{"id":120392,"date":"2022-05-10T11:45:55","date_gmt":"2022-05-10T06:15:55","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=120392"},"modified":"2022-05-10T11:45:55","modified_gmt":"2022-05-10T06:15:55","slug":"chapter-6-electromagnetic-induction-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-6-electromagnetic-induction-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 6 \u2013 Electromagnetic Induction Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"<h2>1. Predict the direction of induced current in the situations described by the following figures (a) to (f).<\/h2>\n<h3>a)<br \/>\nAns: The direction of the induced current in a closed loop could be given by Lenz\u2019s law. The following pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.<br \/>\nNow, by using Lenz\u2019s rule, the direction of the induced current in the given situation is found to be along qrpq.<\/h3>\n<h3>b)<br \/>\nAns: On using Lenz\u2019s law, we find the direction of the induced current here to be along prqp.<\/h3>\n<h3>c)<br \/>\nAns: Using Lenz\u2019s law, we find the direction of the induced current to be along yzxy.<\/h3>\n<h3>d)<br \/>\nAns: Using Lenz\u2019s law, we find the direction of the induced current to be along zyxz.<\/h3>\n<h3>e)<br \/>\nAns: Using Lenz\u2019s law, we found the direction of the induced current to be along xryx.<\/h3>\n<h3>f)<br \/>\nAns: Here we find that, no current is induced since the field lines are lying in the same plane as that of the closed loop.<\/h3>\n<h2>2. Use Lenz\u2019s law to determine the direction of induced current in the situations described by Figure:<br \/>\na) A wire of irregular shape turning into a circular shape;<\/h2>\n<h3>Ans: According to Lenz&#8217;s law, the direction of the induced emf is such that it tends to produce a current that would oppose the change in the magnetic flux that produced it.<br \/>\nThe wire is expanding to form a circle, which means that force would be acting outwards on each part of the wire because of the magnetic field (acting in the downwards direction). Now, the direction of induced current should be such that it will produce magnetic field in the upward direction (towards the reader). Therefore, the force on wire will be towards the inward direction, i.e., induced current would be flowing in anticlockwise direction in the loop from cbad.<\/h3>\n<h2>b) A circular loop being deformed into a narrow straight wire.<\/h2>\n<h3>Ans: On deforming the shape of a circular loop into a narrow straight wire, the flux piercing the surface decreases. Therefore, the induced current flows along abcd according to Lenz\u2019s law.<\/h3>\n<h2>3. A long solenoid with 15 turns per cm has a small loop of area 2.0cm22.0cm2placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from<br \/>\n2.0A2.0A<br \/>\nto<br \/>\n4.0A4.0A<br \/>\nin<br \/>\n0.1s0.1s<br \/>\n, what is the induced emf in the loop while the current is changing?<\/h2>\n<h3>Ans: We are given the following information:<br \/>\nNumber of turns on the solenoid<br \/>\n=15turns\/cm=1500turns\/m=15turns\/cm=1500turns\/m<br \/>\nNumber of turns per unit length,<br \/>\nn=1500 turnsn=1500 turns<br \/>\nThe solenoid has a small loop of area,A=2.0cm2=2\u00d710\u22124m2A=2.0cm2=2\u00d710\u22124m2<br \/>\nCurrent carried by the solenoid changes from<br \/>\n2A to 4A2A to 4A<br \/>\nNow, the change in current in the solenoid,<br \/>\ndi=4\u22122=2Adi=4\u22122=2A<br \/>\nChange in time,<br \/>\ndt=0.1 sdt=0.1 s<br \/>\nInduced emf in the solenoid could be given by Faraday\u2019s law as:<br \/>\n\u03b5=d\u03d5dt\u03b5=d\u03d5dt\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (1)<br \/>\nWhere, induced flux through the small loop, \u03d5=BA\u03d5=BA\u2026\u2026\u2026\u2026\u2026 (2)<br \/>\nEquation (1) would now reduce to:<br \/>\n\u03b5=ddt(BA)=A\u03bc0n\u00d7(didt)\u03b5=ddt(BA)=A\u03bc0n\u00d7(didt)<br \/>\nSubstituting the given values into this equation, we get,<br \/>\n\u03b5=2\u00d710\u22124\u00d74\u03c0\u00d710\u22127\u00d71500\u00d720.1\u03b5=2\u00d710\u22124\u00d74\u03c0\u00d710\u22127\u00d71500\u00d720.1<br \/>\n\u2234\u03b5=7.54\u00d710\u22126V\u2234\u03b5=7.54\u00d710\u22126V<br \/>\nTherefore, the induced voltage in the loop is found to be, \u03b5=7.54\u00d710\u22126V\u03b5=7.54\u00d710\u22126V.