ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Moving Charges and Magnetism Fastracl« Revision » Magnetic Field: Just as static charge produces an electric field, 2 moving charge or, current through conductor produces a magnetic field (8) > Itis the space around a conductor carrying current or the pace around a magnet In hich itsmagnetc effect canbe felt. > Magnetic field is also called as magnetic induction or magnetic Fiyx dens. > Si unit of B is Tesla (1) or weber/metre® and CGS unit Is gauss. 4 1 Tesla = 108 gauss > The direction oF 8 is determined by Fleming's left hand rule. > Magnetic Field due to Moving Charges (Oersted's experiment): Oersted's experiment shows the relation between the magnetic field and electric current. A electric current means moving charges, Oersted also concluded that moving charges also produce magnetic field in their surroundings. > Magnetic Force: It 's experienced by a single charge particle qmoving with velocity vin uniform magnetic Field at an angle. F=qv8sind or IF [=alvx8| > Fleming's Left Hand Rule: It states that if we stretch the first finger (Fore finger), the middle finger and thumb of left hand are mutually perpendicular to each other such that the first finger points in the directlon oF magnetic field and second finger shows the direction of electric current then the thumb represents the direction of Force experienced by the charged particle Lorentz Force: The Force experienced bya chargedpartile moving ina space where both electric and magnetic Fields exist scaled Lorentz Force. > Magnetic Force on a Current Carrying Conductor: ‘The magnetic force experienced by the current carrying conductor placed in uniform magnetic Field is F =16lsin0 then F=0 and iF 0 ie, force is maximum. » Motion in a Magnetic Field: The particle will describe a ircle if vand Bare perpendicular to each other. IF velocky has @ component along 8, the particle will produce helical motion 0, then F= 184 > Blot-Savart Law: According to BlotSavart’s law, the magnitude of the magnetic field is proportional to the current /, the element length and inversely proportional to the square of the distance where, to = permeability of free space and fig =4mx1077 Wb/Am > Magnetic Field on the Axis of a Current Carrying Circular Loop iol TRE > Magnetic Field at the Centre of the Coll 3 oN = ook » Similarities and _Dis-similarities Biot-Savart’s Law and Coulomb's Law > Similarities {) Both the laws are for long range, since in both the laws, the field at a point varies inversely as the square of distance from the source to point of observation. (il) Both the fields obey superposition principle. between the > Dis-similarities () The electrostatic field is produced by a scalar source, Le,’q’ and the magnetic field s produced byavectorsource di” (i) The electrostatic field Is acting along the displacement vector. The magnetic eld isacting perpendicular tothe plane i, along direction of idtxe, (il) Coulomb's law is independent of angle, whereas the Biot-Savart’s laws angle dependent » Relation between jlo, 59 and ¢ wl) Ga)” > Ampere's Circuital Law: It states that the line integral of magnetic field around a closed path is equal to pig times the total current /threading the closed path. 1a Gxt? c? oho §B-dl ug!When there is a system with a symmetry such as for a straight infinite current carrying wire, the Ampere's law enables an easy evaluation ofthe magnetic field The magnetic Field at a distance r outside the wire Is tangential and given by the Following Formula: bo! nr > Solenoid: A solenoid consists of an insulated long wire closely wound in the form of a helix. its lengthis very large as compared toits diameter. » Magnetic Field due to a Straight Solenoid B= 100! » Force between Two Parallel Current Carrying Conductors: Magnetic Force per unt length between two stralght parallel current carrying conductors is Ho faly feb ond d =distance between two conductors {and current of conductors aand brespectively The above expression is used to define the ampere (A), Which Is one of the seven Si base units. The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section and placed cone metre apart in vacuum, would produce on each of these conductors a Force equal to 2 = 10-7 newtons per metre of lencth, where, current carrying loop placed in uniform magnetic field Bis = NIBASin® Further, + © mB sin, where m Is the magnetic dipole moment of the current loop. » Current Loop as a Magnetic Dipole: A current loop or 3 solenoid or a coll behaves like a bar magnet. The magnetic dipole moment of current loop is determined by the Following Formula MIA where, M= magnetic dipole moment Jecurrent ‘A= area of current loop The direction of dipole moment can be determined by right hand thumb rule. The SI unit of dipole moment Is A-m? and dimension Is (av?) » Moving Coll Galvanometer: Moving coll galanometer isan instrument used for detection and measurement of small electric currents Principe: s based onthe principle that when a current, cartyng cols placed in 2 magnetlc ld, experiences 2 vorque leo Nag (Ae)! where, 4 = angle of deflection and of the sping > Current Sensitivity: It Is defined as the deflection produced inthe galvanometer when a unit current Flows through it. torsional constant » Voltage Sensitivity > A galvanometer can be converted into an ammeter by using a low resistance wire in parallel with the galvanometer (R= 0} Ammeter Now, effective resistance of ammeter will be vad mG Ss os Gs > A galvanometer can be converted into a voltmeter by connecting 3 high resistance wire in series with the galvanometer (R= =). Valimeier Effective resistance of voltmeter will be Ry= G+ R. Practice Exercise - Multiple choice Questions y QL An electron is moving along positive X-axis in a magnetic field which is parallel to the positive Y-axis. In what direction will the magnetic force be acting on the electron? a. Along ~ X-axls b. Along - Z-axis Along + Zaxis d. Along - Y-axis, (case sgp 2023-24) - ) » Torque on a Current Loop: The torque experienced by a ee Q2. Which one of the following is not correct about Lorentz force? aa In presence of electric field € (r) and magnetic field 6 (P) the force on a moving electric charge is, P=) +¥ x80). b.The force due to magnetic field on a negative charge Is opposite to that on a positive charge.ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Qa. a4. Qs. Q6. c. The force due to magnetic field becomes zero IF velocity and magnetic field are parallel or antl parallel d. For a static charge the magnetic force is Which of the following graphs correctly represents the variation of the magnitude of the magnetic field outside a straight infinite current carrying wire of radius ‘a’, as a function of distance 'r from the centre ofthe wire? (cosea0as) Ep a 2 5 oli S, a En a ef ah i? ost ‘Two horizontal thin long parallel wires separated by a distance r carry current / each in the opposite directions. The net magnetic field at a point midway between them will be: (cos€2023) a. zero wd o. ($57) vertically downward « (222) ersaty vara 4. (22) vera ownerd Three infinitely tong parallel straight current carrying wires A,B and C are kept at equal distance from each other as shown in the figure. The wire C experiences net force F. What will be the net force ‘on wire C, when the current in wire A is reversed: (CBSE SQP 2021 Term-1) A 8 c br ck 4.26 Biot-Savart’s law indicates that the moving electrons (velocity yj) produce a magnetic field B such that (NCERT EXEMPLAR) asiv b. Biv a Zero Q7. Qe. Qo. Quo. qu que. qua. «. itobeys inverse cube law 4. it along the Une jaining the electron and polit of observation ‘An element of 0.05i m is placed at the origin as shown in figure which carries a large current of 10 A. The magnetic field at a distance of 1 m in perpendicular direction is: Pp? 1m x= 0.05 im a 45 x108T b.55x10%T © 50x10°T 6. 75x 108 T The magnitude of the magnetic field of a tong straight wire carrying a current of 1 Ata distance of 2 emis: awst pw’t c2no7T d 20%T Acurrent loop in a magnetic field: 2a. experiences a torque whether the field Is uniform €or nonuniform in all orlentatlons. b. can be in equilibrium in one orientation. can be equilibrium in two orientations, both the equilibrium states are unstable d. can be in equilibrium in two orientations, one stable while the other is unstable {A circular loop of radius 3 cm is having a current of 12.5 A. The magnitude of magnetic field at a distance of 4 cm on its axis I: a. 565x107 b. 527x107 © 654x107 T 6, 9.20 «10° T A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic. field 8. The work done to rotate the loop by 30° about an axis perpendicular to its plane is: (CERT EXERIPLAR) MB, z ‘Acurrent carrying circular loop of radius R is placed in the X-¥ plane with centre at the origin. Half of the loop with x> 0 Is now bent so that it now ties di zero Blo ane» sai in the ¥-Z plane. (NCERT EXEMPLAR) a. The magnitude of magnetic moment now diminishes. b. The magnetic moment does nat change c, The magnitude of B at (0. 0, z). z>> R increases. d. The magnitude of 8 at (0.0. 2).2>>R Is unchanged. ‘Ampere’scircuital law is given by: a GH dls talane b. §B-dl=glene © fii-dl=yg 4. §B-dl=ng!ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Q14. Along straight wire in the horizontal plane carries a current of 75 A in north to south direction, magnitude and direction of field B at a point 5 m east of the wire is: a 4x10 T vertical up b. 5x10 Tyertical down c 5x10 T.vertical up d. 4107. vertical down 15. An electron is projected with uniform velocity along the axis of a current carrying long solenoid. ‘Which of the following is true? (CERTEXEMPLAR) a. The electron will be accelerated atong the axis, b. The electron path will be circular about the axis. . The electron will experience a force at 45° to the axis and hence execute a helical path d. The electran will continue to move with uniform velocity along the axis of the solenoid. Q16. Asolenoid coil of 200 turns/m is carrying a current of 3 A. The length of the solenoid is 0.2 m and has a diameter of 1 cm. The magnitude of the magnetic field inside the solenoid is: a. 12e x10 T b. 2ax «1ST © Wax 10 T 4. 48n x 10° T QY7. An air-cored solenoid with length 30 cm, area of cross-section 25 cm? and number of turns 800, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10° s. ignoring the variation in magnetic field near the ends of the solenoid, the average back emf induced across the ends of the open switch in the circuit would be: a. zero b. 3125 654 0.1674 18. The two parallel conductors carry equal currents in the same direction. What is the nature of the force acting between them? a, Repulsive «. Cannot predict , Attractive 4. No force Q19. A small circular flexible loop of wire of radius r carries a current /.It is placed in a uniform magnetic field 8. The tension in the loop will be doubled if: a. Mis doubled b. Bis halved rls doubled d. Both Band |are doubled Q20. Two concentric and coplanar circular loops P and Q have their radii in the ratio 2 : 3. Loop Q carries a current 9 A in the anticlockwise direction. For the magnetic field to be zero at the common centre, loop P must carry: (cose 50p 2022.23) a. 3 Ain clockwise direction b. 9.Ain clockwise direction c GAin anti-clockwise direction d. 6 Ain the clackwise direction ga. Qe. Q23. gaa. Qa. 26. 27. Qe. ‘A circular coil of 25 turns and radius 12 em is placed in a uniform magnetic field of 0.5 T narmal to the plane of the coil. f the current in the coil is 6 A, then total torque acting on the coil is: a zero b34Nm .38Nm d44Nm ‘An ammeter of resistance 0.81 ohm reads up to 4/A.The value of the required shunt to increase the range to 10 Ais: (case sap 2023-24) a. 09 ohm b. 0.03 chm ©.0.03 chm 1.03 ohm Which ofthe following expressions is applicable to the moving coll galvanometar? 2. Fy=a(vx8) b. B= Bptano d, None of these In a moving coil galvanometer, the deflecting torque + acting on the coil is related to the current I flowing through it as: (cese 2023) arel beet? cred dead Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity is 20 div/V. ‘The resistance of the galvanometer is: a250 b2509 c 40a 4.5009 Agalvanometer of resistance 10.0 gives full-scale deflection when imA current passes through it.The resistance required to convert it into a voltmeter reading up to 2.5 Vis: A290 b29Q 2902 4, 249000 The coil of a moving coil galvanometer is wound over a metal frame in order to: (c8s€S0°2021 Term-1) a. reduce hysteresis b. increase sensitivity © Increase moment of inertia 4d. provide electromagnetic damping ‘The current sensitivity of a galvanometer increased by 20%. If its resistance also increases by 25%, the voltage sensitivity will: (case Sop 2021 Term-1) a, decrease by 1% b. increase by 5% © increase by 10% d decrease by 4% D Assertion & Reason type Questions w Directions (QNos. 29-37): in the following questions, @ statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: ‘a. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). b. Both Assertion (A) and Reason (R) are true but Reason (R) Is not the correct explanation of Assertion (A). . Assertion (A) Is true but Reason (A) Is False d. Both Assertion (A) and Reason (R) are false. elee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Qs. 30. Qa. Q32. 933. Q34. 