{"id":118547,"date":"2022-04-27T17:05:29","date_gmt":"2022-04-27T11:35:29","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=118547"},"modified":"2022-04-27T17:05:29","modified_gmt":"2022-04-27T11:35:29","slug":"chapter-4-moving-charges-and-magnetism-questions-and-answers-ncert-solutions-for-class-12-physics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-4-moving-charges-and-magnetism-questions-and-answers-ncert-solutions-for-class-12-physics","title":{"rendered":"Chapter 4 &#8211; Moving Charges And Magnetism Questions and Answers: NCERT Solutions for Class 12 Physics"},"content":{"rendered":"<h2>1. A Circular Coil of Wire Consisting of 100100 Turns, Each of Radius 8.0cm8.0cm Carries a Current of 0.40A0.40A. What is the Magnitude of the Magnetic Field B at the Centre of the Coil?<\/h2>\n<h3>Ans: We are given:<br \/>\nNumber of turns on the circular coil, n=100n=100<br \/>\nRadius of each turn, r=8.0cm=0.08mr=8.0cm=0.08m<br \/>\nCurrent flowing in the coil is given to be, I=0.4AI=0.4A<br \/>\nWe know the expression for magnetic field at the centre of the coil as,<br \/>\n|B|=\u03bc04\u03c02\u03c0nIr|B|=\u03bc04\u03c02\u03c0nIr<br \/>\nWhere, \u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121is the permeability of free space.<br \/>\nOn substituting the given values we get,<br \/>\n|B|=4\u03c0\u00d710\u22127\u00d72\u03c0\u00d7100\u00d70.44\u03c0\u00d70.08|B|=4\u03c0\u00d710\u22127\u00d72\u03c0\u00d7100\u00d70.44\u03c0\u00d70.08<br \/>\n\u21d2|B|=3.14\u00d710\u22124T\u21d2|B|=3.14\u00d710\u22124T<br \/>\nClearly, the magnitude of the magnetic field is found to be 3.14\u00d710\u22124T3.14\u00d710\u22124T.<\/h3>\n<h2>2. A Long Straight Wire Carries a Current of 35A35A. What Is the Magnitude of Field B at a Point 20cm from the Wire?<\/h2>\n<h3>Ans: We are given the following:<br \/>\nCurrent in the wire, I=35AI=35A<br \/>\nDistance of the given point from the wire, r=20cm=0.2mr=20cm=0.2m<br \/>\nWe know the expression for magnetic field as,<br \/>\nB=\u03bc04\u03c02IrB=\u03bc04\u03c02Ir<br \/>\nWhere, \u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121is the permeability of free space.<br \/>\nOn substituting the given values, we get,<br \/>\nB=4\u03c0\u00d710\u22127\u00d72\u00d7354\u03c0\u00d70.2B=4\u03c0\u00d710\u22127\u00d72\u00d7354\u03c0\u00d70.2<br \/>\n\u21d2B=3.5\u00d710\u22125T\u21d2B=3.5\u00d710\u22125T<br \/>\nThus, we found the magnitude of the magnetic field at the given point to be 3.5\u00d710\u22125T3.5\u00d710\u22125T.<\/h3>\n<h2>3. A Long Straight Wire in the Horizontal Plane Carries a Current of 50A50A in North to South Direction. Give the Magnitude and Direction of B at a Point 2.5m2.5m East of the Wire.