Dhananjay Parkar – Physics has made the life easy



UNIT: 3 ( MAGNETIC EFFECTS OF CURRENT AND MAGNETISM ) ONE MARK QUESTIONS1. An electric current of 0.25 A flows in a loop of radius 0.2 cm. Calculate the magnitude of the magnetic dipole moment of the dipole formed. 2. Why electrons can’t be accelerated by using a cyclotron?3. A given galvanometer is to be converted into (i) an ammeter (ii)a milliammeter (iii) a voltmeter. In which case will the required resistance be (i)least (ii)highest 4. Why phosphor bronze alloy is preferred for the suspension wire of moving coil galvanometer? 5. A circular coil of ‘N’ turns and diameter ‘d’ carries a current ‘I’. It is unwound and rewound to make another coil of diameter ‘2d’, current ‘I’ remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil. 6. What will be the path of a charged particle moving in a uniform magnetic field at any arbitrary acute angle? 7. An electron and a proton, moving parallel to each other in the same direction with equal momenta, enter into a uniform magnetic field (perpendicular to the plane of the paper)which is at right angles to their velocities. Trace their trajectories in the magnetic field. 8. What types of force is acting between two parallel wires carrying current in the same direction? 9. What is the function of the radial magnetic field in the moving coil galvanometer? 10. Two parallel wires carrying currents in the same direction attract each other, while the two beam of electrons travelling in the same direction repel each other. Give reason.ANSWER OF ONE MARK QUESTIONS Ans 1. 3.14 x 10 - 6 Am2 Ans 2. Because relativistic mass of electrons become very high due to its very high velocity , which violates the condition of magnetic resonance. Ans3. The required resistance has least value in case of an ammeter and maximum value in case of voltmeter. Ans 4. Current sensitivity is more as it is given by k n BA I ; Because k is small for phosphor bronze. Ans 5. Magnetic moment (m) = NIA , mB/mA = 2/1Ans 6. Helical Ans 7. Both proton and electron will make circular path of equal radii in anticlockwise as well as clockwise direction respectively. Ans 8. Force of attraction will act between these two current carrying wires.Ans. 9. The arrangement provides linear scale to the galvanometer, because the plane of the coil always gets aligned parallel to the direction of the applied magnetic field.Ans 10. In case of parallel wires, only attractive magnetic interaction acts. In case of electron beams, the electrostatic repulsion is stronger than the attractive magnetic interaction. --------------------------xxxxxxxxxxxxxx------------------------------- TWO MARKS QUESTIONS:A solenoid of length 50 cm having 100 turns carrying a current of 4.5 A find the magnetic field (i) in the Interior of solenoid (ii) outside the solenoid.How will the magnetic field intensity at the centre of a circular coil carrying current change if the current through the coil is doubled and the radius of the coil is halved.3. State Ampere’s circuital law. 4. A milliammeter whose resistance is 120 has full scale deflection with a current of 5mA . How will you use it to measure a maximum current of 100 A? 5. A solenoid is 1m long and 3 cm in mean diameter. It has 5 layers of windings of 800 turns each and carries a current of 5 A. Find Magnetic Field Induction at the center of the solenoid.Define current sensitivity and voltage sensitivity of a galvanometer.What is the working principle of a moving coil galvanometer ?Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle.State Biot – Savart law and express this law in the vector form.Write the expression, in a vector form, for the magnetic Lorentz force ? ? experienced by a charge q moving with velocity ?? in a magnetic field ??. What is the direction of the magnetic force? -----------------------------xxxxxxxxxxxxx---------------------------------ANSWER OF TWO MARKS QUESTIONS Ans 1. (i) B = ?0ni = 6.28 x 10- 4 T (ii) B is almost zero. Ans 2. B = ?0n x 2I / 2 x (R/2) = 4B Ans 3. It states that the line integral of magnetic field around any closed path in vacuum/air is ?0 times the total current threading the closed path. Mathematically , we can express as Ans 4. S = 0.006 in parallel . Ans 5. 