3s. 936. Assertion (A): If an electron is not deflected while passing through a certain region of space, then only possibility is that there is no magnetic region. Reason (R): Force is inversely proportional to the magnetic field applied. Assertion (A): The magnetic field intensity at the centre of a circular coil carrying current changes, if the current through the coil is doubled. Reason (R): The magnetic field intensity is dependent on current in conductor. Assertion (A): Ampere's circuital law holds for steady currents which do not fluctuate with time. Reason (R): Ampere’ circuital law is similar to that of Biot-Savart’s law. ‘Assertion (A): Magnetic field due to current carrying solenoid is independent of its tength and cross-sectional area Reason (R): The magnetic field inside the solenoid is uniform. Assertion (A): The magnetic field at the ends of a verylong current carrying solenoid is half of that at the centre. Reason (R): If the solenoid is sufficiently Long, the field within itis uniform. Assertion (A): The deflecting torque on a current carrying loop Is zero when its planes perpendicular to the direction of magnetic field. Reason (R): The deflecting torque acting on a Loop of magnetic moment min a magnetic field B is iven by the dot product of m and B .(cBs€2023) Assertion (A): When current is represented by a straight line, the magnetic field will be circular. Reason (R): According to Fleming's left hand rule, direction of force Is parallel to the magnetic field. Assertion (A): When radius of a circular top carrying a steady current is doubled, its magnetic moment becomes four times. Reason (R): The magnetic moment of a circular loop carrying a steady current is proportional to the area of the loop. (c0se2023) Qa7. ‘Assertion (A): On increasing the current sensitivity of a galvanometer by increasing the number of turns, may not necessarily increase its voltage sensitivity. Reason (R): The resistance of the coil of galvanometer increases on increasing the number of turns, in the Blanks Type Questions Qa. ae, 43, gaa. 4s, 46. Answers] v (b) Along - Zaxis (¢) Fora static charge, the magnetic force is maximum. IF charge is not moving, then the magnetic force 1s zero. since Fx qv x8) Asi =0.for stationary charge f=0 ry \ © _ oF * 4 5 Electric current flows through a thick wire, Magnetic field at a point on its surface is (B= 1f/2nR) and is on its axis. 1. When a coil carrying current is set with its plane perpendicular to the direction of magnetic field, the torque on the col ‘An electron passes undeflected when passes through region with electric and magnetic fields. When electric field is switched off, its path will change to ‘A linear conductor carrying current is placed parallel to the direction of magnetic field, then it experiences. force When a magnetic dipole of moment M rotates freely about its axis from unstable equilibrium to stable equilibrium in a magnetic B, the rotational Kinetic energy gained by its. ‘Two linear parallel conductors carrying currents in the opposite direction. each other. Thepath ofchargedparticlemoving perpendicularly with Bis Torque on a current carrying rectangular coil inside galvanometer is maximum and constant orientation as it is suspended ‘magnetic field. inside. ‘To convert galvanometer into a voltmeter of given range, suitable high resistance should be connected n.. with the galvanometer. (@) (22) verticety downward {a} Zero, Let Fis force per unit length between Aand C. f Ha zhxt ax Or Let F; is Force per unit length between Band C. bet pall 2 aRee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET =F A ce Now net force on C’ per unit length is bg Fer, Hoh (ey 2 2" _p iver an (ver) Now fig A B c 2 1 Fj Bi 5 5 Ag " My 2 Fj=Repulsive force between A and C= 725— "=A reactive force between B and C 2 i rope HP Net force on (C.F For Net force on Cis ero, eis Magnetic field produced by charges moving with ¢ 5 (in) gi? velodiy Vat adistanceris 8-[2) «¥ Therefore B LV. (9) 50<10°T | Hellsing oa fats Here dl=4x=0.05 m.J=10A.r=1m sin0 = sin90° =1 ga 1077 2x008%1 a? =050% 107 =50x108T (a) 10-1 Here 1A.x=2em =0.02m igh Magnetic field, 8 = Ho " nx e107 t waster eer? eT M0107 =r 9. (d) can be in equilibrium In two orientations. one stable while the ather Is unstable 10. (@) 565 «10° T sd 2R? + 2p? Here, 1=125 A R=3.cm=3 10% x=4oms4c0? m 4x10"? x125x(3x102)? aeny 4x10F] = 5.65 «10° T ML. (d) zero Rotation of loop by 30° about an axis perpendicular to its plane does not change the angle between magnetic moment and magnetic field. Hence. no work Is dane, 12 (a) The magnitude of magnetic moment now diminishes, B. (0) fF dls igha 14, (¢) 510 T. vertical up From Ampere circultal lav, $8 dl hoon B+ 22 = Hole Holene 2nR -2007 8-5 x10°T ‘The direction afield a the given point vil be verticat Up determined by the screw rule or right hand rue. 1. (¢) The electron wl continue to move with uniform \elocty along the axis ofthe solenoid 16. (0) 2451077 Be ponl = es « 200 «3 = 2a 10 T 1 (16 74V Given length of solenoid 1= 30 cm ‘area of cross-section A= 25 cm? umber of turns N current Time dt =1075 ‘Magnetic field inside a sotenait. N, Bap HoT Flux linked with 'N turns, Initial Fux 9, = NBA = Nig w eng A HoT 4n x10" x800x800x25%25%10~ 030 =16.74x 107 Woee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Final flux, @=0 co _1674x10°3-0 at ww Averagebackemf,| 18. (b) Attractive TiP Polel cuments atvact ond ontprall current rept in tar 19. (a) lis doubled Magnetic mament = 1A When is doubled, Mis also doubled. Also, x= MB'sin0, when Mis doubled, torque Is also doubled, hence tension is doubled. 20. (d) 6 A in the clockwise direction We know that. |B = 184) Given, H/ry=2/2.h=9A Then, Wh= rt or kh 28 =6/A inthe clockwise direction 21 (2) zero The torque acting on the coil |c|=IMxel=mBsine Here the circular coll is placed normal to the direction of magnetic field. then the angle between the direction of magnetic moment (M) and magnetic field (6) Is zero, then MB sind = MB sind"= 0 r=0 22. (b) 0.03 ohm, [6 1x0.81 i=], 10-1 oo9a 23. () Come 24, (leet 25. (b) 2500 Resistance = Substituting the given values. we get 25v Re oA, 108 = 2500 2-10.09» 24902 (2) provide electromagnetic damping The coll of a moving coil galvanometer is wound over metallic frame to provide electromagnetic damping, soit becomes dead beat galvanometer. (@) decrease by 4% According to the question, Gi “o (24-25) 100 =sbx100=-4% B ‘Thus, the voltage sensitivity will decrease by 4%. (4) in this case. we cannot be sure about the absence of the magnetic field because ifthe electron moving parallel to the direction of magnetic field, the angle between velocity and applied magnetic field is zero (F=0), Then. also electron passes without deflection. Also F = eVB5INO => Fx 8 (a) The magnetic field at the centre of circular calls, given by, gato Zr! ano So if current through coil is doubled, then magnetic fields & =28.ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET The magnetic field also gets doubled. The magnetic field Is directly proportional to the current in conductor. 