<\/h3>\n<h2>4. A rectangular wire loop of sides 8cm and 2cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is in a direction normal to the:<br \/>\na) longer side? For how long does the induced voltage last in this case?<\/h2>\n<h3>Ans: We are given the following,<br \/>\nLength of the rectangular wire,<br \/>\nl=8cm=0.08 ml=8cm=0.08 m<br \/>\nWidth of the rectangular wire,<br \/>\nb=2cm=0.02mb=2cm=0.02m<br \/>\nNow, the area of the rectangular loop,<br \/>\nA=lb=0.08\u00d70.02=16\u00d710\u22124m2A=lb=0.08\u00d70.02=16\u00d710\u22124m2<br \/>\nMagnetic field strength,<br \/>\nB=0.3TB=0.3T<br \/>\nVelocity of the loop,<br \/>\nv=1 cm\/s=0.01 m\/sv=1 cm\/s=0.01 m\/s<br \/>\nEmf developed in the loop could be given as:<br \/>\n\u03b5=Blv\u03b5=Blv<br \/>\nSubstituting the given values,<br \/>\n\u03b5=0.3\u00d70.08\u00d70.01=2.4\u00d710\u22124V\u03b5=0.3\u00d70.08\u00d70.01=2.4\u00d710\u22124V<br \/>\nTime taken to travel along the width, t=Distance travelledvelocity=bvt=Distance travelledvelocity=bv<br \/>\n\u21d2t=0.020.01=2s\u21d2t=0.020.01=2s<br \/>\nTherefore, the induced voltage is found to be 2.4\u00d710\u22124V2.4\u00d710\u22124Vwhich lasts for 2s.<\/h3>\n<h2>b) shorter side of the loop? For how long does the induced voltage last in this case?<\/h2>\n<h3>Ans: We know that, Emf developed in the loop could be given as:<br \/>\n\u03b5=Blv\u03b5=Blv<br \/>\nSubstituting the given values,<br \/>\n\u03b5=0.3\u00d70.02\u00d70.01=0.6\u00d710\u22124V\u03b5=0.3\u00d70.02\u00d70.01=0.6\u00d710\u22124V<br \/>\nTime taken to travel along the width, t=Distance travelledvelocity=lvt=Distance travelledvelocity=lv<br \/>\n\u21d2t=0.080.01=8s\u21d2t=0.080.01=8s<br \/>\nTherefore, the induced voltage is found to be 0.6\u00d710\u22124V0.6\u00d710\u22124Vwhich lasts for 8s.<\/h3>\n<h2>5. A<br \/>\n1.0m1.0m<br \/>\nlong metallic rod is rotated with an angular frequency of 400rads\u22121400rads\u22121about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of<br \/>\n0.5T0.5T<br \/>\nparallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nLength of the rod,<br \/>\nl=1ml=1m<br \/>\nAngular frequency, \u03c9=400rad\/s\u03c9=400rad\/s<br \/>\nMagnetic field strength,<br \/>\nB=0.5TB=0.5T<br \/>\nOne end of the rod has zero linear velocity, while the other end has a linear velocity of l\u03c9l\u03c9.<br \/>\nAverage linear velocity of the rod, \u03bd=l\u03c9+02=l\u03c92\u03bd=l\u03c9+02=l\u03c92<br \/>\nEmf developed between the centre and the ring,<br \/>\n\u03b5=Blv=Bl(l\u03c92)=(Bl2\u03c92)\u03b5=Blv=Bl(l\u03c92)=(Bl2\u03c92)<br \/>\nOn substituting the given values,<br \/>\n\u2234\u03b5=0.5\u00d7(1)2\u00d74002=100V\u2234\u03b5=0.5\u00d7(1)2\u00d74002=100V<br \/>\nTherefore, the emf developed between the centre and the ring is<br \/>\n100V100V<\/h3>\n<h2>6. A circular coil of radius<br \/>\n8.0cm8.0cm<br \/>\nand 20 turns is rotated about its vertical diameter with an angular speed of 50rads\u2212150rads\u22121in a uniform horizontal magnetic field of magnitude 3.0\u00d710\u22122T3.0\u00d710\u22122T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance<br \/>\n10\u03a910\u03a9<br \/>\n, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?