2.5 x 10 -2 T, parallel to the axis of the solenoid . Ans 6. Current sensitivity : It is defined as the deflection of coil per unit current flowing in it. Current Sensitivity, S=I=NAB C . Voltage sensitivity : It is defined as the deflection of coil per unit potential difference across its ends.Voltage Sensitivty, SV=V=NABGC ,where G is the resistance of galvanometer. Ans 7. The current carrying coil kept in a magnetic field always experiences a torque. Ans 8. Expression for Period of Revolution : Suppose the positive ion with charge q moves in a dee with a velocity v, then, Ans 9. dB = (?0/ 4π) (I dl x r)/r3Ans 10. ? ? = q (??x ?) -----------------------------xxxxxxxxxxxxx---------------------------------THREE MARKS QUESTIONSTwo straight long parallel conductors carry currents I 1 and I 2 in the same direction. Deduce the expression for the force per unit length between them. Depict the pattern of magnetic field lines around them.2. Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus.Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis.(a) Using Biot-Savart’s law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop.a) Why is the magnetic field radial in a moving coil galvanometer? Explain how it is achieved.(b) A galvanometer of resistance ‘G’ can be converted into a voltmeter of range (0-V) volts by connecting a resistance ‘R’ in series with it. How much resistance will be required to change its range from 0 to V/2 ?1.2.3.A solenoid is a long insulated copper wire wound in the form of a coil. If the solenoid is long 4.5.(a) Need for a radial magnetic field: The relation between the current (i) flowing through the galvanometer coil, and the angular deflection () of the coil (from its equilibrium position), is ?=NABI.sinθkwhere is the angle between the magnetic field B and the equivalent magnetic moment ?m of the current carrying coil. Thus I is not directly proportional to . We can ensure this proportionality by having =90. This is possible only when the magnetic field, B, is a radial magnetic field. In such a field, the plane of the rotating coil is always parallel to B. To get a radial magnetic field, the pole pieces of the magnet, are made concave in shape. Also a soft iron cylinder is used as the core. [Alternatively : Accept if the candidate draws the following diagram to achieve the radial magnetic field.] FIVE MARKS QUESTIONS1. Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle. (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles. [5,2014]Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working. Answer the following : Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer ? Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason. [5,2014](a)Using Biot-Savart's law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop [5,2013] (b) What does a toroid consist of ? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside exterior to the toroid is zero. 4. (a)Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by the particles. (b)An α-particle and a proton are released from the centre of the cyclotron and made to accelerate. (i)Can both be accelerated at the same cyclotron frequency ? Give reason to justify your answer. (ii)When they are accelerated in turn, which of the two will have higher velocity at the exit slit of the dees ? [5,2013] ANSWERS OF FIVE MARKS QUESTIONS1. (a) If particle is performing circular motion due to magnetic force then centripetal force = Magnetic force mv2/r = q ν B sin 90° ? r = mν/qB ∴Time period = 2πr/v T = 2πm/qB T = qB m2 π ∝ ν° ∴Frequency f = T 1 = m2 qB π ∝ ν° (b) Cyclotron is a device to accelerate ions to extremely high velocities, by accelerating them repeatedly through high voltages. Principle – A positive ion can acquire sufficiently large energy with a small alternating potential difference by making the ion cross the same electric field time and again by making use of a strong magnetic field. Construction- It consists of a pair of hollow metal cylindrical chambers shaped like D, and called the Dees ; Both the Dees are placed under the circular pole pieces of a very strong electromagnet. The two Dees are connected to the terminals of a very high frequency and high voltage oscillator whose frequency is of the order of a few million cycles per second. Working - Charged ion is passed through electric field again & again to be energised. Inside dee's strong ⊥ magnetic field turns the particle towards gap. So radius of semi circular path increase continuously.2. (a) Principle. When a current carrying coil is placed in magnetic