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nThe current in the wire, I=50AI=50A<br \/>\nThe distance of the given point from the wire, r=2.5mr=2.5m<br \/>\n(Image will be Uploaded Soon)<br \/>\nWe have the expression for magnetic field as,<br \/>\nB=2\u03bc0I4\u03c0rB=2\u03bc0I4\u03c0r<br \/>\nWhere, \u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121is the permeability of free space.<br \/>\nSubstituting the given values, we get,<br \/>\nB=4\u03c0\u00d710\u22127\u00d72\u00d7504\u03c0\u00d72.5B=4\u03c0\u00d710\u22127\u00d72\u00d7504\u03c0\u00d72.5<br \/>\n\u21d2B=4\u00d710\u22126T\u21d2B=4\u00d710\u22126T<br \/>\nNow from Maxwell\u2019s right hand thumb rule we have the direction of the magnetic field at the given point B to be vertically upward.<\/h3>\n<h2>4. A Horizontal Overhead Power Line Carries a Current of 90A90A in East to West Direction. What is the Magnitude and Direction of the Magnetic Field Due to the Current 1.5m1.5m Below the Line?<\/h2>\n<h3>Ans: We are given the following:<br \/>\nCurrent in the power line, I=90AI=90A<br \/>\nDistance of the mentioned point below the power line, r=1.5mr=1.5m<br \/>\nNow, we have the expression for magnetic field as,<br \/>\nB=2\u03bc0I4\u03c0rB=2\u03bc0I4\u03c0r<br \/>\nWhere, \u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121is the permeability of free space.<br \/>\nOn substituting the given values, we get,<br \/>\nB=4\u03c0\u00d710\u22127\u00d72\u00d7904\u03c0\u00d71.5B=4\u03c0\u00d710\u22127\u00d72\u00d7904\u03c0\u00d71.5<br \/>\n\u21d2B=1.2\u00d710\u22125T\u21d2B=1.2\u00d710\u22125T<br \/>\nWe found the magnitude of the magnetic field to be 1.2\u00d710\u22125T1.2\u00d710\u22125Tand it will be directed towards south as per Maxwell\u2019s right hand thumb rule.<\/h3>\n<h2>5. What is the Magnitude of Magnetic Force Per Unit Length on a Wire Carrying a Current of 8 and Making an Angle of 30\u221830\u2218 with the Direction of a Uniform Magnetic Field of 0.15T0.15T?<\/h2>\n<h3>Ans: Given that,<br \/>\nCurrent in the wire, I=8AI=8A<br \/>\nMagnitude of the uniform magnetic field, B=0.15TB=0.15T<br \/>\nAngle between the wire and magnetic field, \u03b8=30\u2218\u03b8=30\u2218<br \/>\nWe have the expression for magnetic force per unit length on the wire as,<br \/>\nF=BIsin\u03b8F=BIsin\u2061\u03b8<br \/>\nSubstituting the given values, we get,<br \/>\nF=0.15\u00d78\u00d71\u00d7sin30\u2218F=0.15\u00d78\u00d71\u00d7sin\u206130\u2218<br \/>\n\u21d2F=0.6Nm\u22121\u21d2F=0.6Nm\u22121<br \/>\nThus, the magnetic force per unit length on the wire is found to be 0.6Nm\u221210.6Nm\u22121<\/h3>\n<h2>6. A 3.0cm3.0cm Wire Carrying a Current of 10A10A is Placed Inside a Solenoid Perpendicular to Its Axis. The Magnetic Field Inside the Solenoid Is Given to Be 0.27T0.27T. What is the Magnetic Force on the Wire?