Moving Charges And Magnetism

Transcription

1 AIND SINGH ACADEMY Moving Charges An Magnetism Solution of NCET Exercise Q -.: A circular coil of wire consisting of turns, each of raius 8. cm carries a current of. A. What is the magnitue of the magnetic fiel at the centre of the coil? Number of turns on the circular coil, n = aius of each turn, r = 8. cm =.8 m Current flowing in the coil, I =. A, Magnitue of the magnetic fiel at the centre of the coil is given by the relation, πnl π r Where, = Permeability of free space = π 7 T m A 7 π π. -. T π.8 Hence, the magnitue of the magnetic fiel is. T. Q -.: A long straight wire carries a current of 5 A. What is the magnitue of the fiel at a point cm from the wire? Current in the wire, I = 5 A Distance of a point from the wire, r = cm =. m Magnitue of the magnetic fiel at this point is given as: I Where, = Permeability of free space = π 7 T m A π r Q -.: Q -.: 7 π T π. Hence, the magnitue of the magnetic fiel at a point cm from the wire is.5 5 T. A long straight wire in the horizontal plane carries a current of 5 A in north to south irection. Give the magnitue an irection of at a point.5 m east of the wire. Current in the wire, I = 5 A, A point is.5 m away from the East of the wire. Magnitue of the istance of the point from the wire, r =.5 m. I Magnitue of the magnetic fiel at that point is given by the relation, πr Where, = Permeability of free space = π 7 T m A π T π.5 The point is locate normal to the wire length at a istance of.5 m. The irection of the current in the wire is vertically ownwar. Hence, accoring to the Maxwell s right han thumb rule, the irection of the magnetic fiel at the given point is vertically upwar. A horizontal overhea power line carries a current of 9 A in east to west irection. What is the magnitue an irection of the magnetic fiel ue to the current.5 m below the line? Current in the power line, I = 9 A Point is locate below the power line at istance, r =.5 m Hence, magnetic fiel at that point is given by the relation, -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,99966

2 AIND SINGH ACADEMY I, Where, =Permeability of free space = π 7 T m A πr π T π.5 The current is flowing from East to West. The point is below the power line. Hence, accoring to Maxwell s right han thumb rule, the irection of the magnetic fiel is towars the South. Q -.5: What is the magnitue of magnetic force per unit length on a wire carrying a current of 8 A an making an angle of º with the irection of a uniform magnetic fiel of.5 T? Current in the wire, I = 8 A, Magnitue of the uniform magnetic fiel, =.5 T Angle between the wire an magnetic fiel, θ =. Magnetic force per unit length on the wire is given as: f = I sinθ =.5 8 sin =.6 N m Hence, the magnetic force per unit length on the wire is.6 N m. Q -.6: A. cm wire carrying a current of A is place insie a solenoi perpenicular to its axis. The magnetic fiel insie the solenoi is given to be.7 T. What is the magnetic force on the wire? Length of the wire, l = cm =. m Current flowing in the wire, I = A Magnetic fiel, =.7 T Angle between the current an magnetic fiel, θ = 9 Magnetic force exerte on the wire is given as: F = Il sinθ =.7. sin9 = 8. N Hence, the magnetic force on the wire is 8. N. The irection of the force can be obtaine from Fleming s left han rule. Q -.7: Two long an parallel straight wires A an carrying currents of 8. A an 5. A in the same irection are separate by a istance of. cm. Estimate the force on a cm section of wire A. Current flowing in wire A, I A = 8. A Current flowing in wire, I = 5. A Distance between the two wires, r =. cm =. m Length of a section of wire A, l = cm =. m Q -.8: Force exerte on length l ue to the magnetic fiel is given as: Where, µ = Permeability of free space = π 7 T m A 7 IAIl πr π N π. The magnitue of force is 5 N. This is an attractive force normal to A towars because the irection of the currents in the wires is the same. A closely woun solenoi 8 cm long has 5 layers of winings of turns each. The iameter of the solenoi is.8 cm. If the current carrie is 8. A, estimate the magnitue of insie the solenoi near its centre. Length of the solenoi, l = 8 cm =.8 m There are five layers of winings of turns each on the solenoi. Total number of turns on the solenoi, N = 5 = Diameter of the solenoi, D =.8 cm =.8 m Current carrie by the solenoi, I = 8. A NI Magnitue of the magnetic fiel insie the solenoi near its centre is given by the relation, l Where, µ = Permeability of free space = π 7 T m A 7 π 8.8 8π.5 T -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,99966

