: (A) 22. A loop of area 0.5 m2 is placed in a magnetic field of strength 2 Tesla in direction making an angle of 60o with the field. The magnetic flux linked with the loop will be- (A) (B) (C) 2 Weber (D) Solution: (A) 23. The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet of length 5 cm and pole strength of 200A-m on its axial line will be (A) 2.5×10–2 N (B) 2.5×10–3N (C) 5.0×10–2N (D) 5.0×10–3N. Solution: (A) = 2.5×10–2 N 24. The magnetic susceptibility of a paramagnetic substance is 3×10–4. It is placed in a magnetising field of 4×104 amp/m. The intensity of magnetisation will be- (A) 3 × 108 A/M (B) 12 × 108 A/M (C) 12 A/M (D) 24 A/M Solution: (C) I = XH = 3×10–4×4×103 = 12 A/m 25. Volt-second is the unit of- (A) B (B) φ (C) I (D) x Solution: (B) 26. The volume susceptibility of a magnetic material is 30×10–4. Its relative permeability will be- (A) 31 × 10–4 (B) 1.003 (C) 1.0003 (D) 29 × 10–4 Solution: (B) 27. A current of 2 ampere is passed in a coil of radius 0.5 m and number of turns 20. The magnetic moment of the coil is- (A) 0.314 A–m2 (B) 3.14 A–m2 (C) 314 A–m2 (D) 31.4 A–m2 Solution: (D) M = iR2N = 2×3.14×0.25×20 = 31.4 A–m2 28. The value of magnetic susceptibility for super-conductors is- (A) zero (B) infinity (C) +1 (D) –1 Solution: (D) For superconductor μr = 1 + X = 0  X = –1 29. A current of 1 ampere is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be- (A) 0.314 A-m2 (B) 3140 A-m2 (C) 100 A-m2 (D) μ0 A-m2 Solution: (A) M = iA = πR2 Ni = 3.14×0.01×10×1=0.314 Am2 30. The magnetic moment of a magnet of mass 75 gm is 9×10–7 A-m2. If the density of the material of magnet is 7.5×103 kg/m3 then intensity of magnetisation will be- (A) 0.9 A/m (B) 0.09 A/m (C) 9 A/m (D) 90 A/m Solution: (B) For test 1. The area of hysteresis loop of a material is equivalent to 250 Joule. When 10 kg material is magnetised by an alternating field of 50Hz then energy lost in one hour will be if the density of material is 7.5 gm/cm3. (A) 6×104 Joule (B) 6×104 Erg (C) 3×102 Joule (D) 3×102 Erg. Solution: (A) 2. The corecivity of a bar magnet is 100 A/m. It is to be demagnetised by placing it inside a solenoid of length 100 cm and number of turns 50. The current flowing the solenoid will be- (A) 4A (B) 2A (C) 1A (D) zero. Solution: (B) H = ni 3. A current i is flowing in a conductor of length l. When it is bent in the form of a loop then its magnetic moment will be- (A) (B) (C) (D) 4πl2i. Solution: (A) M = iA = i π r2 But 2πr = l 4. A rod of ferromagnetic material with dimensions 10 cm×0.5 cm×0.2 cm is placed in a magnetic field of strength 0.5×104 amp/m as a result of which a magnetic moment of 5 amp-m2 is produced in the rod. The value of magnetic induction will be- (A) 0.54 Tesla (B) 0.358 Tesla (C) 2.519 Tesla (D) 6.28 Tesla. Solution: (D) . 5. The ratio of intensities of magnetic field in the axial and equatorial positions of a magnet will be- (A) 1 : 4 (B) 4 : 1 (C) 1 : 2 (D) 2 : 1 Solution: (D) 6. The resultant magnetic moment due to two current (i) carrying concentric coils of radius r, mutually perpendicular to each other will be- (A) (B) (C) (D) Solution: (A) M' = = 7. The magnetic susceptibility of a paramagnetic material at –73oC is 0.0075 then its value at –173oC will be- (A) 0.0045 (B) 0.0030 (C) 0.015 (D) 0.0075. Solution: (C) For paramagnetic materials 8. The area of cross-section of three magnets of same length are A, 2A and 6A respectively. The ratio of their magnetic moments will be- (A) 6 : 2 : 1 (B) 1 : 2 : 6 (C) 1 : 4 : 36 (D) 36 : 4 : 1 Solution: (B) (Area of cross-section) 9. If the radius of a circular coil is doubled and the current flowing in it is halved then the new magnetic moment will be if its initial magnetic moment is 4 units- (A) 8 units (B) 4 units (C) 2 units (D) zero. Solution: (A) 10. Newton/Weber is the unit of- (A) (B) (C) (D) φ Solution: (A) 11. The inner and the outer radii of a toroid are 9cm and 11cm respectively and the number of turns in it is 3140. A magnetic field of 2.5 Tesla is produced in it when a current of 0.5 ampere is passed in it. The permeability of core material is (in Henry/meter)- (A) 10–1 (B) 10–2 (C) 10–3 (D) 10–4 Solution: (C) 12. In the above problem the relative permeability of the core will be- (A) 684.5 (B) 864.7 (C) 369.4 (D) 796.2 Solution: (D) 13. A magnetising field of 5000 A/m produces a magnetic flux of 5×10–5 Weber in an iron rod. If the area of cross-section of the rod is 0.5 cm2, then the permeability of the rod will be (in henry/m)- (A) 1 × 10–3 (B) 2 × 10–4 (C) 