31. (b) Both Assertion (A) and Reason (R) are true bout Reason (R) is not the correct explanation of Assertion (A), 32. (b) Magnetic field due to a solenoid having n number of turns per metre and carrying current /s 8 = gal which Is independent of the length and area of cross-section of the solencia. ‘The magnetic field Inside the sotenold Is uniform, 33. (b) The magnetic field of a solenoid is given by, 1 8 = sngnfcos0, ~c0st,] For a long current carrying solenoid, the magnetic field at the ends of a very long solenoid Is glven by, 1 1 8 = « Magnetic eld at the centr Jon! = x Magnetic Feld atthe centre ( 6,90", 0)» 1607] 34, (o) When a plane of loop Is perpendicular to the direction of the magnetic field, the angle between the magnetic moment vector and the magnetic field vectar is 90°. In this case. the vector product of the two vectors becomes maximum in magnitude and deflecting torque becomes maximum. However. the dot product of two vectors becames zero. 35, (c) When current Is straight. it means the current is, passing through a straight conductor. the magnetic field produced due to current through a straight conductor Is In the form of concentric circular magnetic lines of force whose centres Ue on the linear conductor and are in a plane perpendicular to the plane of linear conductor. It means the magnetic fields circular. 36. (b) Magnetic dipole moment of the current Loop = Ampere turns x Area of the colt intially magnetic moment = x2 New magnetic moment Af = [n(21}2=4 in2= 44 Thus, magnetic moment becomes four times when radius Is doubled. 37. (a) When we increase current sensitivity by increasing ‘umber of turns. thenresistance of call sisoinereases So. Increasing current seneitity does not necessarily imply that voltage sensitivity vil increase because |, =n Mg a If Jy increase and R increase by different amounts. then Vj may increase or decrease. 38 zero 38. zero 40. circular 41. no 42. 2B 43. repel 4b, circular 45. radial 6. series “@ Case Study Based Questions Case Study 1 QL Qe. 3. a4. Q5. In 1820, A Danish physicist, Hans Christian Cersted, discovered that there is a relationship between electricity and magnetism. By setting up a compass through a wire carrying an electric current, Oersted showed that moving electrons can create a magnetic field. Oersted found that, for 2 straight wire carrying a steady current (DC), the magnetic field lines encircle the current-carrying wire. ‘The magnetic field lines lie ina plane perpendicular tothe wire. Ifthe direction of the currents reversed, the direction of the magnetic force reverses. The strength of the field is directly proportional to the magnitude of the current. The strength of the field at any point is inversely proportional to the distance af the point from the wire. ealieny Wee Tape Compass Read the given passage carefully and give the answer of the following questions: Who was the first to discover the relation between electric and magnetic fields? a. HC Oersted Charles Wiliam Oersted Charles Maxwell d. Andre Marie Ampere If magnitude of the current in the wire strength of magnetic field: 3. Increases b. decreases remains unchanged — d. None of these Which of the following statements is true? a. There Is no relationship between electricity and magnetism b, Anelectric current produces a magnetic Feld . Acompass is not affected by electricity 4. A compass is not affected by a magnet ‘A compass needle is placed below a straight conducting wire. If current is passing through the conducting wire from north to south, then the deflection of the compass is: a, towards west b. towards east . keeps oscillating in east-west direction 4. no deflection Charges at rest can produce: a static electric field —_b. magnetic field . induced current d. conventional current reases,ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET (@) HC. Oersted Hans Christian Oersted discovered that there is a relationship between electricity and magnetism. (2) increases Ba! (b) An electric current produces a magnetic Feld Magnetism is related to electricity according to Oersted (b) towards east {@) static electric field Charges at rest can produce static electric Feld Case Study 2 a. ge. A magnetic field can be produced by moving, charges or electric currents. The basic equation governing the magnetic field due to a current distribution is the Biot-Savart’s law. Finding the magnetic field resulting fom a current distribution involves the vector product and is inherently a calculus problem when the distance from the current (othe field point is continuously changing, According to this law, the magnetic field at a point due to a current element of length di'carrying current [,ata distance r from the element is _ bo Ha 1x P) np aB Biot-Savart’s law has certain similarities as well as difference with Coulomb's law for electrostatic field. e.g, There is an angle dependence in Biot- Savant’s law which is not present in electrostatic case Read the given passage carefully and give the answer of the following questions: The direction of magnetic field dB’ due to a current element dt’ at a point of distance r from it, when a current passes through a long conductor is in the direction: 2. of position vector” ofthe point bof current element dT” & perpendicular to both dt and” 4. perpendicular to dT” ‘The magnetic field due toa currentina straightwire segment of length L ata point on its perpendicular bisector at a distance r (r>>L): 1 a. decreases as 1 b. decreases as —> i decreases as —> d. approaches a finite limit as r—>0 Q3. Two Long straight wires are set parallel to each other. Each carries a current iin the same direction, and the separation between them is 2r. The intensity of the magnetic field midway between il, fl 4. A Long straight wire carries a current along the Z-axis for any two points in the x-y plane. Which of the following is always false? a The magnetic flelds are equal. b. The directions of the magnetic fields are the . The magnitudes of the magnetic fields are equal d. The field at one point is opposite to that at the other point QS. Biot-Savart's law can be expressed alternatively as: a. Coulomb's Law ¢ Ohms Law b. Ampere’s circuital law d, Gauss's Law —— Answers }-—— 1. (O) perpendicular to both dT” and i? According to Biot-Savart’s law. the magnetic Induction due to a current element is given by bpd tar pare Thus. itis perpendicular to bath dl” and 1 2 (b) decreases as — From Blot-Savart'’s law, 4, (a) The magnetic fields are equal 5. (b) Ampere’s circuital law Case Study 3 Moving coil galvanometer operates on Permanent Magnet Moving Coil (PMMC) mechanism and ‘was designed by the scientist d’Arsonval. Moving coil galvanometers are of two types: (i) Suspended coil (i) Pivoted coil type or tangent galvanometer.Softiron core Its working is based on the fact that when a ‘current carrying coil is placed in a magnetic field, it experiences a torque. This torque tends to rotate the coil about its axis of suspension in such a way that the magnetic flux passing through the coil is, maximum. Read the given passage carefully and give the answer of the following questions: ‘A moving coil galvanometer is an instrument which: 2. is used to measure emf bis used to measure potential difference c isused to measure resistance d. isa deflection instrument wihich gives a deflection when a current flows thraugh its coil To make the field radial in a moving coil ‘galvanometer: a number of turns of coils kept smal bb