field, it experiences a torque. Construction. It consists of a narrow rectangular coil PQRS consisting of a large number of turns of fine insulated copper wire wound over a frame made of light, non-magnetic metal. A soft iron cylinder known as the core is placed symmetrically within the coil and detached from it. The coil is suspended between the two cylindrical pole pieces (N and S of a strong permanent horse shoe magnet) by a thin flat phosphor bronze strip the upper end of which is connected to a movable torsion head T. The lower end of the coil is connected to a hair spring s of phosphor bronze having only a few turns. Redial magnetic field. The magnetic field in the small air gap between the cylindrical pole pieces is radial. The magnetic lines of force within the air gap are along the radii. On account of this, the plane of the coil remains always parallel to the direction of the magnetic field The magnetic field in the small air gap between the cylindrical pole-pieces is radial. On account of this, the plane of the coil remains always parallel to the direction of the magnetic field. Theory : Let I = current flowing through the coil B = magnetic field induction l = length of the coil ; b = breadth of the coil N = number of turns in the coil A (= l × b) = area of the coil Moment of deflecting couple = NBIl × b = NBIA When the coil deflects, the suspension fibre gets twisted. On account of elasticity, a restoring couple is set up in the fibre. This couple is proportional to the twist. If α be the angular twist then Moment of restoring couple = kα For equilibrium of the coil, NBIA = kα or α = NBAI/ k or α =KI where K= NBA/ k is the galvanometer constant. Now, α ∝I or I ∝α (b) (i) By using soft iorn core . magnetic field is increased so sensitivity increases and mag. field becomes radial So angle between plane of coil & magnetic line of force is zero in all orientations of coil. (b) (i) By using soft iorn core . magnetic field is increased so sensitivity increases and mag. field becomes radial So angle between plane of coil & magnetic line of force is zero in all orientations of coil. (ii) Voltage sensitivity = Current sensitivity/ RIt means that V S increases if C S is increased. But if resistance of coil is also increases in same ratio then V S may be constant 3. a. According to Biot-Savart's law, magnetic field due to a small element XY at point P is Resolving dB into two components : (i) dB cos θ, which is perpendicular to the axis of the coil (ii) dB sin θ which is along the axis of the coil and away from the centre of the coil. b For any point inside the empty space surrounded by toroid and outside the toroid, magnetic field B is zero because the net current enclosed in these spaces is zero. But magnetic field is not exactly zero.4. Electric field accelerate the charged particle where as magnetic field makes its path circular. vα/ vp = 2 PREVIOUS YEAR’S QUESTIONS AND ANSWERS1.A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency ? (1, 2018)Ans - υ = qB/ 2πm, νe > νp as q is same for both and me < mp2. (a) An iron ring of relative permeability μr has windings of insulated copper wire of n turns per metre. When the current in the windings is I, find the expression for the magnetic field in the ring.[1, 2018 Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current[1,2014]Ans. One ampere is that current which if passed in each of two parallel conductors of infinite length and one meter apart in vacuum, causes each conductor to experience a force of 2×10–7 Newton per meter of length of conductor. What can be the cause of helical motion of a charged particle ? [1,2017]Ans. The perpendicular component of velocity is responsible for circular motion where as the parallel component makes it move forward. Hence a helical motion is found.Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed. [2,2017]Ans- qvB = qE, when v, B and E are mutually perpendicularV = E/BWrite two properties of a material suitable for making (a) a permanent magnet, and (b) an electromagnet.[2,2017]Two properties of material used for making permanent magnets are (a) high coercivity (b) high retentivity and high hystresis loss Two properties of material used for making electromagnets are (a) high permeability (b) low coercivity(a)State Biot – Savart law and express this law in the vector form.