3 × 10–5 (D) 4 × 10–6 Solution: (B) 14. In the above problem the magnetic susceptibility of the rod will be- (A) 158.2 (B) 198.0 (C) 295.3 (D) 343.6 Solution: (A) 15. In the above problem the intensity of magnetisation will be (in A/m)- (A) 7.9×102 (B) 7.9×102 (C) 7.9×102 (D) 7.9×105 Solution: (D) 16. At any place on earth, the horizontal component of earth's magnetic field is times the vertical component. The angle of dip at that place will be- (A) 60o (B) 45o (C) 90o (D) 30o Solution: (D) 17. The horizontal component of earth's magnetic field at any place is 0.36×10–4 Weber/m2. If the angle of dip at that place is 60o then the value of vertical component of earth's magnetic field will be-(in Wb/m2)- (A) 0.12 × 10–4 (B) 0.24 × 10–4 (C) 0.40 × 10–4 (D) 0.62 × 10–4 Solution: (D) 18. The value of angle of dip at a place on earth is 45o. If the horizontal component of earth's magnetic field is 5×10–5 Telsa then the total magnetic field of earth will be- (A) Tesla (B) Tesla (C) Tesla (D) zero. Solution: (A) 19. The ratio of intensities of magnetic field, at distances x and 2x from the centre of a magnet of length 2cm on its axis, will be- (A) 4 : 1 (B) 4 : 1 approx (C) 8 : 1 (D) 8 : 1 approx. Solution: (D) 20. The length of a bar magnet is 10 cm and its pole strength is 10–3 Weber. It is placed in a magnetic field of induction 4π × 10–3 Tesla in a direction making an angle of 30o with the field direction. The value of orque acting on the magnet will be- (A) 2π × 10–7 N-m (B) 2π × 10–5 N-m (C) 0.5 × 102 N-m (D) 0.5 N-m. Solution: (A) 21. A magnetic needle of magnetic moment 60 amp-m2 experiences a torque of 1.2×10–3 N-m directed in geographical north. If the horizontal intensity of earth's magnetic field at that place is 40Wb/m2, then the angle of declination will be- (A) 30° (B) 45° (C) 60° (D) 90° Solution: (A) 22. The ratio of total intensities of magnetic field at the equator and the poles will be- (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 1 : 4 Solution: (A) Total intensity of magnetic field remains constant 23. The magnetic flux density of magnetic field is 1.5 Wb/m2. A proton enters this field with a velocity of 2×107 m/s in a direction making an angle of 30o with the field. The force acting on the proton will be- (A) 2.4 × 10–12 Newton (B) 0.24 × 10–12 Newton (C) 24 × 10–12 Newton (D) 0.24 × 10–12 Newton Solution: (A) 24. The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial axis. The ratio of distances of X and Y from the centre of the magnet will be- (A) (B)–3 (B) (B)–3 (C) 23 (D) 21/3 Solution: (D) 25. Two magnets A and B are equal in length, breadth and mass, but their magnetic moments are different. If the time period of B in a vibration magnetometer is twice that of A, then the ratio of magnetic moments will be- (A) (B) 2 (C) 4 (D) 12 Solution: (C) 26. The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts lengthwise then the time period of each part will be- (A) 4 sec. (B) 2 sec. (C) 0.5 sec. (D) 0.25 sec. Solution: (A) 27. The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnets is the same but the magnetic moment of B is four times that of A. The time period of B will be- (A) (B) (C) 2T (D) 4 T Solution: (B) 28. The value of current enclosed by a circular path of radius 0.30 cm is 9.42 ampere. The value of magnetic field along the path will be- (A) 500 amp/m (B) 1000 Amp/m (C) 5 × 104 Amp/m (D) Zero Solution: (A) 29. A magnetic wire is bent at its midpoint at an angle of 60o. If the length of the wire is l and its magnetic moment is M then the magnetic moment of new shape of wire will be- (A) 2 M (B) M (C) (D) Solution: (C) 30. A current of 2 ampere is flowing in a coil of radius 50 cm and number of turns 20. The magnetic moment of the coil will be- (A) 3.14 amp-m2 (B) 31.4 amp-m2 (C) 314 amp-m2 (D) 0.314amp-m2 Solution: (B) 1. 1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising force generated in the material will be- (A) 1.2 × 10–3 A/m (B) 2.6 × 10–3 A/m (C) 0.6 × 10–4 A/m (D) 2 × 10–3 A/m Solution: (D) 2. A magnet makes 10 oscillation per minute at a place where the horizontal component of earth's magnetic filed (H) is 0.33 oersted. The time period of the magnet at a place where the value of H is 0.62 oersted will be- (A) 4.38 S (B) 0.38 S (C) 2.38 S (D) 8.38 S Solution: (A) 3. The magnetic induction inside a solenoid is 6.5×10–4T. When it is filled with iron medium then the induction becomes 1.4T. The relative permeability of iron will be- (A) 1578 (B) 2355 (C) 1836 (D) 2154. Solution: (D)