magnet is taken In the form of horse-shoe poles are of very strong magnets d. poles are cylindrically cut The deflection in a moving coil galvanometer is: a. directly proportional to torsional constant of spring b. directly proportional to the number of turns in the coil inversely proportional to the area of the coll d. inversely proportional to the current in the coil Ina moving coil galvanometer, a coil of N-turns of area A and carrying current J is placed in a radial field of strength 8 is: a. NA“ b. NAB «NABI d. NABI To increase the current sensitivity of a moving coil galvanometer, we should decrease: a, strength of magnet b. torsional constant of spring number of turns in coll d. area of coll - Answers * 1. (@) sadeftection instrurnent which glves.a deflection when a current flows through its coil Amoving coll galvanometer isa sensitive Instrument which is used to measure a deflection when a current flows through its coll QL gz. Qs. Q4 Qs. EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET ee 2. (d) poles are cylindrically cut Uniform field is made racial by cutting pole pieces cylindrically 3. (b) directly proportional to the number of turns in the call The deflection in @ moving coil galvanometer. nag ara or @ & N.where Nis number of turns in coil is magnetic field and As area of cross-section 4 (@) NABI The deflecting torque acting an the cols Xeahecen NAB 5. (b) torsional constant of spring Current sensitivity of galyanometer Nea oe é 1 Hence, to increase (current sensitivity) 5, (torsional constant of spring). k must be decreased Case Study 4 Ampere’s law gives a method to calculate the magnetic field due to given current distribution According to this, the circulation 8-0 of the resultant magnetic field along a closed plane curve is equal to iy times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant Ampere’s law is more useful under certain symmetrical conditions. Consider one such case of a long straight wire with circular cross-section (radius A) carrying current J uniformly distributed across this cross-section. Read the given passage carefully and give the auswer of the following questions: QL. What is the magnetic field at a radial distance r from the centre of the wire in the region r> R? Q2. What is the magnetic field at a distance r in the region r< R? Q3. A long straight wire of a circular cross-section (radius a) carries a steady current land the current is uniformly distributed across this cross-section. Plot the graph which represent the variation of magnitude of magnetic field 8 with distance r from the centre of the wire.Q4. A long straight wire of radius R carries a steady current /.The current i uniformly distributed across its cross-section. What Is the ratio of magnetic field at R/2 and 28? QS. If a long straight wire in the horizontal plane carries a current of 40 A, calculate the magnitude of the field 8 at a point 15 cm away from the wire. = Answers = 1. Magnetic feld due to a long current carrying wire at ¢ is p-to! 2 Let "be the current in the region r< R Then. 3. Magnetic field due to a long straight wire of radius o carrying current lat a point distant r from the centre of the wire Is given as Fallows: igh ar a2 torrno 2na forr0 2ar ‘The variation of magnetic field B with distance r from the centre of wire is shown in the figure. 4 Let the magnetic Fields due to a lang straight wire of radius R carrying a steady current lat a distance r from the centre of the wire are at Bye riae lforrR) R So, the magnetic field at r=2 is e Z oe (A) oh Dept 2)4ak Hol Hol (2k) aR and atr=2R B, EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET ee Thelr corresponding ratio is ight Le t= nl 8,” ugl /4eR) By Be 5. Given that 1 <40A r= 15em =15 «10% m Ho! _ 4x 1077 40 _ a0 10° 2nr nx 15x10 5 = 5.34% 10°T Case Study 5 A solenoid is a long coil of wire tightly wound in the helical form. Solenoid consists of closely stacked rings electrically insulated from each other wrapped around a non-conducting cylinder. Figure below shows the magnetic field lines of 2 solenoid carrying a steady current /. We see that if the tums are closely spaced, the resulting magnetic field inside the solenoid becomes fairly uniform, provided that the length of the solenoid is much greater than its diameter. For an ideal solenoid, which is infinitely long with tums tightly packed, the magnetic field inside the solenoid is uniform ‘and parallel to the axis, and vanishes outside the solenoid. Read the given passage carefully and give the answer of the following questions: ‘A long solenoid has 800 turns per metre length of solenoid. A current of 1.6 A flows through Calculate the magnetic induction at the end of the solenoid on its axis. What is the nature of magnetic field lines passing through the centre of current carrying solenoid? Whatis the magnetic field (B) inside. Long solenoid having 1 turns per unit length and carrying current J when iron core is kept in it (jip = permeability of vacuum, 7 = magnetic susceptibility)? 4. A solenoid of length { and having 1 turns carries a current / in anti-clockwise direction. What is the magnetic field? QL Qe. 93.ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Answers Hohl _ (dxx10"7) x80 «16 2 2 LAs B ex1o4T 2. Magnetic Field tines atthe centre ofthe solenoid are straight lines 2s magnetic Feld inside a solenoid is uniform, 3. Magnetic Field Inside 2 tong sotenald with an lon core inside tis B= unt aut weg) Bo yf ag)nt 4, A solenoid of length | and having a turns carries a current in anti-clockwise direction, The magnetic feld ts HOM" ns arection wil be along the axis of solenoid. Very Short Answer tupe Questions QL Write the relation for the force acting on a charged particle q moving with velocity in the presence of a magnetic field B. (case2019) ‘Ans. When a charged particle qmoves with velocity vin a uniform magnetic 8. then the fore acting on itiagven by Feqlvx8) Q2. Write the expression, in vector form, for the Lorentz magnetic force F_ due toa charge moving with velocity v in a magnetic eld B . What is the direction of the magnetic force? (case 2016) ‘Ans. Expression is F= q (v x8). The direction af magnetic farce is perpendicular to the plane containing velocity and magnetic field vectors Q3. Under what condition is the force acting on a charge moving through a uniform magnetic field ‘minimum? Ans. When it moves parallel or anti-parallel to the direction of magnetic fet. gTiP Swhen 0=0°or@=1B0%F =D= Fix, Q4. Under what condition is the force acting on a charge (or an electron) moving through a uniform magnetic feld maximum? Ans. When it moves perpendicular to the direction of magnetic Fel g@TiP— — F=Gqvsing = when 9=90°F =F, Qs. ‘Ans. Qo. Ans. Q7. ‘Ans. 08 Ans. Q9. Ans, Quo. Ans. Qu. Ans, State the condition under which a charged particle moving with velocity ¥ goes undeflected in 2 magnetic field B (c0se2017) Fy -alve8) The charge will go undeflected when Fj, =0. ie. if V Is parallel or