Torque on a coil

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  • 1. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 246.6 Torque on a coil6.6.1 Formula of torque Consider a rectangular coil (loop) of wire with side lengths a andb that it can turn about axis PQ. The coil is in a magnetic field offlux density B and the plane of the coil makes an angle withthe direction of the magnetic field. A current I is flowing roundthe coil as shown in Figure 6.31. 1

2. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 BB F1 FI IP BA bB B F1 a IQ F I Figure 6.31a 2 3. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24B F1BA b sin rotationB 2brotation2 Q b sin F1 2Figure 6.31b: side view From the Figure 6.31b, the magnitude of the force F1 is given by F1 = IlB sin 90 and l = a F1 = IaB From the Figure 6.31a, the forces F lie along the axis PQ.3 4. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 From the Figure 6.31a, the forces F lie along the axis PQ. The resultant force on the coil is zero but the nett torque isnot zero because the forces F1 are perpendicular to the axis PQas shown in Figure 6.31a. The forces F1 cause the coil to rotate in the clockwisedirection about the axis PQ as shown in Figure 6.31b. The magnitude of the nett torque about the axis PQ (refer toFigure 6.31b) is given bybb = F1 sin F1 sin 22 b and F = IaB = 2 F1 sin 1 2 b = 2( IaB ) sin 2 = IabB sin and ab = A(area of coil) 4 5. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 = IAB sin since = 90 thus ( = IAB sin 90 ) = IAB cos For a coil of N turns, the magnitude of the torque is given by = NIAB sin (6.14)OR = NIAB cos (6.15) where : torque on the coil B : magnetic flux density I : current flows in the coil : angle between vector area A and B : angle between the plane of the coil and B N : number of turns (coils) 5 6. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 From the eq. (6.14), thus the formula of torque in the vector formis given by ( = NI A B)(6.16) The torque is zero when = 90 or = 0 and is maximumwhen = 0 = 90 as shown in Figures 6.32a and 6.32b.orA BB = 90 = 90A =0 = 0Figure 6.32a Figure 6.32b = NIAB sin 0 plane of the coil = NIAB sin 90OROR = NIAB cos 90 = NIAB cos 0 =0 max = NIAB6 7. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 In a radial field, the plane of the coil is always parallel to thefieldmagnetic field for any orientation of the coil about the verticalaxis as shown in Figure 6.33. = 0 NSOR = 90 radial field fixed soft coiliron cylinder Figure 6.33: Plan view of moving coil meter Hence the torque on the coil in a radial field is always constant and maximum given by = NIAB sin 90 OR = NIAB cos 0 = NIAB maximum Radial field is used in moving coil galvanometer.7 8. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Example 13 :A 50 turns rectangular coil with sides 10 cm 20 cm is placedvertically in a uniform horizontal magnetic field of magnitude 2.5 T.If the current flows in the coil is 7.3 A, determine the torque actingon the coil when the plane of the coil isa. perpendicular to the field,b. parallel to the field,c. at 75 to the field.Solution : N = 50 turns; B = 2.5 T; I = 7.3 AThe area of the coil is given by A = (10 10 2 )( 20 10 2 ) = 2.0 10 2 m 2a.B From the figure, = 90 and = 0 , thus the = 90 torque on the coil isA = NIAB cos OR = NIAB sin = NIAB cos 90 = NIAB sin 0 =08 9. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Solution : N = 50 turns; B = 2.5 T; I = 7.3 A b. A From the figure, = 0 and = 90 , thus the B torque on the coil is = 90 = NIAB cos ( = ( 50 )( 7.3) 2.0 10 2 )( 2.5) cos 0 = 18.3 N mc.B From the figure, = 75 and = 15,thusthe torque on the coil is = 75 = 15 = NIAB cos = ( 50 )( 7.3) ( 2.0 10 2 )( 2.5) cos 75 A = 4.72 N m9 10. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 246.6.2 Moving-coil galvanometer A galvanometer consists of a coil of wire suspended in themagnetic field of a permanent magnet. The coil is rectangularshape and consists of many turns of fine wire as shown inFigure 6.34. 10 Figure 6.34 11. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 When the current I flows through the coil, the magnetic fieldcoilexerts a torque on the coil as given by = NIAB This torque is opposed by a spring which exerts a torque, sgiven by s = k (6.17) where k : torsional constant : rotation angle of the coil in radian The coil and pointer will rotate only to the point where thespring torque balances the torque due to magnetic field, fieldthus = sNIAB = kk I= (6.18) NAB 11 12. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Example 14 :A rectangular coil of 10 cm 4.0 cm in a galvanometer has 50 turnsand a magnetic flux density of 5.0 102 T. The resistance of thecoil is 40 and a potential difference of 12 V is applied across thegalvanometer, calculate the maximum torque on the coil.2Solution : N = 50 turns; B = 5.0 10 T; R = 40 ; V = 12 VThe area of the coil is given by A = (10 10 2 )( 4.0 10 2 ) = 4.0 10 3 m 2The current through the galvanometer isV = IR12 = I ( 40 ) I = 0.3 ATherefore the maximum torque on the coil is max = NIAB= ( 50)( 0.3) ( 4.0 10 3 )( 5.0 10 2 ) max = 3.0 10 3 N m 12 13. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 246.6.3 Electrical instrumentsOhmmeter It is used to measure the unknown resistance of the resistor.resistor Figure 6.35 shows the internal connection of an Ohmmeter.where0RM : meter (coil) resistanceRS : variable resistance RX : unknown resistanceRMRSPRXQFigure 6.35 13 14. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 When nothing is connected to terminals P and Q, so that theQcircuit is open (that is, when R ), there is no current andno deflection.deflection When terminals P and Q are short circuited (that is whenR = 0), the ohmmeter deflects full-scale. 0 full-scale For any value of RX the meter deflection depends on the value ofRX.Ammeter It is used to measure a current flows in the circuit. circuit Ammeter is connected in series with other elements in thecircuit because the current to be measured must pass directlythrough the ammeter. ammeter An ammeter should have low internal resistance (RM) so thatthe current in the circuit would not affected. affected 14 15. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 The maximum reading from the ammeter is known as full scaledeflection (fs). If the full scale current passing through the ammeter then thepotential difference (p.d.) across that ammeter is given byVfs = I fs RMwhere RM : meter(coil) resistanceI fs : full scale currentVfs : full scale potential difference (p.d.) If the meter is used to measure currents that are larger than itsfull scale deflection (I >Ifs), some modification has to be done. A resistor has to be connected in parallel with the meter (coil) resistance RM so that some of the current will bypasses the meter (coil) resistance.resistance This parallel resistor is called a shunt denoted as RS. Figure 6.36 shows the internal connection of an ammeter with a shunt in parallel. 15 16. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 0maxA I I fs I IS RMRS Figure 6.36 Since shunt is connected in parallel with the meter (coil) resistance thenVRM = VRS I fs RM = I S RS and I S = I I fs I fs I fs RM = ( I I fs ) RS RS = I I RM (6.19) 16 fs 17. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Voltmeter It is used to measure a potential difference (p.d.) acrosselectrical elements in the circuit.circuit Voltmeter is connected in parallel with other elements in thecircuit therefore its resistance must be larger than theresistance of the element so that a very small amount ofcurrent only can flows through it. An ideal voltmeter hasitinfinity resistance so that no current exist in it. To measure a potential difference that are larger than its fullscale deflection (V > Vfs), the voltmeter has to be modified. A resistor has to be connected in series with the meter(coil) resistance RM so that only a fraction of the total p.d.appears across the RM and the remainder appears acrossthe serial resistor. This serial resistor is called a multiplier OR bobbindenoted as RB.17 18. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Figure 6.37 shows the internal connection of a voltmeter with a multiplierin series.0maxV I fsRBRM VI I1Electrical element Since the multiplier is connected in series with the meter (coil) resistanceFigure 6.37then the current through them are the same, Ifs.18 19. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 The p.d. across the electrical element is given byV = VRB + VRM Hence the multiplier resistance isV = I fs RB + I fs RM V I fs RM RB = (6.20)I fsNote:To convert a galvanometer to ammeter, a shuntammeter (parallel resistor) is used.To convert a galvanometer to voltmeter, a multipliervoltmeter (serial resistor) is used. 19 20. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Example 15 :A milliammeter with a full scale deflection of 20 mA and an internalresistance of 40 is to be used as an ammeter with a full scaledeflection of 500 mA. Calculate the resistance of the shuntrequired. I = 20 10 3 A; R = 40 ; I = 500 10 3 Afs MSolution :By applying the formula of shunt resistor, thusI fs RS = I I RMfs 20 10 3 = ( 40) 500 10 20 10 33 RS = 1.67 20 21. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Example 16 :A galvanometer has an internal resistance of 30 and deflects fullscale for a 50 A current. Describe how to use this galvanometerto makea. an ammeter to read currents up to 30 A.b. a voltmeter to give a full scale deflection of 1000 V.6Solution : I fs = 50 10 A; RM = 30 a. We make an ammeter by putting a resistor in parallel ( RS) withthe internal resistance, RM of the galvanometer as shown infigure below.RMI I fsISGR