[3,2017](b)Two identical circular coils, P and Q each of radius R, carrying currents 1A and 3 A respectively, areAns. - placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils.Ans-The strength of magnetic field dB due to a small current element dl carrying a current I at a point P distant r from the element is directly proportional to I, dl, sin θ and inversely proportional to the square of the distance (r2) where θ is the angle between dl and r. dB α I ii) dB α dl iii) dB α sin θ iv) dB α 1 / r2 B1 = μ0I1/2R, B2 = μ0I2/2R, B = √( B12 + B22)State Ampere's circuital law. Use this law to find magnetic field due to straight infinite current carrying wire. How are the magnetic field lines different from the electrostatic field lines?ORState the principle of a cyclotron. Show that the time period of revolution of particles in a cyclotron is independent of their speeds. Why is this property necessary for to operation of a cyclotron?[3,2016] The line integral of magnetic field over a closed loop is ?0 times the total current passing through the loop. Consider an amperian loop of radius r and applying ampere's circuital law The magnetic field lines form continuous closed loops, where as electrostatic field lines never forms closed loop.ORPrinciple. The frequency of revolution of the charged particle in magnetic field is independent of its energy. If particle is performing circular motion due to magnetic force then centripetal force = Magnetic force The time period of revolution of particles in a cyclotron is independent of their speeds. This property is necessary for the operation of a cyclotron to attain the resonance condition (νa = νc) where νa is the frequency of the applied voltage and νc is cyclotron frequency.Deduce the expression for the torque ? acting on a planar loop of area A and carrying current I in a uniform magnetic field B If the loop is free to rotate, what would be its orientation in stable equilibrium?[3,2015] We consider the case when the plane of the loop, is not along the magnetic field, but makes an angle θ with it. Fig.(a) illustrates this general case. The forces on the arms BC and DA are equal, opposite, and act along the axis of the coil, which connects the centres of mass of BC and DA. Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms AB and CD are F1 and F2. They too are equal and opposite, with magnitude, F1 = F2 = I b B But they are not collinear! This results in a couple Fig.(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is,τ = F1 (a/2) sinθ + F2 (a/2) sinθ = I ab B sin θ= I A B sin θ ... (i) As θ → 0, the perpendicular distance between the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero. The torques in above equation can be expressed as vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as, m = I A where the direction of the area vector A is given by the right-hand thumb rule and is directed into the plane of the paper in Fig.(a). Then as the angle between m and B is θ, equation (i) can be expressed by one expression τ = m × B we see that the torque τ vanishes when m is either parallel or antiparallel to the magnetic field B . This indicates a state of equilibrium as there is no torque on the coil (this also applies to any object with a magnetic moment m r ). When m r and B r are parallel the equilibrium is a stable one. A cyclotron's oscillator frequency is 10 MHz What should be the operating magnetic field for accelerating protons? If the radius of its 'dees' is 60 cm, calculate the kinetic energy (in MeV) of the proton beam produced by the accelerator.[3, 2015 A wire AB is carrying a steady current of 6A and is lying on the table. Another wire CD carrying 4A is held directly above AB at a height of 1mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms–2] [3,2013] TOPIC- MAGNETISM ONE MARK QUESTIONS1. Relative permeability of a material ?r = 0.5. Identify the nature of the magnetic material and write its relation of magnetic susceptibility.2. What are permanent magnets? Give one example?3. Where on the surface of earth’s vertical component of earth’s magnetic field zero?4. The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°.what is the value of vertical component of the earth’s magnetic field at equator?5. What is the angle of dip at a place where the horizontal component and vertical components of the earth’s magnetic field are equal?6. A magnetic needle free to rotate in vertical plane and tends itself vertically at a certain