antl-parallel to 8, e. elther @=0° or O~ 180° When a charge q 1s moving in the presence of electric (E) and magnetic fields (B) which are perpendicular to each other and also perpendicular to the velocity v of the particle, write the relation expressing () in terms of (E) and (8). F. Fiorente = Fetwciie + Fmognetic »g& +q(v«B)=q(E »(VxB)) The force between the electric and magnetic Feld is opposite Now if we adjust the force such thatthe force due to electric and magnetic fleld are same. then the particle moves undeflected = qe € = vad 6 Write the mathematical form of Ampere-Maxwell circuital law. (c8s€2020) Or State Ampere'scircuital law. (cose2016) Ampere’s circuttal law states that the line integral of the magnetic field, around a closed loop. equals ug times the total current passing through the surface enclosed by that loop. tg! What do you mean by solenoid? Asolenold consists of an insulating long wire closely ‘would in the form of a helix ts length is very large a8 compared toits diameter. Write the formula for magnetic field due to a solenoid. Magnetic Held due toa solenold is glven by B=hon! ie, Bal ‘where, n= number of turns per unit length vig = permeability of free space. Define moving coil galvanometer. Moving coll galvanometer Is a device used to detect the small electrical current in the circuit Current sensitivity and voltage sensitivity of galvanometer depend on which parameters? The current sensitivity and voltage sensitivity of galvanometer depend on number of turns of coll, magnetic field, area of the coil and torsion constant Kof the spring or suspension wireee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET ~ Short Answer type-I Questions y QL Two tong straight parallel wires A and B separated by a distance d, carry equal current / flowing in same direction as shown in the figure. é ()) Find the magnetic field at a point P situated between them at a distance x from one wire. (li) Show graphically the variation of the magnetic field with distance xfor0 M= Nine? (i) According to Biot-Savarts law. y higtdixr ax gl Gar? a8, components due to diametrically opposite components cancel out. Only 48, components refrain. b= a= onde, oR ton ar eexon Q3. Two identical circular Loops (1) and (2) of radius R and carrying the same current are kept in Perpendicular planes such that they have a common centre at P as shown in the figure. Find the magnitude and direction of the net magnetic field at the point P due to the loops.ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Sol According to the question, igh? WR P= aH = 0- (aR +eF 02 8, tano= 2-4 &, => q= 45" with either 8 or 8 Q4. Using Biot-Savart’s law, deduce the expression for the magnetic field at a point (x) on the axis of a circular current carrying loop of radius R. How is the direction of the magnetic field determined at this point? (c8S€2023, 17, 16) Sol Magnetic field due to a current carrying loop at a point on its axis: According to Biot-Savert’s law. the magnetic field at Pdue to current element Jdl_ at C. big! dlsin 90° oa 2 ato Ia Ong ! dosing st eae @ Resolving d8 into horizontal and vertical components, resultant magnetic field at F ip Lal Be fdbsino= Nox zone 1 = Haein Tsing a gi? Rep? rave) Qs. Sol. 0) Qe. Sol, Fora coll of N turns, ofl? Re x2 P? Direction of the magnetic field at this point Pcan be determined by the right hand thumb rule. The given figure shows a long straight wire of a circular cross-section (radius a) carrying steady current /. The current 1 is uniformly distributed across this cross-section. Calculate the magnetic field in the region ra. (case sop 2023-24) e ou 1 ® (Consider the case ro. The Amperian loop. labelled 2, Is a circle concentric with the cross- section. For this loop. L=2xr Using Ampere circultal law. we can write, Hol ‘nr Consider the case r< 0. The Amperian loop is circle labelled 1.For this toop. taking the radius of the circle tober = L=2nr Now the current enclosed fis not J but Is less than this value. Since the current distribution is uniform, the current enclosed is, xP) ie xo") go? f Using Ampere’ law B(2ar) eng B(2n)= pL B = Bat (r>0) ! 6 (4) Ber {r<0) Two infinitely long straight wires A, and A, carrying currents / and 2/ flowing in the same directions are kept d distance apart. Where should a third stralght wire A; carrying current 1.5/ be placed between A, and A;, so that it experiences no net force due to Ay and A;? Does the net force acting on As depend on the current flowing through it? (case2019) Let the current in the third wire A, be in same direction as that of A, and Ay. So, It vill experience attractive force due to both 151 Fait] Fa Ay hye t+ 1ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Q7. Sal The force on A; due to A, is Hg. PxLSbxl “Rx x Fy where, ls unit length of conductor wire A, and x= distance between A, and Ay Similarly, force on A3 due to Az is olin 15be2txl 2° 2n (d-x) According ta question Fy, = 4 Hg SPL tg 3 2x x 2n(d-x) 63 ~ x d-x = da-x02% = soa 3 Yes. the net force acting on As depends on the current flowing through it Two long straight parallel current carrying conductors are kept ‘a’ distant apart in air. The direction of current in both the conductors is same Find the magnitude of force per unit length and direction of the force between them. Hence define ‘one ampere. (CBSE Sop 2023-24) The magnetic induction 8, set up by the current | flowing in first conductar at a point somewhere in the middle of second conductor is Hoh hse ) ® Bi 2 eho eee ' e mR, The magnetic force acting on the portion P,Q: of length |; of second conductor is f= (8, sn 30" 2) From eqs. {I) and (2), 7 Palle towards first conductor Fgh Stok 5 1 ano 8) The magnetic induction 8, set up by the current f flowing in second conductor at point somewhere in the middle of frst conductor is Hole (4) 2° 2na The magnetic force acting on the portion P,Q, of length ly of first conductor is F.=hhB, sin90" 45) From eqs. (3) and (5). Hol toh Fatah 2x0 tole (6) y 2n0 ‘The standard definition of 14 Ifthe hala h=b=Im: then Fifi Hox) > grim G& Qnxa ‘Thus one ampere is that electric current which when flows in each one of the two infinitely long straight Barallel_conductors placed 1m apart In vacuum causes each one of them to experience a force of 2107 Nim. Q8. (i) State the underlying principle of a moving coil galvanometer. (ii) Give two reasonsto explain whya galvanometer cannot as such be used to measure the value of the current in a given circuit. (ili) Define the terms: (a) voltage sensitivity and (b) current sensitivity of a galvanometer. (c8SE2020, 19) Sol. (i) Working Principle: if 2 current carrying cal is, freely suspended or pivoted in a uniform magnetic field, it experiences a deflecting torque As pivoted coil is placed in a radial magnetic field, hence on passing current Jthrough it a deflecting torque acts on the coll which is given by vo NAB ‘where. N= total number of turns in the coll. ‘A-area of the coll. = magnetic field (ii) Reasons: (2) Galvanometer is a very sensitive device and gives a full scale deflection for a current of the oder of a few uA Hence. ft cannot be ‘Used to measure current. (b) Resistance of galvanometer isnot very small hence it will change the value of current In the edu Branch shen connected Wr se that branch i) (a) Voltage sensitivity: It is defined as the deflection produced in the galvanometer when unit voltage Is applied across the col of the galvanometer. @., [NBA SUK where, R= resistance of the coll (b) Current sensitivity: It is defined as the deflection produced per unit current in the galvanometer when , .