3 AIND SINGH ACADEMY Hence, the magnitue of the magnetic fiel insie the solenoi near its centre is.5 T. Q -.9: A square coil of sie cm consists of turns an carries a current of A. The coil is suspene vertically an the normal to the plane of the coil makes an angle of º with the irection of a uniform horizontal magnetic fiel of magnitue.8 T. What is the magnitue of torque experience by the coil? Length of a sie of the square coil, l = cm =. m Current flowing in the coil, I = A Number of turns on the coil, n = Angle mae by the plane of the coil with magnetic fiel, θ = Strength of magnetic fiel, =.8 T Magnitue of the magnetic torque experience by the coil in the magnetic fiel is given by the relation, τ = n IA sinθ Where, A = Area of the square coil l l =.. =. m τ =.8. sin =.96 N m Hence, the magnitue of the torque experience by the coil is.96 N m. Q -.: Two moving coil meters, M an M have the following particulars: = Ω, N =, A =.6 m, =.5 T, = Ω, N =, A =.8 m, =.5 T (The spring constants are ientical for the two meters). Determine the ratio of (a) current sensitivity an (b) voltage sensitivity of M an M. For moving coil meter M : esistance, = Ω, Number of turns, N =, Area of cross-section, A =.6 m, Magnetic fiel strength, =.5 T, Spring constant K = K For moving coil meter M : esistance, = Ω, Number of turns, N =, Area of cross-section, A =.8 m, Magnetic fiel strength, =.5 T, Spring constant, K = K N A (a) Current sensitivity of M is given as: Isl K NA An, current sensitivity of M is given as: Is K l atio I s s NA K K N A.5.8 K K.5.6 Hence, the ratio of current sensitivity of M to M is.. NA (b) Voltage sensitivity for M is given as: Vs K N A An, voltage sensitivity for M is given as: Vs K Vs NA K atio Vs K N A..5.8 K K.5.6 Hence, the ratio of voltage sensitivity of M to M is. Q -.: In a chamber, a uniform magnetic fiel of 6.5 G ( G = T) is maintaine. An electron is shot into the fiel with a spee of.8 6 m s normal to the fiel. Explain why the path of the electron is a circle. Determine the raius of the circular orbit. (e =.6 9 C, m e = 9. kg) Magnetic fiel strength, = 6.5 G = 6.5 T Spee of the electron, v =.8 6 m/s Charge on the electron, e =.6 9 C Mass of the electron, m e = 9. kg Angle between the shot electron an magnetic fiel, θ = 9 Magnetic force exerte on the electron in the magnetic fiel is given as: F = ev sinθ -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,99966

4 AIND SINGH ACADEMY This force provies centripetal force to the moving electron. Hence, the electron starts moving in a circular path of raius r. Hence, centripetal force exerte on the electron, Fc r In equilibrium, the centripetal force exerte on the electron is equal to the magnetic force i.e., Fc F 9..8 ev sinθ r. m.cm r e sinθ sin9 Hence, the raius of the circular orbit of the electron is. cm. Q -.: In Exercise. obtain the frequency of revolution of the electron in its circular orbit. Does the answer epen on the spee of the electron? Explain. Magnetic fiel strength, = 6.5 T Charge of the electron, e =.6 9 C Mass of the electron, m e = 9. kg Velocity of the electron, v =.8 6 m/s aius of the orbit, r =. cm =. m Frequency of revolution of the electron = ν Angular frequency of the electron = ω = πν Velocity of the electron is relate to the angular frequency as: v = rω In the circular orbit, the magnetic force on the electron is balance by the centripetal force. Hence, we can write: -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,99966 m m e ev e (rω) rπv v r r r πm This expression for frequency is inepenent of the spee of the electron. On substituting the known values in this expression, we get the frequency as: v 8. Hz 8 MHz. 9. Hence, the frequency of the electron is aroun 8 MHz an is inepenent of the spee of the electron. Q -.: (a) A circular coil of turns an raius 8. cm carrying a current of 6. A is suspene vertically in a uniform horizontal magnetic fiel of magnitue. T. The fiel lines make an angle of 6º with the normal of the coil. Calculate the magnitue of the counter torque that must be applie to prevent the coil from turning. (b) Woul your answer change, if the circular coil in (a) were replace by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltere.) (a) Number of turns on the circular coil, n = aius of the coil, r = 8. cm =.8 m Area of the coil πr 6 π.8.m Current flowing in the coil, I = 6. A Magnetic fiel strength, = T Angle between the fiel lines an normal with the coil surface, θ = 6 The coil experiences a torque in the magnetic fiel. Hence, it turns. The counter torque applie to prevent the coil from turning is given by the relation, τ = n IA sinθ (i) = 6. sin6 =. N m (b) It can be inferre from relation (i) that the magnitue of the applie torque is not epenent on the shape of the coil. It epens on the area of the coil. Hence, the answer woul not change if the circular coil in the above case is replace by a planar coil of some irregular shape that encloses the same area. Q -.: Two concentric circular coils X an Y of raii 6 cm an cm, respectively, lie in the same vertical plane containing the north to south irection. Coil X has turns an carries a current of 6 A; coil Y has 5 turns an carries a current of 8 A. The sense of the current in X is anticlockwise, an clockwise in Y, for an observer looking at the coils facing west. Give the magnitue an irection of the net magnetic fiel ue to the coils at their centre. aius of coil X, r = 6 cm =.6 m aius of coil Y, r = cm =. m Number of turns of on coil X, n = Number of turns of on coil Y, n = 5