place on the earth what are the values ofa. Horizontal component of the earth’s magnetic field andb. angle of dip at this place ?7. Where on the surface of the earth is the angle of dip 90°?8. The permeability of a magnetic material is 0.9983. Name the type of magnetic material, it represents.9. The susceptibility of a magnetic material is 1.9 * 10??. Name the magnetic material, it represents.10. The susceptibility of a magnetic material is -4.2 * 10??.Name the type of magnetic material, it represents.11. What is the characteristics property of a diamagnetic material?12. Define the term magnetic declination.ANSWERS OF ONE MARK QUESTIONSANS: The nature of Magnetic Material is Diamagnetic.?r=1+χmANS: Permanent Magnet are those magnets which have high retentivity and coercivity. The magnetision of permanent magnet is not easily destroyed even if it is handled roughly or exposed in stray, reverse magnetic field ,e.g: steel.ANS: At equator, vertical component will be zeroANS: {The vertical and horizontal components of the earth’s magnetic fields are perpendicular to each other}Horizontal component of earth’s magnetic field,H = Be cos 60°= B{given}= Be * ? = BOr Be = 2BVertical component of earth’s magnetic field,V = Be sin 60°V =2B *√3/2V = √3BANS: The angle of dip is given byΔ = tan?1 {B? / BH}B? = vertical component of the earth’s magnetic field.BH= horizontal component of the earth’s magnetic field.So, as B?= BH Then, δ = tan-1[1] = 45°Therefore, The angle of dip will be δ = 45°.ANS: a. The coil is free to move in vertical plane , it means that there is no component A the earth’s magnetic field in horizontal direction, so the horizontal component of the earth’s magnetic field is 0b. The angle of dip is 90°ANS : At the poles, the angle o dip is 90°ANS: The magnetic material is Diamagnetic substance for which ?r<1.ANS: The small and positive susceptibility of 1.9 *10?? represents paramagnetic substance.ANS: Negative susceptibility diamagnetic substanceANS: Diamagnetic material acquires magnetization in the opposite direction of the magnetic field when they are placed in an external magnetic field.ANS: Magnetic declination the angle between geographical meridian and magnetic meridian at any place of the earth is known as magnetic declination.TWO MARKS QUESTIONS13. An electron moves around the nucleus in a hydrogen atom of radius 0.5 A?, with a velocity of 2 * 10?m/s. Calculate the following(1) the equivalent current due to orbital motion of electron(2) the magnetic field produced at the centre of the nucleus(3) the magnetic moment associated with the electron.14. Out of the two magnetic materials, A has relative permeability slightly greater than unity while B has less than unity. Identify the nature of the material A and B. Will their susceptibilities be positive or negative?15. Give two points to distinguish between a paramagnetic and diamagnetic substance.16. Write two properties of a material which makes it suitable for making electromagnet?17. The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties.18. a. How does a diamagnetic material behave when it is cooled to very low temperature? b. Why does a paramagnetic sample display greater magnetisation when cooled? Explain.19.A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its North tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place.20. Distinguish between diamagnetic and ferromagnetic materials in terms of(I) Susceptibility and (II) Their behavior in a non uniform magnetic field.21 (a) Write two characteristics of a material used for making permanent magnets?(b) Why is core of electromagnets made of ferromagnetic materials?22. The horizontal component of the earth’s magnetic fields at a place is 3 times its vertical component there. Finds the value of the angle of dip at that place? What is the ratio of the horizontal component to the total magnetic field of the earth at that place?23. The horizontal component of the earth’s magnetic field at a place equals to its vertical component there. Find the value of the angle of dip at that place. what is the ratio of the horizontal component to the total magnetic field of the earth at that place?24. Draw magnetic field lines when a (1) diamagnetic, (2) paramagnetic substance is placed in an external magnetic field. Which magnetic property distinguished this behavior of the field lines due to the two substances?25. State