@_ (NEA) radian STK ampere" towards second conductor Am in V/A x radian vot J Ry>G (always) (i) Conversion of Galvanometer into Ammeter: A galvanometer_is converted Into_an_ammeter_by connecting 2 very small resistance (called shunt resistance) in parallel with it <= ey Amineter IyxG =1,) Effective resistance of ammeter 11d 56 G > te (-I))x = Pa< G (always) ‘ Long Answer type Questions y QL State Biot-Savart’s law and give the mathematical expression for it. Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis. How does a circular loop carrying current behave as a magnet? Sol. According to Biot Savart’s law, the magnetic field due to small current carrying element dl at any nearby point Pis given by big fl sine an 2 ig leloer an ep a. or 8 where, Hy /4x © 10-7 Tm/A=107H/m v It x Here Hg © permeability of free space or vacuum and ‘re distance af point P from current carrying element, Its direction is given by right hand thumb rule. ‘Magnetic field due to a circular coll carrying current ata point along its axis: Let us consider a circular loop. of radius r with centre C Let the plane of the coll be perpendicular to the plane of the paper and current The flawing in the direction as shawn in the figure. Suppose P is any point on the axis at a direction x from the centre. Now. consider a current element fl on top (L) where current comes out of paper normally, whereas at bottom (i) enters into the plane of paper normally. LP Lit Atco, MP Ia rempavy + ‘The magnetic field at polnt P due to current element lat: According to Biot-Savart’s law, ‘where, r= radius of circular loop. x= distance of point P from centre alang the axis. The direction of d8 Is perpendicular to LP and along PQ where PQ. LP. Similarly, the same magnitude of ‘magnetic field is obtained due to current element lal at the bottom and direction is along PQ. where, POLMP. Now. resolving df due to current element at L and 'M dB cos components balance each other and net magnetic field is given by a gaasing«90( 4) 2 tear) eae InaPcesingee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET ge. Ans. lig fo area For Nturns, B= AAs current carrying loop has the magnetic field lines around it which exerts a force on a maving charge. Thus, it behaves as a magnet with two mutually opposite poles. QO ‘The anti-clockwise flow of current behaves Uke 2 North pole. whereas clockwise flow as South pole. Hence. loop behaves as a magnet. Derive an expression for the force per unit length between the two infinitely long straight parallel current carrying conductors. Hence define SI unit of current. (case 2016, 15) Magnetic field due to conductor ‘' at any point on conductar'2' is By right hand rule, 8, will act perpendicular to canductar 2’ and into the plane of the paper. ue to this magnetic field, farce an length (of wire 2's (2 Je bghht 2nr Similarly, force on length Lof wire ‘Tis 8), lsin90" iol J, _Boh’ot facBion att S Hohlal og Q3. Ans. Hence. force per unit length F Hohl fet ottt 0 By Fleming's left hand rule, Fy will act towards conductor, ‘1! and Fy. will act towards conductor ‘2. Obviously the two conductors will attract each other. If the currents are in opposite directions, then there will be repulsion between the two conductors. The Sl unit of the current is ampere One ampere is the current which when flowing through each of the two infinite long straight parallel conductors placed one metre apart from the each other in free space will exert a force of 2x 1077 N per metre of their length State the principle of working of a galvanometer. Agalvanometer of resistance Gis converted into 2 voltmeter to measure up to Vvolts by connecting a resistance R, in series with the coll. Ifa resistance Ry is connected in series with it,then it can measure up to ¥/2 volts. Find the resistance, in terms of Ry ‘and R required to be connected to convert it into ‘voltmeter that can read up to 2 V. Also, find the resistance G of the galvanometer in terms of Ry and R,. Principle of Galvanometer: The principle of moving coll galvanometer is based on the fact that when a current carrying coil is placed in a magnetic Field, it experiences 3 torque A high resistance is connected in series with the galvanometer to convert into voltmeter. The value ofthe restanceispen by Ra where, V = potential difference across the terminals of the voltmeter. J,= current through the galvanometer and G = resistance of the galvanometer When resistance Ris connected in series with the galvanometer, then Ay =(V/l,)-6 0) When resistance R) is connected in series with the v jalvanometer. then Ry =—-6. 2) 8 te 2a, (2) From eqs. (I) and (2), we get R\-R=V/ 2), and G=R,-28, ‘The resistance R, required to convert the given galvanometer Into voltmeter of range 0 to 2V is iven by ® Ry =(V/|,)-6 = Ry © A(R, Rp) ~(R, ~2Rp) 3R,—2R, Gin terms of Rand Ris given by G =A,-2R,ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET 4. (i) Write the principle and explain the working of ‘a moving coil galvanometer. A galvanometer as. such cannot be used to measure the current in a circuit. Why? (ii) Why is the magnetic field made radial in a moving coil galvanometer? How Is it achieved? (case 2023) ‘Ans. (i) Principle of Galvanometer: A curvent carrying coil placed in a magnetic field experiences a torque. the magnitude of which depends on the strength of current, Working of Galvanometer: Consider a single turn of the rectangular coll PORS whose length be ( and breadth b PQ = RS= and QR = SP = b. Let /be the electric current flowing through the rectangular coil PORS. The horse-shoe magnet has hemispherical magnetic poles which produces a radial magnetic _— tonic e t sa ° ae ) : Magnolc fats Force downwards ve to this radial field, the sides QR and SPare always parallel to the B+ield (magnetic field) and experience no force. The sides PQ and R5 are always parallel to the B-field and experience force and due to this. torque is produced. For single tur, the deflection couple is += b= bBll = (Ib) 61 = AB! since, area of the coll Az Ibfor coil with Nturns. we getr=NABL (1) Oue to this deflecting torque. the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence, the magnitude of restoring couple 's proportional to the amount of twist 8 Thus ¢= KO 2) where, K is the restoring couples per unit twist or torsional constant of the spring At equilibrium. the