5 Q -.5: Q -.6: Current in coil X, I = 6 A AIND SINGH ACADEMY Current in coil Y, I = 8 A n I Magnetic fiel ue to coil X at their centre is given by the relation, r Where, µ = Permeability of free space = π x -7 T m A - 7 π 6 π T towars East.6 Magnetic fiel ue to coil Y at their centre is given by the relation, πni r 7 π 5 8. Hence, net magnetic fiel can be obtaine as: 9π - - π 5π 9π T T towars.57 West T towars West A magnetic fiel of G ( G = T) is require which is uniform in a region of linear imension about cm an area of cross-section about m. The maximum current-carrying capacity of a given coil of wire is 5 A an the number of turns per unit length that can be woun roun a core is at most turns m. Suggest some appropriate esign particulars of a solenoi for the require purpose. Assume the core is not ferromagnetic Magnetic fiel strength, = G = T Number of turns per unit length, n = turns m Current flowing in the coil, I = 5 A Permeability of free space, µ = π x -7 T m A - Magnetic fiel is given by the relation, = µ ni nl π A/m If the length of the coil is taken as 5 cm, raius cm, number of turns, an current A, then these values are not unique for the given purpose. There is always a possibility of some ajustments with limits. For a circular coil of raius an N turns carrying current I, the magnitue of the magnetic fiel at a point on its axis at a istance x from its centre is given by, x I N (a) Show that this reuces to the familiar result for fiel at the centre of the coil. (b) Consier two parallel co-axial circular coils of equal raius, an number of turns N, carrying equal currents in the same irection, an separate by a istance. Show that the fiel on the axis aroun the mi-point between the coils is uniform over a istance that is small as compare to, an is given by, NI.7, approximately. [Such an arrangement to prouce a nearly uniform magnetic fiel over a small region is known as Helmholtz coils.] aius of circular coil = Number of turns on the coil = N Current in the coil = I Magnetic fiel at a point on its axis at istance x is given by the relation, Where, µ = Permeability of free space (a) If the magnetic fiel at the centre of the coil is consiere, then x =. l N IN x I N -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,

6 AIND SINGH ACADEMY -/, SECTO-8, OHINI, NEW DELHI-89, Phone , This is the familiar result for magnetic fiel at the centre of the coil. (b) aii of two parallel co-axial circular coils = Number of turns on each coil = N Current in both coils = I Distance between both the coils = Let us consier point Q at istance from the centre. Then, one coil is at a istance of from point Q. Magnetic fiel at point Q is given as: NI Also, the other coil is at a istance of from point Q. Magnetic fiel ue to this coil is given as: NI Total magnetic fiel, IN.7 IN I I : we get,, neglecting the factor For I 5 5 I I Hence, it is prove that the fiel on the axis aroun the mi-point between the coils is uniform. Q -.7: A toroi has a core (non-ferromagnetic) of inner raius 5 cm an outer raius 6 cm, aroun which 5 turns of a wire are woun. If the current in the wire is A, what is the magnetic fiel (a) outsie the toroi, (b) insie the core of the toroi, an (c) in the empty space surroune by the toroi. Inner raius of the toroi, r = 5 cm =.5 m Outer raius of the toroi, r = 6 cm =.6 m Number of turns on the coil, N = 5 Current in the coil, I = A (a) Magnetic fiel outsie a toroi is zero. It is non-zero only insie the core of a toroi. (b) Magnetic fiel insie the core of a toroi is given by the relation, = l NI