briefly an efficiency way of making a paramagnet magnet. Write two properties to select suitable materials for making permanent magnets.26. Out of the following, identify the material which can be classified as (1)paramagnetic (2)Diamagnetica) Aluminium b) Bismuth c) Copper d) SodiumWrite one property to distinguish between paramagnetic and diamagnetic materials.27. The following figure shows the variation of intensity of magnetisation versus the applied magnetic field intensity H for two magnetic materials A and B, I BAHIdentify the material A and B.Why does the material B have a large susceptibility than A for a given field at constant temperature?28. The following fig shows the variation of intensity of magnetisation versus the applied magnetic field intensity H for two magnetic materials A and B.AH-IBIdentify the material A and B.Draw the variation of susceptibility χm with temperature for B.29. The following Fig shows the variation of intensity of magnetisation versus the applied magnetic field intensity H for two magnetic materials A and B .BIA HIdentify the material A and B.For the material A, plot the variation of intensity of magnetisation.30. Define magnetic susceptibility of a material. Name two element, one having positive susceptibility and other having negative susceptibility. What does negative susceptibility signify?31. Draw a plot showing the variation of intensity of magnetisation with the applied magnetic field intensity for bismuth. Under what condition a diamagnetic material exhibits perfect conductivity and perfect diamagnetism? ANSWER OF TWO MARKS QUESTIONSANS: Here r = 0.51 * 10?10 m, v = 2 * 105 ms-1I = e/T = ev/2πr = 1.6 * 10-19 * 2 *105 /2π * 0.51 * 10-10 = 10-4AB = μ0 I/2r = 4π * 10-7 * 10-4 /2 *0.51 *10-10 =1.23 TM = IA =ev/2πr * πr = evr /2 1.6 * 10-19*2*105*0.51*10-10 =8.16 * 10-25 Am2 ANS : The nature of material A is paramagnetic and its susceptibility ?M is positiveThe nature of the material B is Diamagnetic and its susceptibility ?M is negative. ANS: PARAMAGNETIC SUBSTANCE DIAMAGNETIC SUBSTANCE1. A paramagnetic substance is feebly attracted by magnet.1. A diamagnetic substance is feebly repelled by a magnet.2. For a paramagnetic substance, the intensity of magnetization has a small positive value.2. For a diamagnetic substance, the intensity of magnetism has a small negative value.ANS: An electromagnet consists of a core, made of a ferromagnetic material placed inside a solenoid. it behaves like a strong magnet when current flows through the solenoid and effectively loses its magnetism when the current is switched off.a. A permanent magnet is also made up of a ferromagnetic material but it retains its magnetism at room temperature for a long time after being magnetized one.b. Properties of material are as below:(i) High permeability(ii) Low permeability(iii) Low coercivityANS: Ferromagnetic substances as these substances have very high magnetic permeability.Properties (1) High retentivity (2) High susceptibilityANS: a. As the resistance (electrical of metal decreases with decrease in temperature) nut for diamagnetic substances, the variation of susceptibility is very small, i.e. diamagnetic material are unaffected by the change in temperature (except bismuth).b. Paramagnetic materials when cooled due to thermal agitation tendency alignment of magnetic dipoles decreases. Hence, they show greater magnetisation.ANS: Angle of dip, δ = 60° = π/3Horizontal component of the earth’s magnetic field, H =0.4 GEarth magnetic field (B?) =? Horizontal component of the earth’s magnetic field, H = B? cos δ B? = H/cos δ = 0.4 G /cos 60° = 0.4 G /? = 0.8 GTherefore, B? = 0.8 GANS : a. Susceptibility for diamagnetic material it is independent of magnetic field and temperature (except for bismuth at low temperature)Susceptibility for ferromagnetic material The susceptibility of ferromagnetic materials decreases steadily with increase in temperature. At the Curie temperature, the ferromagnetic materials become paramagnetic.b. Behavior in non uniform magnetic field Diamagnets are feebly repelled ,whereas ferromagnets are strongly attracted by non uniform field, i.e. diamagnets move in the direction of decreasing field, whereas ferromagnet feels force in the direction of increasing field intensity.ANS : (a) Two characteristics of material used for making permanent magnets are (i) high coercivity (ii) high retentivity (b) Core of an electromagnet made of ferromagnetic material because of its (i) low coercvity (ii) low hysterisis loss (ii) low retentivityANS:As vertical and horizontal components of magnetic fields are perpendicular to each other,when their magnitudes are equal,resultant will divide their angle equally.According to the question,H =√3VWhere, H and V are the horizontal and vertical components of the earth’s magnetic field. if angle of dip at the place is δ, thenTan δ = V/H = V/√3VTan δ = 1/√3δ = π/6Horizontal component of the earth’s magnetic field,H =B? cos δB? =Earth’s magnetic field H/B? = cos δ= cos π/6=√3/2H : B? = √3 :2ANS:BH / B = 1/√2And cos δ = BH /B Cos δ = 1/√2i.e. Angle of dip ,δ = 45°ANS: (i) Behaviour of magnetic field; lines when diamagnetic substance is placed in an external field.Behaviour of magnetic field lines when paramagnetic substances is placed in a external fieldBehaviour of magnetic field lines when paramagnetic substances is placed in a external fieldMagnetic Susceptibility: distinguishes the behaviour of the field lines due to diamagnetic and paramagnetic substances.ANS: Permanent magnet can be made by putting a steel rod inside the solenoid and a strong current is allowed to pass through solenoid. The strong magnetic field inside the solenoid magnetise the rod.For making permanent magnets, the material used should have (I) high residual magnetism, and (II) high coercitivity.ANS: Paramagnetic Substance : Aluminium , SodiumDiamagnetic Substance : Bismuth , Copper , the susceptibility of the diamagnetic materials is small and negative , i.e. -1 <?m < 0, whereas for paramagnetic substance the susceptibility is small and positive, i.e. 0 < χm < a, where ‘ a’ is a small number. The slope of I-H curve gives the magnetic susceptibility,?m =I /HMATERIAL A: paramagnetic substance (χm is small but positive)MATERIAL B: Ferromagnetic substance (χm is larger and positive)The susceptibility of material B is larger than that of A in a given magnetic field because ferromagnetic material gets strongly magnetised and hence, produces larger intensity of magnetisation in a comparison to paramagnetic substance, therefore it is strongly magnetised.ANS: Material A : Paramagnetic substance (χm is small and positive )Material B : Diamagnetic substance (χm is small but negative)The susceptibility of a diamagnetic substance is independent of temperature and megnetising field. magnetic susceptibility (χm) Temperature (-ve)This is the graph for variation of χm with temperature ( T ) of diamagnetic substance for material B.ANS: (1) Material A is paramagnetic and B is ferromagnetic. (2)The magnetic susceptibility (χm) of paramagnetic substance (i.e. material A) varies inversely with the absolute temperature.ANS : Magnetic susceptibility The magnetic susecptibility (?m) of a magnetic material is equal to the ratio of intensity of magnetisation and magnetising field i.e.χm = I/HWhere,I = intensity of magnetisation H =magnetising fieldIt has no unit.Positive magnetic susceptibility possesses by paramagnetic substances, e.g. aluminium, sodium.Negative magnetic susceptibility possesses by diamagnetic substance, e.g. bismuth, copper.Negative susceptibility shows that substances gets magnetised in a direction opposite to the direction of magnetising field.ANS: Bismuth, diamagnetic material gets feebly intensity of magnetisation with the applied magnetic field intensity in a direction opposite to it.Intensity of Magnetising field AMagnetisation Curve For BiThe diamagnetic material exhibits perfect conductivity (superconductivity) and perfect diamagnetism when metal is cooled below the critical temperature of the material (Meissner effect)THREE MARKS QUESTIONS32. A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120rev/min in a plane normal to the horizontal component of the earth’s magnetic field. The earth’s magnetic field at the place is 0.4 G and the angle of dip is 60°. Calculate the emf induced between the 3txle and the rim of wheel. How will the value of the emf be affected if the number of spokes were increased?33. Three identical specimens of a magnetic materials nickel, antimony and aluminium are kept in a non-uniform magnetic field. Draw the modification in the field lines in each case. Justify your answer.34. (1) What happens when a diamagnetic substance is placed in a varying magnetic field?