<\/h2>\n<h3>Ans: We are given the following,<br \/>\nLength of the wire, l=3cm=0.03ml=3cm=0.03m<br \/>\nCurrent flowing in the wire, I=10AI=10A<br \/>\nMagnetic field, B=0.27TB=0.27T<br \/>\nAngle between the current and magnetic field, \u03b8=90\u2218\u03b8=90\u2218<br \/>\n(Since the magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis)<br \/>\nThe magnetic force exerted on the wire is given as,<br \/>\nF=BIlsin\u03b8F=BIlsin\u2061\u03b8<br \/>\nSubstituting the given values,<br \/>\nF=0.27\u00d710\u00d70.03sin90\u2218F=0.27\u00d710\u00d70.03sin\u206190\u2218<br \/>\n\u21d2F=8.1\u00d710\u22122N\u21d2F=8.1\u00d710\u22122N<br \/>\nClearly, the magnetic force on the wire is found to be 8.1\u00d710\u22122N8.1\u00d710\u22122N. The direction of the force can be obtained from Fleming\u2019s left-hand rule.<\/h3>\n<h2>7. Two Long and Parallel Straight Wires A and B Carrying Currents of 8.0A8.0Aand 5.0A5.0A in the Same Direction are Separated by a Distance of 4.0cm4.0cm. Estimate the Force on a 10cm10cm Section of Wire A.<\/h2>\n<h3>Ans: We are given:<br \/>\nCurrent flowing in wire A, IA=8.0AIA=8.0A<br \/>\nCurrent flowing in wire B, IB=5.0AIB=5.0A<br \/>\nDistance between the two wires, r=4.0cm=0.04mr=4.0cm=0.04m<br \/>\nLength of a section of wire A, l=10cm=0.1ml=10cm=0.1m<br \/>\nForce exerted on length ll due to the magnetic field is given as,<br \/>\nB=2\u03bc0IAIBl4\u03c0rB=2\u03bc0IAIBl4\u03c0r<br \/>\nWhere, \u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121is the permeability of free space.<br \/>\nOn substituting the given values, we get,<br \/>\nB=4\u03c0\u00d710\u22127\u00d72\u00d78\u00d75\u00d70.14\u03c0\u00d70.04B=4\u03c0\u00d710\u22127\u00d72\u00d78\u00d75\u00d70.14\u03c0\u00d70.04<br \/>\n\u21d2B=2\u00d710\u22125N\u21d2B=2\u00d710\u22125N<br \/>\nThe magnitude of force is 2\u00d710\u22125N2\u00d710\u22125N. This is an attractive force that is normal to A towards B because the direction of the currents in the wires is the same.<\/h3>\n<h2>8. A Closely Wound Solenoid 80cm80cm Long has 55 Layers of Windings of 400400 Turns Each. The Diameter of the Solenoid is 1.8cm1.8cm. If the Current Carried is 8.0A8.0A, Estimate the Magnitude of B Inside the Solenoid Near its Centre.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nLength of the solenoid, l=80cm=0.8ml=80cm=0.8m<br \/>\nSince there are five layers of windings of 400 turns each on the solenoid.<br \/>\nTotal number of turns on the solenoid would be, N=5\u00d7400=2000N=5\u00d7400=2000<br \/>\nDiameter of the solenoid, D=1.8cm=0.018mD=1.8cm=0.018m<br \/>\nCurrent carried by the solenoid, I=8.0AI=8.0A<br \/>\nWe have the magnitude of the magnetic field inside the solenoid near its centre given by the relation,<br \/>\nB=\u03bc0NIlB=\u03bc0NIl<br \/>\nWhere, \u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121is the permeability of free space.<br \/>\nOn substituting the given values we get,<br \/>\nB=4\u03c0\u00d710\u22127\u00d72000\u00d780.8B=4\u03c0\u00d710\u22127\u00d72000\u00d780.8<br \/>\n\u21d2B=2.512\u00d710\u22122T\u21d2B=2.512\u00d710\u22122T<br \/>\nClearly, the magnitude of the magnetic field inside the solenoid near its centre is found to be 2.512\u00d710\u22122T2.512\u00d710\u22122T.