deflection couples is equal to the restoring couple. Therefore, by comparing eqs. (1) and (2). we get NABI) => to Tap! or I= (3) where, G= is called galvanometer constant K ‘NAB or current reduction factor of the galvanometer, Since. suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. Most of the galuanometer, we use arc painter type moving coil galvanometer. (i), The relation between the current Flowing through the galvanometer coll and the angular deflection (6) of the coll (From its equilibrium position). is . (j2etene) * K where is the angle between the magnetic field Hm vector Band the equivalent magnetic moment vector of the current carrying coil Thus 1 is not directly proportional to 6. We can ensure this proportionality by having 0 = 90°. This is possible only when the ‘magnetic field vector 8. is a radial magnetic field. In such a field, the plane of the rotating coil is always parallel to vector & 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, Softiron core Chapter Test Multiple Choice Questions QL A long straight wire of radius ‘a’ carries a steady current The current is uniformly distributed across its area of cross-section. The ratio of magnitude of "at a/2 and B, at distance 2a is: (case 2023) aw bl 2 a4 ‘magnetic field Q2. Beams of electrons and protons move parallel to each other in the same direction. They: (case 2023) a. attract each other b. repel each other neither attract nor repel di force of attraction or repulsion depends upon ‘speed af beams.Assertion and Reason Type Questions Directions (Q.Nos. 3-4): In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mork the correct choice as: a. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). b. Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). Assertion (A) is true but Reason (A) is false 4d. Both Assertion (A) and Reason (R) are false. Q3. Assertion (A): Increasing the current sensitivity of a galvanometer necessarily increases the voltage sensitivity. Reason (R): Voltage sensiti proportional to current senst wersely ity. Q4. Assertion (A): Two parallel conducting wires carrying currents in opposite direction, come close to each other. Reason (R): Parallel currents repel and anti-parallel currents attract. Fill in the blanks QS. The magnetic effect of electric current was first noticed by... Q6. Agalvanometer acting as a voltmeter should have : swim series with its coll Case Study Based Question Q7. The path of a charged particle in magnetic field depends upon angle between velocity and magnetic field If velocity ¥ is at angle 8 to B, component gf velocity parallel to magnetic field (v cos®) is responsible for circular motion, thus the charge particle moves in a helical path. ‘The plane of the circle is perpendicular to the magnetic field and the axis of the helix is parallel (o the magnetic field. The charged particle moves along helical path touching the line parallel to the magnetic field passing through the starting point after each rotation Radius of circular path is mvsin® 4B Hence, the resultant path of the charged particle will bea helix, with its axis along the direction of 'B as shown in figure. v sino Helical path EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET ee Read the given passage carefully and give the answer of the following questions: () When a positively charged particle enters into 2 uniform magnetic field with uniform velocity, its trajectory can be (i) a straight line, (ii) a circle, (ii) a helix. a. Only (i) b. (or (il) c (or (ii) dd Anyone of (9) and (i) (ii) Two charged partictes A and 8 having the same charge, mass and speed enter into a magnetic field in such a way that the initial path of A makes an angle of 30° and that of B makes an angle of 90° with the field. Then the trajectory of: a. B will have smaller radius of curvature than that of A b. both will have the same curvature . Auill have smaller radius of curvature than that of B 4. both will move along the direction of thelr original velocities ‘An electron having momentum 2.410"? kg m/s enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of 30° with the initial velocity vector of the electron. The radius of the helical path of the slectronin the field shall be: 3 az2mm bimm c Ymm 05mm (iv) The magnetic field in a certain region of space is given by B=8.35%10"7 T.A proton shot into the field with velocity y = (2108 i+ 4%105j) m/s. ‘The proton follows a helical path in the field. The distance moved by proton in the X-direction during the period of one revolution in the ¥Z-plane will be (Mass proton = 1.67 x 10°” kg): 2.0053m b.0136m <.0157m 4. 0236m (¥) The frequency of revolution ofthe particle ist az Be Rh 4g te @ 2nm vcoso sind Very Short Answer Type Questions Q8. A straight wire carrying a current of 13 A is bent into a semi-circular arc of radius 2 cm as shown in figure. The magnetic field is 1.5 x 10*Tat the centre of arc, then what is the magnetic field due to straight segment? os Q9. Is the steady electric current the only source of ‘magnetic field? 10. A tightly wound 90 turn coil of radius 15 cm has a magnetic field of 4 x 10 T at its centre. What will be the current flowing through it?ee EERE EEE IEEE EE EID EI EI IDI IDI EI IDI ID II II III II II IID IEEE EO ET Short Answer Type-I Questions QUL An element Al = AX is placed at the origin and carries a large current | = 10 A. What is the magnetic field on the Y-axis at a distance of 0.5m and AX= tem? y 12. Briefly explain why and how a galvanometer is converted into an ammeter. (case 2023) Short Answer Type-I! Questions Q13. (i) Two long straight parallel conductors a and carrying steady currents |, and /, respectively ‘are separated by a distance d. Write its ‘magnitude and direction. What is the nature and magnitude of the force between the two conductors? Gi) Show with the help of a diagram, how the force between the two conductors would change when the currents in them flow in the opposite directions? Q14. (i) State Ampere's circuital Law. (i) Derive an expression for magnetic field inside along the axis of an air cored solenoid. Long Answer Type Questions Q15. Figure shows a long straight wire of a circular ‘cross-section (radius 0) carrying steady current J. The current / is uniformly distributed across this cross-section. Calculate the magnetic field in the regions r a. Q16. A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2. (0) What is the field at the centre of the coil? (i) What is the magnetic moment of this coil? The coll is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 27 in the horizontal direction exists such that initially the axis of the coll is in the direction of the fied. ‘The coil rotates through an angle of 90° under the influence of the magnetic field. What are the magnitudes of the torques on the coil in the initial and final position? (iv) What is the angular speed acquired by the coil when it has rotated by 90°? The moment of inertia of the coil is 0.1 kg m?.