(2)Which of the substances are diamagnetic?Bi, Al, Na, Cu, Ca and Ni35. (1) How does angle of dip change as line goes from magnetic pole to magnetic equator of the earth?(2) A uniform magnetic field gets modified as shown in fig .when two specimens X and Y are placed in it. Identify whether specimens X and Y are diamagnetic, paramagnetic of ferromagnetic.36. When two materials are placed in an external magnetic field, the behaviour of magnetic field lines is as shown in the fig. Identify the magnetic nature of these two materials.37. Name the three elements required to specify the earth’s magnetic field at a given place. Draw a labeled diagram to define these elements. Explain briefly how these elements are determining to find out the magnetic field at a given place on the surface of the earth.38. If χm stands for the magnetic susceptibility of a given material, identify the class of materials for whicha. -1 < χm < 0b.0 < χm < ? (where ? stands for a small positive number)(I) Write the range of relative magnetic permeability of the materials.(II) Draw the pattern of magnetic field lines when the materials are placed in external magnetic field.39. Name and define the elements of the earth’s magnetic field other than the horizontal component of the earth’s magnetic field. Why do we say that at the place like Mumbai and Delhi, a magnetic needle shows the true north direction quite accurately as compared to other places in India.ANSWER OF THREE MARKS QUESTIONSANS: Horizontal component,H = B cos θ = 0.4 cos 60° = 0.4 * ? = 0.2 G H = 0.2 * 10?? TThe wheel is rotating in a plane normal to the horizontal component, so it will cut the horizontal component only, Vertical component of earth will contribute nothing in emf.Thus, the emf induced is given as e = ? H ω L?,Where ω = 2πN/t andL = length of the spoke =50 cm =0.5 mE = ? * 0.2 * 10?? * (0.5)? * 2 *3.14 * 120 /60E = 3.14 *10??VThe value of emf induced is independent of the number of spokes as the emf’s across the spokes are in parallel. So, the emf will be unaffected with the increase in spokes.ANS: The modifications are shown in the figure.It happens because(1) Nickel is a ferromagnetic substance.(2) Antimony is a diamagnetic substance(3) Aluminium is a paramagnetic substance.ANS : (1) When diamagnetic substance is placed in a varying magnetic field, it tends to move from stronger magnetic field to weaker magnetic field (2) Bi and Cu.ANS(1) The angle of dip decreases from 90° to 0°.(2) For paramagnetic material, no magnetic lines of force enter in it. So the specimen X is paramagnetic. For ferromagnetic materials, all magnetic lines of force prefer to go through it. So specimen Y is ferromagnetic.ANS:(1)Material X is paramagnetic substance. when a specimen of a paramagnetic substance is placed in a magnetizing field, the lines of force prefer to pass through the specimen rather than through air. Thus magnetic induction inside the sample is more than the magnetic intensity.(2)Material Y is a ferromagnetic substance. These are the substance in which a strong magnetism is reduced in the same direction as the applied magnetic field, these are strongly attracted by a magnet.37. ANS: These are three elements determining earth’s magnetic field at any point of the earth.Magnetic declination.Magnetic dip.Horizontal component of earth’s magnetic field.Angle between geographical meridian and the magnetic meridian at any point is known as the magnetic declination at that point. Magnetic dip is the angle between the direction of earth’s magnetic field and the horizontal direction along the magnetic meridian. ANS: a. -1 < χm < 0 = Diamagnetic material0 < χm < ? =Paramagnetic substance,Relative permeability of diamagnetic materialμ = B/H where, 0 < μ < 1For paramagnetic substance ,μ> 1but μ is not very largeANS: The two elements of earth’s magnetism other than horizontal components of magnetic field areMagnetic Declination: The angle between the magnetic meridian and geographical meridian is known as the angle of declination at a given place of earth.Angle of dip: The angle made by resultant of the earth’s magnetic field with the horizontal in magnetic meridian is known as angle of dip at the given place of earth.The angle of dip is 90° at poles and 0°4 NE at Delhi and 0.58° NW at Mumbai. Thus, at both of these laces a magnetic needle shows the true north quite accurately. ................
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