<\/h2>\n<h3>Ans: We are given:<br \/>\nMax induced emf<br \/>\n=0.603V=0.603V<br \/>\nAverage induced emf<br \/>\n=0V=0V<br \/>\nMax current in the coil<br \/>\n=0.0603A=0.0603A<br \/>\nAverage power loss<br \/>\n=0.018W =0.018W<br \/>\n(Power comes from the external rotor)<br \/>\nRadius of the circular coil,<br \/>\nr=8cm=0.08mr=8cm=0.08m<br \/>\nArea of the coil, A=\u03c0r2=\u03c0\u00d7(0.08)2m2A=\u03c0r2=\u03c0\u00d7(0.08)2m2<br \/>\nNumber of turns on the coil,<br \/>\nN=20N=20<br \/>\nAngular speed, \u03c9=50rad\/s\u03c9=50rad\/s<br \/>\nMagnetic field strength, B=3\u00d710\u22122TB=3\u00d710\u22122T<br \/>\nResistance of the loop,<br \/>\nR=10 \u03a9R=10 \u03a9<br \/>\nMaximum induced emf could be given as:<br \/>\n\u03b5=N\u03c9AB=20\u00d750\u00d7\u03c0\u00d7(0.08)2\u00d73\u00d710\u22122\u03b5=N\u03c9AB=20\u00d750\u00d7\u03c0\u00d7(0.08)2\u00d73\u00d710\u22122<br \/>\n\u2234\u03b5=0.603V\u2234\u03b5=0.603V<br \/>\nThe maximum emf induced in the coil is found to be<br \/>\n0.603V0.603V<br \/>\nOver a full cycle, the average emf induced in the coil is found to be zero.<br \/>\nMaximum current is given as:<br \/>\nI=\u03b5RI=\u03b5R<br \/>\n\u21d2I=0.60310\u21d2I=0.60310<br \/>\n\u21d2I=0.0603A\u21d2I=0.0603A<br \/>\nAverage power loss due to joule heating:<br \/>\n\u2234P=eI2=0.603\u00d70.06032=0.018W\u2234P=eI2=0.603\u00d70.06032=0.018W<br \/>\nWe know that the current induced in the coil would produce a torque opposing the rotation of the coil. Since the rotor is an external agent, it must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.<\/h3>\n<h2>7. A horizontal straight wire<br \/>\n10m10m<br \/>\nlong extending from east to west is falling with a speed of 5.0ms\u221215.0ms\u22121, at right angles to the horizontal component of the earth\u2019s magnetic field, 0.30\u00d710\u22124Wbm\u221220.30\u00d710\u22124Wbm\u22122.<br \/>\na) What is the instantaneous value of the emf induced in the wire?<\/h2>\n<h3>Ans: We are given the following:<br \/>\nLength of the wire,<br \/>\nl=10ml=10m<br \/>\nFalling speed of the wire,<br \/>\nv=5.0m\/sv=5.0m\/s<br \/>\nMagnetic field strength, B=0.3\u00d710\u22124Wbm\u22122B=0.3\u00d710\u22124Wbm\u22122<br \/>\nEmf induced in the wire is thus found to be,<br \/>\n\u03b5=Blv\u03b5=Blv<br \/>\n\u21d2\u03b5=0.3\u00d710\u22124\u00d75\u00d710\u21d2\u03b5=0.3\u00d710\u22124\u00d75\u00d710<br \/>\n\u2234\u03b5=1.5\u00d710\u22123V\u2234\u03b5=1.5\u00d710\u22123V<br \/>\nHence, the emf induced in the wire is \u03b5=1.5\u00d710\u22123V\u03b5=1.5\u00d710\u22123V.<\/h3>\n<h2>b) What is the direction of the emf?<\/h2>\n<h3>Ans: Using Fleming\u2019s rule, we find that the direction of the induced emf is from West to East.<\/h3>\n<h2>c) Which end of the wire is at the higher electrical potential?<\/h2>\n<h3>Ans: The eastern end of the wire is the end that is at higher potential.<\/h3>\n<h2>8. Current in a circuit falls from<br \/>\n5.0A to 0.0A5.0A to 0.0A<br \/>\nin<br \/>\n0.1s0.1s<br \/>\n. If an average emf of<br \/>\n200V200V<br \/>\nis induced, give an estimate of the self-inductance of the circuit.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nInitial current, I1=5.0AI1=5.0A<br \/>\nFinal current, I2=0.0AI2=0.0A<br \/>\nChange in current, dI=I1\u2212I2=5AdI=I1\u2212I2=5A<br \/>\nTime taken for the change,<br \/>\nt=0.1st=0.1s<br \/>\nAverage emf, \u03b5=200V\u03b5=200V<br \/>\nFor self-inductance (L) of the coil, we have the relation for average emf that could be given as:<br \/>\n\u03b5=Ldidt\u03b5=Ldidt<br \/>\n\u21d2L=\u03b5(didt)\u21d2L=\u03b5(didt)<br \/>\nSubstituting the given values we get,<br \/>\n\u2234L=200(50.1)=4H\u2234L=200(50.1)=4H<br \/>\nTherefore, we found the self induction in the coil to be 4H.<\/h3>\n<h2>9. A pair of adjacent coils has a mutual inductance of<br \/>\n1.5H1.5H<br \/>\n. If the current in one coil changes from<br \/>\n0 to 20A in 0.5s0 to 20A in 0.5s<br \/>\n, what is the change of flux linkage with the other coil?<\/h2>\n<h3>Ans: We are given the following,<br \/>\nMutual inductance of a pair of coils, \u03bc=1.5H\u03bc=1.5H<br \/>\nInitial current, I1=0AI1=0A<br \/>\nFinal current, I2=20AI2=20A<br \/>\nChange in current, dI=I2\u2212I1=20\u22120=20AdI=I2\u2212I1=20\u22120=20A<br \/>\nTime taken for the change,<br \/>\nt=0.5st=0.5s<br \/>\nInduced emf, \u03b5=d\u03d5dt\u03b5=d\u03d5dt\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (1)<br \/>\nWhere, d\u03d5d\u03d5is the change in the flux linkage with the coil.