<\/h3>\n<h2>9. A Square Coil of Side 10cm10cm Consists of 20 Turns and Carries a Current of 12A12A. The Coil Is Suspended Vertically and the Normal to the Plane of the Coil Makes an Angle of 30\u221830\u2218 with the Direction of a Uniform Horizontal Magnetic Field of Magnitude 0.80T0.80T. What is the Magnitude of Torque Experienced by the Coil?<\/h2>\n<h3>Ans: We are given the following:<br \/>\nLength of a side of the square coil, l=10cm=0.1ml=10cm=0.1m<br \/>\nArea of the square, A=l2=(0.1)2=0.01m2A=l2=(0.1)2=0.01m2<br \/>\nCurrent flowing in the coil, I=12AI=12A<br \/>\nNumber of turns on the coil, n=20n=20<br \/>\nAngle made by the plane of the coil with magnetic field, \u03b8=30\u2218\u03b8=30\u2218<br \/>\nStrength of magnetic field, B=0.80TB=0.80T<br \/>\nMagnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,<br \/>\n\u03c4=nIABsin\u03b8\u03c4=nIABsin\u2061\u03b8<br \/>\nSubstituting the given values, we get,<br \/>\n\u03c4=20\u00d70.8\u00d712\u00d70.01\u00d7sin30\u2218\u03c4=20\u00d70.8\u00d712\u00d70.01\u00d7sin\u206130\u2218<br \/>\n\u21d2\u03c4=0.96Nm\u21d2\u03c4=0.96Nm<br \/>\nThus, the magnitude of the torque experienced by the coil is 0.96 N m.<\/h3>\n<h2>10. Two Moving Coil Meters, M1M1 and M2M2 Have the Following Particulars:<br \/>\nR1=10\u03a9R1=10\u03a9 , N1=30N1=30, A1=3.6\u00d710\u22123m2A1=3.6\u00d710\u22123m2 , B1=0.25TB1=0.25T,R2=14\u03a9R2=14\u03a9 ,N2=42N2=42A2=1.8\u00d710\u22123m2A2=1.8\u00d710\u22123m2 , B2=0.50TB2=0.50T<br \/>\n(The spring constants are identical for the meters).<br \/>\nDetermine the Ratio of:<br \/>\na) Current Sensitivity of M2 and M1M2 and M1<\/h2>\n<h3>Ans: We are given:<br \/>\nFor moving coil meter M1M1,<br \/>\nResistance, R1=10\u03a9R1=10\u03a9<br \/>\nNumber of turns, N1=30N1=30<br \/>\nArea of cross-section, A1=3.6\u00d710\u22123m2A1=3.6\u00d710\u22123m2<br \/>\nMagnetic field strength, B1=0.25TB1=0.25T<br \/>\nSpring constant, K1=KK1=K<br \/>\nFor moving coil meter M2M2:<br \/>\nResistance, R2=14\u03a9R2=14\u03a9<br \/>\nNumber of turns, N2=42N2=42<br \/>\nArea of cross-section, A2=1.8\u00d710\u22123m2A2=1.8\u00d710\u22123m2<br \/>\nMagnetic field strength, B2=0.50TB2=0.50T<br \/>\nSpring constant, K2=KK2=K<br \/>\nCurrent sensitivity of M1M1 is given as:<br \/>\nIS1=N1B1A1K1IS1=N1B1A1K1<br \/>\nAnd, current sensitivity of M2M2 is given as:<br \/>\nIS2=N2B2A2K2IS2=N2B2A2K2<br \/>\nOn taking the ratio, we get,<br \/>\n\u21d2IS2IS1=N2B2A2K2N1B1A1K1\u21d2IS2IS1=N2B2A2K2N1B1A1K1<br \/>\nSubstituting the values we get,<br \/>\n\u21d2IS2IS1=42\u00d70.5\u00d71.8\u00d710\u22123\u00d710\u00d7K14\u00d730\u00d70.25\u00d73.6\u00d710\u22123\u00d7K\u21d2IS2IS1=42\u00d70.5\u00d71.8\u00d710\u22123\u00d710\u00d7K14\u00d730\u00d70.25\u00d73.6\u00d710\u22123\u00d7K<br \/>\n\u21d2IS2IS1=1.4\u21d2IS2IS1=1.4<br \/>\nTherefore, the ratio of current sensitivity of M2 and M1M2 and M1 is 1.4.