7 AIND SINGH ACADEMY Where, µ = Permeability of free space = π x -7 T m A - l = length of toroi r r π 5 π π.5.5π -.6.5π. T (c) Magnetic fiel in the empty space surroune by the toroi is zero. Q -.8: Answer the following Q -s: (a) A magnetic fiel that varies in magnitue from point to point but has a constant irection (east to west) is set up in a chamber. A charge particle enters the chamber an travels uneflecte along a straight path with constant spee. What can you say about the initial velocity of the particle? (b) A charge particle enters an environment of a strong an non-uniform magnetic fiel varying from point to point both in magnitue an irection, an comes out of it following a complicate trajectory. Woul its final spee equal the initial spee if it suffere no collisions with the environment? (c) An electron travelling west to east enters a chamber having a uniform electrostatic fiel in north to south irection. Specify the irection in which a uniform magnetic fiel shoul be set up to prevent the electron from eflecting from its straight line path. (a) The initial velocity of the particle is either parallel or anti-parallel to the magnetic fiel. Hence, it travels along a straight path without suffering any eflection in the fiel. (b) Yes, the final spee of the charge particle will be equal to its initial spee. This is because magnetic force can change the irection of velocity, but not its magnitue. (c) An electron travelling from West to East enters a chamber having a uniform electrostatic fiel in the North-South irection. This moving electron can remain uneflecte if the electric force acting on it is equal an opposite of magnetic fiel. Magnetic force is irecte towars the South. Accoring to Fleming s left han rule, magnetic fiel shoul be applie in a vertically ownwar irection. Q -.9: An electron emitte by a heate cathoe an accelerate through a potential ifference of. kv, enters a region with uniform magnetic fiel of.5 T. Determine the trajectory of the electron if the fiel (a) is transverse to its initial velocity, (b) makes an angle of º with the initial velocity. Magnetic fiel strength, =.5 T Charge on the electron, e =.6 9 C Mass of the electron, m = 9. kg Potential ifference, V =. kv = V Thus, kinetic energy of the electron = ev ev v ev m...() Where, v = velocity of the electron (a) Magnetic force on the electron provies the require centripetal force of the electron. Hence, the electron traces a circular path of raius r. Magnetic force on the electron is given by the relation, ev Centripetal force r ev r r e From equations () an (), we get...() 7 -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,

8 AIND SINGH ACADEMY r m e ev m m mm Hence, the electron has a circular trajectory of raius. mm normal to the magnetic fiel. (b) When the fiel makes an angle θ of with initial velocity, the initial velocity will be, v l = v sin θ From equation (), we can write the expression for new raius as: r. e sinθ e m.5 mm Hence, the electron has a helical trajectory of raius.5 mm along the magnetic fiel irection. Q -.: A magnetic fiel set up using Helmholtz coils (escribe in Exercise.6) is uniform in a small region an has a magnitue of.75 T. In the same region, a uniform electrostatic fiel is maintaine in a irection normal to the common axis of the coils. A narrow beam of (single species) charge particles all accelerate through 5 kv enters this region in a irection perpenicular to both the axis of the coils an the electrostatic fiel. If the beam remains uneflecte when the electrostatic fiel is 9. 5 V m, make a simple guess as to what the beam contains. Why is the answer not unique? Magnetic fiel, =.75 T Accelerating voltage, V = 5 kv = 5 V Electrostatic fiel, E = 9 5 V m Mass of the electron = m Charge of the electron = e Velocity of the electron = v Kinetic energy of the electron = ev -/, SECTO-8, OHINI, NEW DELHI-89, Phone , sin e v ev...() m V Since the particle remains uneflecte by electric an magnetic fiels, we can infer that the electric fiel is balancing the magnetic fiel. E ee ev v...() Putting equation () in equation (), we get E m V E V C / kg e 7 5 (.75) This value of specific charge e/m is equal to the value of euteron or euterium ions. This is not a unique answer. Other possible answers are He ++, Li ++, etc. Q -.: A straight horizontal conucting ro of length.5 m an mass 6 g is suspene by two vertical wires at its ens. A current of 5. A is set up in the ro through the wires. (a) What magnetic fiel shoul be set up normal to the conuctor in orer that the tension in the wires is zero? (b) What will be the total tension in the wires if the irection of current is reverse keeping the magnetic fiel same as before? (Ignore the mass of the wires.) g = 9.8 m s. Length of the ro, l =.5 m Mass suspene by the wires, m = 6 g = 6 kg Acceleration ue to gravity, g = 9.8 m/s Current in the ro flowing through the wire, I = 5 A (a) Magnetic fiel () is equal an opposite to the weight of the wire i.e., ll mg T