<br \/>\nEmf is related with mutual inductance could be given as:<br \/>\n\u03b5=\u03bcdIdt\u03b5=\u03bcdIdt\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (2)<br \/>\nEquating equations (1) and (2), we get,<br \/>\nd\u03d5dt=\u03bcdIdtd\u03d5dt=\u03bcdIdt<br \/>\n\u21d2d\u03d5=1.5\u00d7(20)\u21d2d\u03d5=1.5\u00d7(20)<br \/>\n\u2234d\u03d5=30Wb\u2234d\u03d5=30Wb<br \/>\nHence, we found the change in the flux linkage to be 30Wb.<\/h3>\n<h2>10. A jet plane is travelling towards west at a speed of<br \/>\n1800km\/h1800km\/h<br \/>\n. What is the voltage difference developed between the ends of the wing having a span of<br \/>\n25m,25m,<br \/>\nif the Earth\u2019s magnetic field at the location has a magnitude of 5\u00d710\u22124T5\u00d710\u22124Tand the dip angle is 30\u221830\u2218.<\/h2>\n<h3>Ans: Speed of the jet plane,<br \/>\nv=1800 km\/h=500 m\/sv=1800 km\/h=500 m\/s<br \/>\nWingspan of jet plane,<br \/>\nl=25ml=25m<br \/>\nEarth\u2019s magnetic field strength, B=5.0\u00d710\u22124TB=5.0\u00d710\u22124T<br \/>\nAngle of dip, \u03b4=30\u2218\u03b4=30\u2218<br \/>\nVertical component of Earth\u2019s magnetic field could be given by,<br \/>\nBV=Bsin\u03b4BV=Bsin\u2061\u03b4<br \/>\n\u21d2BV=5\u00d710\u22124sin30\u2218=2.5\u00d710\u22124T\u21d2BV=5\u00d710\u22124sin\u206130\u2218=2.5\u00d710\u22124T<br \/>\nVoltage difference between the ends of the wing can be calculated as,<br \/>\n\u03b5=BV\u00d7l\u00d7v\u03b5=BV\u00d7l\u00d7v<br \/>\nSubstituting the given values,<br \/>\n\u21d2\u03b5=2.5\u00d710\u22124\u00d725\u00d7500\u21d2\u03b5=2.5\u00d710\u22124\u00d725\u00d7500<br \/>\n\u2234\u03b5=3.125V\u2234\u03b5=3.125V<br \/>\nHence, the voltage difference developed between the ends of the wings is<br \/>\n3.125V3.125V<\/h3>\n<h2>11. Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of<br \/>\n0.3T0.3T<br \/>\nat the rate of<br \/>\n0.02Ts\u221210.02Ts\u22121<br \/>\n. If the cut is joined and the loop has a resistance of<br \/>\n1.6\u03a91.6\u03a9<br \/>\nhow much power is dissipated by the loop as heat? What is the source of this power?<\/h2>\n<h3>Ans: We are given,<br \/>\nSides of the rectangular loop are 8cm and 2cm. Hence, area of the rectangular wire loop would be,<br \/>\nA = length \u00d7 widthA = length \u00d7 width<br \/>\nInitial value of the magnetic field,<br \/>\nB\u2032=0.3TB\u2032=0.3T<br \/>\nRate of decrease of the magnetic field, dBdt=0.02T\/sdBdt=0.02T\/s<br \/>\nEmf developed in the loop is given as:<br \/>\n\u03b5=d\u03d5dt\u03b5=d\u03d5dt<br \/>\nWhere,<br \/>\n\u03b5=d(AB)dt=AdBdt\u03b5=d(AB)dt=AdBdt<br \/>\n\u21d2\u03b5=16\u00d710\u22124\u00d70.02=0.32\u00d710\u22124V\u21d2\u03b5=16\u00d710\u22124\u00d70.02=0.32\u00d710\u22124V<br \/>\nResistance of the loop, R=1.6\u03a9R=1.6\u03a9<br \/>\nThe current induced in the loop could be given as:<br \/>\ni=\u03b5Ri=\u03b5R<br \/>\nSubstituting the given values,<br \/>\n\u21d2i=0.32\u00d710\u221241.6=2\u00d710\u22125A\u21d2i=0.32\u00d710\u221241.6=2\u00d710\u22125A<br \/>\nPower dissipated in the loop in the form of heat could be given as:<br \/>\nP=i2RP=i2R<br \/>\n\u21d2P=(2\u00d710\u22125)2\u00d71.6\u21d2P=(2\u00d710\u22125)2\u00d71.6<br \/>\n\u2234P=6.4\u00d710\u221210W\u2234P=6.4\u00d710\u221210W<br \/>\nThe source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.<\/h3>\n<h2>12. A square loop of side<br \/>\n12cm12cm<br \/>\nwith its sides parallel to X and Y axes is moved with a velocity of<br \/>\n18cm18cm<br \/>\nin the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10\u22123Tcm\u2212110\u22123Tcm\u22121 along the negative x-direction (that is it increases by 10\u22123Tcm\u2212110\u22123Tcm\u22121as one moves in the negative x-direction), and it is decreasing in time at the rate of 10\u22123Ts\u2212110\u22123Ts\u22121. Determine the direction and magnitude of the induced current in the loop if its resistance is<br \/>\n4.50m\u03a94.50m\u03a9<\/h2>\n<h3>Ans: We are given,<br \/>\nSide of the square