<\/h3>\n<h2>b) Voltage Sensitivity of M2 and M1M2 and M1<\/h2>\n<h3>Ans: Voltage sensitivity for M2M2is given is:<br \/>\nVS2=N2B2A2K2R2VS2=N2B2A2K2R2<br \/>\nAnd, voltage sensitivity for M1M1is given as:<br \/>\nVS1=N1B1A1K1R1VS1=N1B1A1K1R1<br \/>\nOn taking the ratio we get,<br \/>\n\u21d2VS2VS1=N2B2A2K1R1K2R2N1B1A1\u21d2VS2VS1=N2B2A2K1R1K2R2N1B1A1<br \/>\nSubstituting the given values, we get,<br \/>\n\u21d2VS2VS1=42\u00d70.5\u00d71.8\u00d710\u22123\u00d710\u00d7KK\u00d714\u00d730\u00d70.25\u00d73.6\u00d710\u22123=1\u21d2VS2VS1=42\u00d70.5\u00d71.8\u00d710\u22123\u00d710\u00d7KK\u00d714\u00d730\u00d70.25\u00d73.6\u00d710\u22123=1<br \/>\nThus, the ratio of voltage sensitivity of M2 and M1M2 and M1is 1.<\/h3>\n<h2>11. In a Chamber, a Uniform Magnetic Field of 6.5G(1G=10\u22124T)6.5G(1G=10\u22124T)is Maintained. An Electron Is Shot Into the Field With a Speed Of 4.8\u00d7106ms\u221214.8\u00d7106ms\u22121 Normal to the Field. Explain Why the Path of the Electron is a Circle. Determine the Radius of the Circular Orbit.(e=1.6\u00d710\u221219C,me=9.1\u00d710\u221231kg)(e=1.6\u00d710\u221219C,me=9.1\u00d710\u221231kg)<\/h2>\n<h3>Ans: Magnetic field strength, B=6.5G=6.5\u00d710\u22124TB=6.5G=6.5\u00d710\u22124T<br \/>\nSpeed of the electron, V=4.8\u00d7106m\/sV=4.8\u00d7106m\/s<br \/>\nCharge on the electron, e=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C<br \/>\nMass of the electron, me=9.1\u00d710\u221231kgme=9.1\u00d710\u221231kg<br \/>\nAngle between the shot electron and magnetic field, \u03b8=90\u2218\u03b8=90\u2218<br \/>\nMagnetic force exerted on the electron in the magnetic field could be given as:<br \/>\nF=evBsin\u03b8F=evBsin\u2061\u03b8<br \/>\nThis force provides centripetal force to the moving electron and hence, the electron starts moving in a circular path of radius r.<br \/>\nHence, centripetal force exerted on the electron would be,<br \/>\nFC=mv2rFC=mv2r<br \/>\nHowever, we know that in equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,<br \/>\nFC=FFC=F<br \/>\n\u21d2mv2r=evBsin\u03b8\u21d2mv2r=evBsin\u2061\u03b8<br \/>\n\u21d2r=mvBesin\u03b8\u21d2r=mvBesin\u2061\u03b8<br \/>\nSubstituting the given values we get,<br \/>\n\u21d2r=9.1\u00d710\u221231\u00d74.8\u00d71066.5\u00d710\u22124\u00d71.6\u00d710\u221219\u00d7sin90\u2218\u21d2r=9.1\u00d710\u221231\u00d74.8\u00d71066.5\u00d710\u22124\u00d71.6\u00d710\u221219\u00d7sin\u206190\u2218<br \/>\n\u21d2r=4.2cm\u21d2r=4.2cm<br \/>\nClearly, we found the radius of the circular orbit to be 4.2cm.<\/h3>\n<h2>12. In Exercise 4.11 Obtain the Frequency of Revolution of the Electron in Its Circular Orbit. Does the Answer Depend on the Speed of the Electron? Explain.