9 AIND SINGH ACADEMY A horizontal magnetic fiel of.6 T normal to the length of the conuctor shoul be set up in orer to get zero tension in the wire. The magnetic fiel shoul be such that Fleming s left han rule gives an upwar magnetic force. (b) If the irection of the current is revere, then the force ue to magnetic fiel an the weight of the wire acts in a vertically ownwar irection. Total tension in the wire = Il + mg =.6 x 5 x.5 + (6 x - ) x 9.8 =.76N Q -.: The wires which connect the battery of an automobile to its starting motor carry a current of A (for a short time). What is the force per unit length between the wires if they are 7 cm long an.5 cm apart? Is the force attractive or repulsive? Current in both wires, I = A Distance between the wires, r =.5 cm =.5 m Length of the two wires, l = 7 cm =.7 m I Force between the two wires is given by the relation, F πr Where, µ = Permeability of free space = π x -7 m A - 7 π F.N/m π.5 Since the irection of the current in the wires is opposite, a repulsive force exists between them. Q -.: A uniform magnetic fiel of.5 T exists in a cylinrical region of raius. cm, its irection parallel to the axis along east to west. A wire carrying current of 7. A in the north to south irection passes through this region. What is the magnitue an irection of the force on the wire if, (a) the wire intersects the axis, (b) the wire is turne from N-S to northeast-northwest irection, (c) the wire in the N-S irection is lowere from the axis by a istance of 6. cm? Magnetic fiel strength, =.5 T aius of the cylinrical region, r = cm =. m Current in the wire passing through the cylinrical region, I = 7 A (a) If the wire intersects the axis, then the length of the wire is the iameter of the cylinrical region. Thus, l = r =. m Angle between magnetic fiel an current, θ = 9 Magnetic force acting on the wire is given by the relation, F = Il sin θ =.5 7. sin 9 =. N Hence, a force of. N acts on the wire in a vertically ownwar irection. (b) New length of the wire after turning it to the Northeast-Northwest irection can be given as: : l l Angle between magnetic fiel an current, θ = 5 Force on the wire, F = Il sin θ Il.5 7..N Hence, a force of. N acts vertically ownwar on the wire. This is inepenent of angle θ because l sinθ is fixe. (c) The wire is lowere from the axis by istance, = 6. cm Let l be the new length of the wire. l r 6 6 l 8 6cm.6m Magnetic force exerte on the wire, F = Il =.5 x 7 x.6 =.68 N Hence, a force of.68 N acts in a vertically ownwar irection on the wire. l sin θ -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,