loop,<br \/>\ns=12cm=0.12ms=12cm=0.12m<br \/>\nArea of the square loop, A=0.12\u00d70.12=0.0144m2A=0.12\u00d70.12=0.0144m2<br \/>\nVelocity of the loop,<br \/>\nv=8cm\/s=0.08m\/sv=8cm\/s=0.08m\/s<br \/>\nGradient of the magnetic field along negative x-direction,<br \/>\ndBdx=10\u22123Tcm\u22121=10\u22121Tm\u22121dBdx=10\u22123Tcm\u22121=10\u22121Tm\u22121<br \/>\nAnd, rate of decrease of the magnetic field,<br \/>\ndBdt=10\u22123Ts\u22121dBdt=10\u22123Ts\u22121<br \/>\nResistance of the loop,<br \/>\nR=4.5m\u03a9=4.5\u00d710\u22123\u03a9R=4.5m\u03a9=4.5\u00d710\u22123\u03a9<br \/>\nRate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:<br \/>\nd\u03d5dt=A\u00d7dBdx\u00d7vd\u03d5dt=A\u00d7dBdx\u00d7v<br \/>\n\u21d2d\u03d5dt=144\u00d710\u22124m2\u00d710\u22121\u00d70.08\u21d2d\u03d5dt=144\u00d710\u22124m2\u00d710\u22121\u00d70.08<br \/>\n\u21d2d\u03d5dt=11.52\u00d710\u22125Tm2s\u22121\u21d2d\u03d5dt=11.52\u00d710\u22125Tm2s\u22121<br \/>\nRate of change of the flux due to explicit time variation in field B is given as:<br \/>\nd\u03d5\u2032dt=A\u00d7dBdxd\u03d5\u2032dt=A\u00d7dBdx<br \/>\n\u21d2d\u03d5\u2032dt=144\u00d710\u22124\u00d710\u22123=1.44\u00d710\u22125Tm2s\u22121\u21d2d\u03d5\u2032dt=144\u00d710\u22124\u00d710\u22123=1.44\u00d710\u22125Tm2s\u22121<br \/>\nSince the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:<br \/>\ne=1.44\u00d710\u22125+11.52\u00d710\u22125e=1.44\u00d710\u22125+11.52\u00d710\u22125<br \/>\n=12.96\u00d710\u22125V=12.96\u00d710\u22125V<br \/>\n\u2234Induced current, i=eR\u2234Induced current, i=eR<br \/>\n\u21d2i=12.96\u00d710\u221254.5\u00d710\u22123\u21d2i=12.96\u00d710\u221254.5\u00d710\u22123<br \/>\n\u2234i=2.88\u00d710\u22122A\u2234i=2.88\u00d710\u22122A<br \/>\nTherefore, the direction of the induced current is such that there is an increase in the flux through the loop along the positive z-direction.<\/h3>\n<h2>13. It is desired to measure the magnitude of field between the poles of a powerful loudspeaker magnet. A small flat search coil of area<br \/>\n2cm22cm2<br \/>\nwith 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick<br \/>\n90\u221890\u2218<br \/>\nturn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is<br \/>\n7.5mC7.5mC<br \/>\n. The combined resistance of the coil and the galvanometer is<br \/>\n0.50\u03a90.50\u03a9<br \/>\n. Estimate the field strength of magnet.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nArea of the small flat search coil,<br \/>\nA=2cm2=2\u00d710\u22124m2A=2cm2=2\u00d710\u22124m2<br \/>\nNumber of turns on the coil,<br \/>\nN=25N=25<br \/>\nTotal charge flowing in the coil,<br \/>\nQ=7.5mC=7.5\u00d710\u22123CQ=7.5mC=7.5\u00d710\u22123C<br \/>\nTotal resistance of the coil and galvanometer,<br \/>\nR=0.50\u03a9R=0.50\u03a9<br \/>\nInduced current in the coil,<br \/>\nI=Induced emf(\u03b5)RI=Induced emf(\u03b5)R<br \/>\n\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (1)<br \/>\nInduced<br \/>\nemfemf<br \/>\nis given as:<br \/>\n\u03b5=\u2212Nd\u03d5dt\u03b5=\u2212Nd\u03d5dt<br \/>\n\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (2)<br \/>\nWhere,<br \/>\nd\u03d5=Induced fluxd\u03d5=Induced flux<br \/>\nCombining equations (1) and (2), we get<br \/>\nI=\u2212Nd\u03d5dtRI=\u2212Nd\u03d5dtR<br \/>\nIdt=\u2212NRd\u03d5Idt=\u2212NRd\u03d5<br \/>\n\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (3)<br \/>\nInitial flux through the coil,<br \/>\n\u03d5i=BA\u03d5i=BA<br \/>\nWhere,<br \/>\nB=Magnetic field strengthB=Magnetic field strength<br \/>\nFinal flux through the coil,<br \/>\n\u03d5f=0\u03d5f=0<br \/>\nIntegrating equation (3) on both sides, we have<br \/>\n\u222bIdt=\u2212NR\u222b\u03d5i\u03d5fd\u03d5\u222bIdt=\u2212NR\u222b\u03d5i\u03d5fd\u03d5<br \/>\nBut total charge could be given as,<br \/>\nQ=\u222bIdtQ=\u222bIdt<br \/>\n\u21d2Q=\u2212NR(\u03d5f\u2212\u03d5i)=\u2212NR(\u2212\u03d5i)=+N\u03d5iR\u21d2Q=\u2212NR(\u03d5f\u2212\u03d5i)=\u2212NR(\u2212\u03d5i)=+N\u03d5iR<br \/>\nQ=NBARQ=NBAR<br \/>\n\u21d2B=QRNA\u21d2B=QRNA<br \/>\nSubstituting the given values, we get,<br \/>\n\u21d2B=7.5\u00d710\u22123\u00d70.525\u00d72\u00d710\u22124\u21d2B=7.5\u00d710\u22123\u00d70.525\u00d72\u00d710\u22124<br \/>\n\u2234B=0.75T\u2234B=0.75T<br \/>\nTherefore, the field strength of the magnet is found to be<br \/>\n0.75T0.75T<\/h3>\n<h2>14. Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod<br \/>\n=15cm=15cm<br \/>\n,<br \/>\nB=0.50TB=0.50T<br \/>\n, resistance of the closed loop containing the rod<br \/>\n=9.0m\u03a9=9.0m\u03a9<br \/>\n. Assume the field to be uniform.<br \/>\n(Image will be uploaded soon)<br \/>\nSuppose K is open and the rod is moved with a speed of<br \/>\n12cm\u22c5s\u2212112cm\u22c5s\u22121<br \/>\nin the direction shown. Give the polarity and magnitude of the induced<br \/>\nemfemf<\/h2>\n<h3>Ans: We are given:<br \/>\nLength of the rod,<br \/>\nl=15cm=0.15ml=15cm=0.15m<br \/>\nMagnetic field strength,<br \/>\nB=0.50TB=0.50T<br \/>\nResistance of the closed loop,<br \/>\nR=9m\u03a9=9\u00d710\u22123\u03a9R=9m\u03a9=9\u00d710\u22123\u03a9<br \/>\nInduced<br \/>\nemf=9mVemf=9mV<br \/>\nHere, polarity of the induced<br \/>\nemfemf<br \/>\nis such that end P shows positive while end Q shows negative ends.<br \/>\nSpeed of the rod,<br \/>\nv=12cm\/s=0.12m\/sv=12cm\/s=0.12m\/s<br \/>\nWe know that the induced<br \/>\nemfemf<br \/>\ncould be given as:<br \/>\n\u03b5=Bvl\u03b5=Bvl<br \/>\nSubstituting the given values, we get,<br \/>\n\u03b5=0.5\u00d70.12\u00d70.15\u03b5=0.5\u00d70.12\u00d70.15<br \/>\n\u21d2\u03b5=9\u00d710\u22123v\u21d2\u03b5=9\u00d710\u22123v<br \/>\n\u2234\u03b5=9mV\u2234\u03b5=9mV<br \/>\nTherefore, the magnitude of the induced emf is found to be<br \/>\n\u03b5=9mV\u03b5=9mV<br \/>\nan =d the polarity of the induced emf is such that end P shows positive while end Q shows negative.<\/h3>\n<h2>a) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?<\/h2>\n<h3>Ans: Yes; when key K is closed, excess charge could be maintained by the continuous flow of current. When key K is open, there is excess charge built up at both rod ends but when key K is closed, excess charge is maintained by the continuous flow of current.<\/h3>\n<h2>b) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.<\/h2>\n<h3>Ans: Magnetic force is cancelled by the electric force that is set-up due to the excess charge of opposite nature at both rod ends. There is no net force on the electrons in rod PQ when key K is open and the rod would move uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.<\/h3>\n<h2>c) What is the retarding force on the rod when K is closed?<\/h2>\n<h3>Ans: We know that the retarding force exerted on the rod could be given by,<br \/>\nF=IBlF=IBl<br \/>\nWhere,<br \/>\nI=Current flowing through the rodI=Current flowing through the rod<br \/>\nSubstituting the given values, we get,<br \/>\nI=eR=9\u00d710\u221239\u00d710\u22123=1AI=eR=9\u00d710\u221239\u00d710\u22123=1A<br \/>\n\u21d2F=1\u00d70.5\u00d70.15\u21d2F=1\u00d70.5\u00d70.15<br \/>\n\u2234F=75\u00d710\u22123N\u2234F=75\u00d710\u22123N<br \/>\nTherefore, we found the retarding force on the rod when the key K is closed to be,<br \/>\nF=75\u00d710\u22123NF=75\u00d710\u22123N<\/h3>\n<h2>d) How much power is required (by an external agent) to keep the rod moving at the same speed<br \/>\n(=12cm\u22c5s\u22121)(=12cm\u22c5s\u22121)<br \/>\nwhen K is closed? How much power is required when K is open?<\/h2>\n<h3>Ans: We are given:<br \/>\nSpeed of the rod,<br \/>\nv=12cm\/s=0.12m\/sv=12cm\/s=0.12m\/s<br \/>\nNow, power could be given as:<br \/>\nP=FvP=Fv<br \/>\nSubstituting the given values, we get,<br \/>\n\u21d2P=75\u00d710\u22123\u00d70.12\u21d2P=75\u00d710\u22123\u00d70.12<br \/>\n\u21d2P=9\u00d710\u22123W\u21d2P=9\u00d710\u22123W<br \/>\n\u2234P=9mW\u2234P=9mW<br \/>\nTherefore, we found the power that is required (by an external agent) to keep the rod moving at the same speed<br \/>\n(=12cm\u22c5s\u22121)(=12cm\u22c5s\u22121)<br \/>\nwhen K is closed to be<br \/>\nP=9mWP=9mW<br \/>\nand when key K is open, no power is expended.