<\/h2>\n<h3>Ans: We are given the following:<br \/>\nMagnetic field strength, B=6.5\u00d710\u22124TB=6.5\u00d710\u22124T<br \/>\nCharge of the electron, e=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C<br \/>\nMass of the electron, me=9.1\u00d710\u221231kgme=9.1\u00d710\u221231kg<br \/>\nVelocity of the electron, v=4.8\u00d7106m\/sv=4.8\u00d7106m\/s<br \/>\nRadius of the orbit, r=4.2cm=0.042mr=4.2cm=0.042m<br \/>\nFrequency of revolution of the electron \u03bd\u03bd<br \/>\nAngular frequency of the electron \u03c9=2\u03c0\u03b8\u03c9=2\u03c0\u03b8<br \/>\nVelocity of the electron is related to the angular frequency as:<br \/>\nv=r\u03c9v=r\u03c9<br \/>\nIn the circular orbit, the magnetic force on the electron provides the centripetal force. Hence,<br \/>\nevB=mv2revB=mv2r<br \/>\n\u21d2eB=mr(r\u03c9)=mr(r2\u03c0\u03bd)\u21d2eB=mr(r\u03c9)=mr(r2\u03c0\u03bd)<br \/>\n\u21d2\u03bd=6.5\u00d710\u22124\u00d71.6\u00d710\u2212192\u00d73.14\u00d79.1\u00d710\u221231\u21d2\u03bd=6.5\u00d710\u22124\u00d71.6\u00d710\u2212192\u00d73.14\u00d79.1\u00d710\u221231\u2234\u03bd=18.2\u00d7106Hz\u224818MHz\u2234\u03bd=18.2\u00d7106Hz\u224818MHz<br \/>\nThus, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.<\/h3>\n<h2>13.<br \/>\na) A Circular Coil of 30 Turns and Radius 8.0cm8.0cm Carrying a Current of 6.0A6.0A is Suspended Vertically in a Uniform Horizontal Magnetic Field of Magnitude 1.0T1.0T The Field Lines Make an Angle of 60\u221860\u2218 with the Normal of the Coil. Calculate the Magnitude of the Counter Torque That Must Be Applied to Prevent the Coil from Turning.<\/h2>\n<h3>Ans: Number of turns on the circular coil, n=30n=30<br \/>\nRadius of the coil, r=8.0cm=0.08mr=8.0cm=0.08m<br \/>\nArea of the coil, A=\u03c0r2=\u03c0(0.08)2=0.0201m2A=\u03c0r2=\u03c0(0.08)2=0.0201m2<br \/>\nCurrent flowing in the coil is given to be,\u00a0 I=6.0AI=6.0A<br \/>\nMagnetic field strength, B=1TB=1T<br \/>\nAngle between the field lines and normal with the coil surface, \u03b8=60\u2218\u03b8=60\u2218<br \/>\nThe coil will turn when it experiences a torque in the magnetic field. The counter torque applied to prevent the coil from turning is given by the relation, \u03c4=nIABsin\u03b8\u03c4=nIABsin\u2061\u03b8<br \/>\n\u21d2\u03c4=30\u00d76\u00d71\u00d70.0201\u00d7sin60\u2218\u21d2\u03c4=30\u00d76\u00d71\u00d70.0201\u00d7sin\u206160\u2218<br \/>\n\u21d2\u03c4=3.133Nm\u21d2\u03c4=3.133Nm<\/h3>\n<h2>b) Would Your Answer Change, If the Circular Coil in (a) Were Replaced by a Planar Coil of Some Irregular Shape that Encloses the Same Area? (all Other Particulars Are Also Unaltered).<\/h2>\n<h3>Ans: From the part(a) we could infer that the magnitude of the applied torque is not dependent on the shape of the coil.<br \/>\nOn the other hand, it is dependent on the area of the coil.<br \/>\nThus, we could say that the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.