10 AIND SINGH ACADEMY Q -.: A uniform magnetic fiel of G is establishe along the positive z-irection. A rectangular loop of sies cm an 5 cm carries a current of A. What is the torque on the loop in the ifferent cases shown in Fig..8? What is the force on each case? Which case correspons to stable equilibrium? Magnetic fiel strength, = G = T =. T Length of the rectangular loop, l = cm With of the rectangular loop, b = 5 cm Area of the loop, A = l b = 5 = 5 cm = 5 m Current in the loop, I = A Now, taking the anti-clockwise irection of the current as positive an vice-versa: (a) Torque, τ A From the given figure, it can be observe that A is normal to the y- z plane an is irecte along the z-axis. Q -.5: τ 5 i.k.8 j N m The torque is.8 x - N m along the negative y-irection. The force on the loop is zero because the angle between A an is zero. (b) This case is similar to case (a). Hence, the answer is the same as (a). (c) Torque τ IA From the given figure, it can be observe that A is normal to the x-z plane an is irecte along the z-axis. τ 5 j.k -.8 i Nm The torque is.8 x - N m along the negative x irection an the force is zero. () Magnitue of torque is given as: τ IA 5..8 Torque is.8 x - N m at an angle of with positive x irection. The force is zero. (e) Torque τ I A 5 k. k Hence, the torque is zero. The force is also zero. (f) Torque τ I A 5 k. k N m Hence, the torque is zero. The force is also zero. In case (e), the irection of I A an is the same an the angle between them is zero. If isplace, they come back to an equilibrium. Hence, its equilibrium is stable. Whereas, in case (f), the irection of I A an is opposite. The angle between them is 8. If isturbe, it oes not come back to its original position. Hence, its equilibrium is unstable. A circular coil of turns an raius cm is place in a uniform magnetic fiel of. T normal to the plane of the coil. If the current in the coil is 5. A, what is the (a) total torque on the coil, (b) total force on the coil, (c) average force on each electron in the coil ue to the magnetic fiel? (The coil is mae of copper wire of cross-sectional area 5 m, an the free electron ensity in copper is given to be about 9 m.) -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,99966

11 AIND SINGH ACADEMY Number of turns on the circular coil, n = aius of the coil, r = cm =. m Magnetic fiel strength, =. T Current in the coil, I = 5. A (a) The total torque on the coil is zero because the fiel is uniform. (b) The total force on the coil is zero because the fiel is uniform. (c) Cross-sectional area of copper coil, A = 5 m Number of free electrons per cubic meter in copper, N = 9 /m Charge on the electron, e =.6 9 C Magnetic force, F = ev Where, v = Drift velocity of electrons Q -.6: Q -.7: I ei. 5. F 5 5 N NeA NeA 9 5 Hence, the average force on each electron is 5 x -5 N. A solenoi 6 cm long an of raius. cm has layers of winings of turns each. A. cm long wire of mass.5 g lies insie the solenoi (near its centre) normal to its axis; both the wire an the axis of the solenoi are in the horizontal plane. The wire is connecte through two leas parallel to the axis of the solenoi to an external battery which supplies a current of 6. A in the wire. What value of current (with appropriate sense of circulation) in the winings of the solenoi can support the weight of the wire? g = 9.8 m s Length of the solenoi, L = 6 cm =.6 m aius of the solenoi, r =. cm =. m It is given that there are layers of winings of turns each. Total number of turns, n = = 9 Length of the wire, l = cm =. m Mass of the wire, m =.5 g =.5 kg Current flowing through the wire, i = 6 A Acceleration ue to gravity, g = 9.8 m/s ni Magnetic fiel prouce insie the solenoi, L Where, µ = Permeability of free space = π x -7 T m A - I = Current flowing through the winings of the solenoi ni Magnetic force is given by the relation, F il il L Also, the force on the wire is equal to the weight of the wire. niil mg L mgl I niil π A Hence, the current flowing through the solenoi is 8 A. A galvanometer coil has a resistance of Ω an the metre shows full scale eflection for a current of ma. How will you convert the metre into a voltmeter of range to 8 V? esistance of the galvanometer coil, G = Ω Current for which there is full scale eflection, I g = ma = A ange of the voltmeter is, which nees to be converte to 8 V. V = 8 V Let a resistor of resistance be connecte in series with the galvanometer to convert it into a voltmeter. This resistance is given as: V G I g Ω -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,99966

12 AIND SINGH ACADEMY Q -.8: Hence, a resistor of resistance 5988Ω is to be connecte in series with the galvanometer. A galvanometer coil has a resistance of 5 Ω an the metre shows full scale eflection for a current of ma. How will you convert the metre into an ammeter of range to 6 A? esistance of the galvanometer coil, G = 5 Ω Current for which the galvanometer shows full scale eflection, I g = ma = A ange of the ammeter is, which nees to be converte to 6 A. Current, I = 6 A A shunt resistor of resistance S is to be connecte in parallel with the galvanometer to convert it into an ammeter. The value of S is given as: IgG S I I g S Ω mω Hence, a mω shunt resistor is to be connecte in parallel with the galvanometer. -/, SECTO-8, OHINI, NEW DELHI-89, Phone ,99966