<\/h3>\n<h2>e) How much power is dissipated as heat in the closed circuit? What is the source of this power?<\/h2>\n<h3>Ans: We know that,<br \/>\nPower dissipated as heat , P=I2RPower dissipated as heat , P=I2R<br \/>\n\u21d2P=(1)2\u00d79\u00d710\u22123\u21d2P=(1)2\u00d79\u00d710\u22123<br \/>\n\u2234P=9mW\u2234P=9mW<br \/>\nThe power dissipated as heat in the closed circuit is found to be<br \/>\nP=9mWP=9mW<br \/>\nand the source of this power is found to be an external agent.<\/h3>\n<h2>f) What is the induced emfemf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?<\/h2>\n<h3>Ans: In this case, no<br \/>\nemfemf<br \/>\nwould be induced in the coil because the motion of the rod does not cut across the field lines.<\/h3>\n<h2>15. An air-cored solenoid with length<br \/>\n30cm30cm<br \/>\n, area of cross-section<br \/>\n25cm225cm2<br \/>\nand number of turns<br \/>\n500500<br \/>\n, carries a current of<br \/>\n2.5A2.5A<br \/>\n. The current is suddenly switched off in a brief time of<br \/>\n10\u22123s10\u22123s<br \/>\n. How much is the average back<br \/>\nemfemf<br \/>\ninduced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nLength of the solenoid,<br \/>\nl=30cm=0.3ml=30cm=0.3m<br \/>\nArea of cross-section,<br \/>\nA=25cm2=25\u00d710\u22124m2A=25cm2=25\u00d710\u22124m2<br \/>\nNumber of turns on the solenoid,<br \/>\nN=500N=500<br \/>\nCurrent in the solenoid,<br \/>\nI=2.5AI=2.5A<br \/>\nCurrent flows for time,<br \/>\nt=10\u22123st=10\u22123s<br \/>\nAverage back<br \/>\nemfemf<br \/>\ne=d\u03d5dte=d\u03d5dt<br \/>\n\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026. (1)<br \/>\nWhere,<br \/>\nd\u03d5=Change in flux=NABd\u03d5=Change in flux=NAB<br \/>\n\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (2)<br \/>\nWhere,<br \/>\nB=Magnetic field strength=\u03bc0NIlB=Magnetic field strength=\u03bc0NIl<br \/>\n\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (3)<br \/>\nWhere,<br \/>\n\u03bc0=Permeability of free space=4\u03c0\u00d710\u22127TmA\u22121\u03bc0=Permeability of free space=4\u03c0\u00d710\u22127TmA\u22121<br \/>\nSubstituting equations (2) and (3) in equation (1), we get,<br \/>\n\u03b5=\u03bc0N2IAlt\u03b5=\u03bc0N2IAlt<br \/>\n\u21d2\u03b5=4\u03c0\u00d710\u22127\u00d7(500)2\u00d72.5\u00d725\u00d710\u221240.3\u00d710\u22123\u21d2\u03b5=4\u03c0\u00d710\u22127\u00d7(500)2\u00d72.5\u00d725\u00d710\u221240.3\u00d710\u22123<br \/>\n\u2234\u03b5=6.5V\u2234\u03b5=6.5V<br \/>\nHence, the average back<br \/>\nemfemf<br \/>\ninduced in the solenoid is found to be<br \/>\n6.5V6.5V<\/h3>\n<h2>16.<br \/>\na) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in below figure.<br \/>\n(Image will be uploaded soon)<\/h2>\n<h3>Ans: Consider a small element dydy<br \/>\nin the loop at a distance y from the long straight wire (as shown in the given figure).<br \/>\n(Image will be uploaded soon)<br \/>\nMagnetic flux associated with element<br \/>\ndy,d\u03d5=BdAdy,d\u03d5=BdA<br \/>\nWhere,<br \/>\ndA=Area of element dy=a dydA=Area of element dy=a dy<br \/>\nB=magnetic field at distance y =\u03bc0I2\u03c0yB=magnetic field at distance y =\u03bc0I2\u03c0y<br \/>\nI=Current in the wireI=Current in the wire<br \/>\n\u03bc0=Permeability of free space=4\u03c0\u00d710\u22127H\/m\u03bc0=Permeability of free space=4\u03c0\u00d710\u22127H\/m<br \/>\nCarrying out the substitutions accordingly, we get,<br \/>\n\u21d2d\u03d5=\u03bc0Ia2\u03c0dyy\u21d2d\u03d5=\u03bc0Ia2\u03c0dyy<br \/>\n\u21d2\u03d5=\u03bc0Ia2\u03c0\u222bdyy\u21d2\u03d5=\u03bc0Ia2\u03c0\u222bdyy<br \/>\nNow, the limit of y will be from x to<br \/>\na+xa+x<br \/>\n, on applying the limits we get,<br \/>\n\u21d2\u03d5=\u03bc0Ia2x\u222bxa+xdyy\u21d2\u03d5=\u03bc0Ia2x\u222bxa+xdyy<br \/>\n\u21d2\u03d5=\u03bc0Ia2\u03c0[logey]a+xx\u21d2\u03d5=\u03bc0Ia2\u03c0[logey]xa+x<br \/>\n\u21d2\u03d5=\u03bc0Ia2\u03c0loge(a+xx)\u21d2\u03d5=\u03bc0Ia2\u03c0loge(a+xx)<br \/>\nFor mutual inductance M, the flux could be given as:<br \/>\n\u03d5=MI\u03d5=MI<br \/>\n\u21d2MI=\u03bc0Ia2\u03c0loge(ax+1)\u21d2MI=\u03bc0Ia2\u03c0loge(ax+1)<br \/>\n\u2234M=\u03bc0a2\u03c0loge(ax+1)\u2234M=\u03bc0a2\u03c0loge(ax+1)<br \/>\nTherefore, the expression for the mutual inductance between the given long straight wire and the square loop of side a is found to be,<br \/>\nM=\u03bc0a2\u03c0loge(ax+1)M=\u03bc0a2\u03c0loge(ax+1)<\/h3>\n<h2>b) Now assume that the straight wire carries a current of<br \/>\n50A50A<br \/>\nand the loop is moved to the right with a constant velocity,<br \/>\nv=10m\/sv=10m\/s<br \/>\n. Calculate the induced<br \/>\nemfemf<br \/>\nin the loop at the instant when<br \/>\nx=0.2mx=0.2m<br \/>\n. Take<br \/>\na=0.1ma=0.1m<br \/>\nand assume that the loop has a large resistance.<\/h2>\n<h3>Ans: We know that, the<br \/>\nEmfEmf<br \/>\ninduced in the loop,<br \/>\n\u03b5=B\u2032av=(\u03bc0I2\u03c0x)av\u03b5=B\u2032av=(\u03bc0I2\u03c0x)av<br \/>\nWe are given the following,<br \/>\nI=50AI=50A<br \/>\nx=0.2mx=0.2m<br \/>\na=0.1ma=0.1m<br \/>\nv=10m\/sv=10m\/s<br \/>\nOn substituting the given values into the equation, we get,<br \/>\n\u03b5=4\u03c0\u00d710\u22127\u00d750\u00d70.1\u00d7102\u03c0\u00d70.2\u03b5=4\u03c0\u00d710\u22127\u00d750\u00d70.1\u00d7102\u03c0\u00d70.2<br \/>\n\u2234\u03b5=5\u00d710\u22125V\u2234\u03b5=5\u00d710\u22125V<br \/>\nTherefore, induced<br \/>\nemfemf<br \/>\nin the loop at the given instant is found to be,<br \/>\n\u03b5=5\u00d710\u22125V\u03b5=5\u00d710\u22125V<\/h3>\n<h2>17. A line charge<br \/>\n\u03bb\u03bb<br \/>\nper unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (figure). A uniform magnetic field extends over a circular region within the rim. It is given by,<br \/>\nA=\u2212B0k(r\u2264a;a&lt;R)A=\u2212B0k(r\u2264a;a&lt;R)<br \/>\nA=0 (otherwise)A=0 (otherwise)<br \/>\nWhat is the angular velocity of the wheel after the field is suddenly switched off?<br \/>\n(Image will be uploaded soon)<\/h2>\n<h3>Ans: We know that, the Line charge per unit length<br \/>\n=\u03bb=Total chargeLength=Q2\u03c0r=\u03bb=Total chargeLength=Q2\u03c0r<br \/>\nWhere,<br \/>\nr = Distance of the point within the wheel<br \/>\nMass of the wheel = M<br \/>\nRadius of the wheel = R<br \/>\nMagnetic field,<br \/>\nB\u20d7 =\u2212B0kB\u2192=\u2212B0k<br \/>\nAt distance r, the magnetic force would be balanced by the centripetal force i.e.,<br \/>\nBQv=Mv2rBQv=Mv2r<br \/>\nWhere,<br \/>\nv=linear velocity of the wheelv=linear velocity of the wheel<br \/>\n\u21d2B2\u03c0r\u03bb=Mvr\u21d2B2\u03c0r\u03bb=Mvr<br \/>\n\u21d2v=B2\u03c0\u03bbr2M\u21d2v=B2\u03c0\u03bbr2M<br \/>\n\u2234Angular velocity,\u03c9=vR=B2\u03c0\u03bbr2M\u2234Angular velocity,\u03c9=vR=B2\u03c0\u03bbr2M<br \/>\nr\u2264ar\u2264a<br \/>\n, and<br \/>\na&lt;Ra&lt;R<br \/>\nwe would get:<br \/>\n\u03c9=2\u03c0B0a2\u03bbMRk\u03c9=2\u03c0B0a2\u03bbMRk<br \/>\nTherefore, we found the angular velocity of the wheel after the field is suddenly switched off to be given as,<br \/>\n\u03c9=2\u03c0B0a2\u03bbMRk\u03c9=2\u03c0B0a2\u03bbMRk<\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Class 12 Physics NCERT book solutions for Chapter 6 &#8211; Electromagnetic Induction Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":118347,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-120392","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/120392","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=120392"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/120392\/revisions"}],"predecessor-version":[{"id":120404,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/120392\/revisions\/120404"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/118347"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=120392"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=120392"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=120392"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}