<\/h3>\n<h2>14. Two Concentric Circular Coils X and Y Radii 16 cm and 10 cm, Respectively, Lie in the Same Vertical Plane Containing the North to South Direction. Coil X has 20 Turns and Carries a Current of 16 A; Coil Y has 25 Turns and Carries a Current Of 18 A. The Sense of the Current in X is Anticlockwise, and Clockwise in Y, for an Observer Looking at the Coils Facing West. Give the Magnitude and Direction of the Net Magnetic Field Due to the Coils at Their Centre.<\/h2>\n<h3>Ans: We are given,<br \/>\nRadius of coil X, r1=16cm=0.16mr1=16cm=0.16m<br \/>\nRadius of coil Y, r2=10cm=0.1mr2=10cm=0.1m<br \/>\nNumber of turns of on coil X,n1=20n1=20<br \/>\nNumber of turns of on coil Y,n2=25n2=25<br \/>\nCurrent in coil X, I1=16AI1=16A<br \/>\nCurrent in coil Y,I2=18AI2=18A<br \/>\nMagnetic field due to coil X at their centre is given by the relation,<br \/>\nB1=\u03bc0n1I12r1B1=\u03bc0n1I12r1<br \/>\nWhere, Permeability of free space, \u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121<br \/>\nB1=4\u03c0\u00d710\u22127\u00d720\u00d7162\u00d70.16B1=4\u03c0\u00d710\u22127\u00d720\u00d7162\u00d70.16<br \/>\n\u21d2B1=4\u03c0\u00d710\u22124T\u21d2B1=4\u03c0\u00d710\u22124T(towards East)<br \/>\nMagnetic field due to coil Y at their centre is given by the relation,<br \/>\nB2=\u03bc0n2I22r2B2=\u03bc0n2I22r2<br \/>\n\u21d2B2=4\u03c0\u00d710\u22127\u00d725\u00d7182\u00d70.10\u21d2B2=4\u03c0\u00d710\u22127\u00d725\u00d7182\u00d70.10<br \/>\n\u21d2B2=9\u03c0\u00d710\u22124T\u21d2B2=9\u03c0\u00d710\u22124T(towards West)<br \/>\nClearly, net magnetic field could be obtained as:<br \/>\nB=B2\u2212B1=9\u03c0\u00d710\u22124\u22124\u03c0\u00d710\u22124B=B2\u2212B1=9\u03c0\u00d710\u22124\u22124\u03c0\u00d710\u22124<br \/>\n\u21d2B=1.57\u00d710\u22123T\u21d2B=1.57\u00d710\u22123T(towards West)<\/h3>\n<h2>15. A magnetic field of 100G100G(where, 1G=10\u22124T)(where, 1G=10\u22124T) is required which is uniform in a region of linear dimension about 10cm10cm and area of cross-section about10\u22123m210\u22123m2. The maximum current carrying capacity of a given coil of wire is 15A15A and the number of turns per unit length that can be wound a core is at most 1000 turns per m1000 turns per m. Suggest some appropriate design particulars to a solenoid for the required purpose. Assume the core is not ferromagnetic.<\/h2>\n<h3>Ans: We are given,<br \/>\nMagnetic field strength,B=100G=100\u00d710\u22124TB=100G=100\u00d710\u22124T<br \/>\nNumber of turns per unit length,n=1000turns per mn=1000turns per m<br \/>\nCurrent flowing in the coil,I=15AI=15A<br \/>\nPermeability of free space, \u03bc0=4\u03c0\u00d710\u22127TmA\u22121\u03bc0=4\u03c0\u00d710\u22127TmA\u22121<br \/>\nMagnetic field is given the relation,<br \/>\nB=\u03bc0nIB=\u03bc0nI<br \/>\n\u21d2nI=B\u03bc0=100\u00d710\u221244\u03c0\u00d710\u22127\u21d2nI=B\u03bc0=100\u00d710\u221244\u03c0\u00d710\u22127<br \/>\n\u21d2nI\u22488000A\/m\u21d2nI\u22488000A\/m<br \/>\nIf the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.<\/h3>\n<h2>16. For a Circular Coil of Radius R and N Turns Carrying Current I, the Magnitude of the Magnetic Field at a Point on Its Axis at a Distance X from Its Centre Is Given By,<br \/>\na) Show that this reduces to the familiar result for the field at the centre of the coil.<\/h2>\n<h3>Ans: We are given,<br \/>\nRadius of circular coil = R<br \/>\nNumber of turns on the coil = N<br \/>\nCurrent in the coil = I<br \/>\nMagnetic field at a point on its axis at distance x is given by the relation,<br \/>\nWhere,\u03bc0=4\u03c0\u00d710\u22124TmA\u22121\u03bc0=4\u03c0\u00d710\u22124TmA\u22121Permeability of free space<br \/>\nIf the magnetic field at the centre of the coil is considered, then x=0x=0<br \/>\nB=\u03bc0IR2N2(x2+R2)32B=\u03bc0IR2N2(x2+R2)32<br \/>\nThis is the familiar result for the magnetic field at the centre of the coil.<\/h3>\n<h2>b) Consider Two Parallel Co-axial Circular Coils of Equal Radius R, and the Number of Turns N, Carrying Equal Currents in the Same Direction, and Separated by a Distance R. Show That the Field on the Axis Around the Mid-point Between the Coils Is Uniform Over a Distance That Is Small as Compared to R, and Is Given By, Approximately, (such an Arrangement to Produce a Nearly Uniform Magnetic Field Over a Small Region Is Known as Helmholtz Coils.)<\/h2>\n<h3>Ans: Given that,<br \/>\nRadii of two parallel co-axial circular coils = R<br \/>\nNumber of turns on each coil = N<br \/>\nCurrent in both coils = I<br \/>\nDistance between both the coils = R<br \/>\nLet us consider point Q at distance d from the centre.<br \/>\nThen, one coil is at a distance of R2+dR2+dfrom point Q.<br \/>\nMagnetic field at point Q could be given as:<br \/>\nB=\u03bc0IR2N2(x2+R2)32B=\u03bc0IR2N2(x2+R2)32<br \/>\nAlso, the other coil is at a distance of R2+dR2+dfrom point Q.<br \/>\nMagnetic field due to this coil is given as:<br \/>\nB2=\u03bc0NIR22[(R2\u2212d)2+R2]32B2=\u03bc0NIR22[(R2\u2212d)2+R2]32<br \/>\nNow we have the total magnetic field as,<br \/>\nB=B1+B2B=B1+B2<br \/>\n\u21d2B=\u03bc0IR22[{{(R2\u2212d)2+R2}\u221232+{(R2+d)2+R2}\u221232\u00d7N]\u21d2B=\u03bc0IR22[{{(R2\u2212d)2+R2}\u221232+{(R2+d)2+R2}\u221232\u00d7N]<br \/>\n\u21d2B=\u03bc0IR22[{{5R24+d2\u2212Rd}\u221232+{5R24+d2+Rd}\u221232\u00d7N]\u21d2B=\u03bc0IR22[{{5R24+d2\u2212Rd}\u221232+{5R24+d2+Rd}\u221232\u00d7N]<br \/>\n\u21d2B=\u03bc0IR22[{{1+4d25R2\u22124d5R}\u221232+{1+4d25R2+4d5R}\u221232\u00d7N]\u21d2B=\u03bc0IR22[{{1+4d25R2\u22124d5R}\u221232+{1+4d25R2+4d5R}\u221232\u00d7N]<br \/>\nNow for d\u226aRd\u226aR, we could neglect the factor d2R2d2R2, we get,<br \/>\nB\u2248\u03bc0IR22\u00d7(5R24)\u221232[(1\u22124d5R)\u221232+(1+4d5R)\u221232]\u00d7NB\u2248\u03bc0IR22\u00d7(5R24)\u221232[(1\u22124d5R)\u221232+(1+4d5R)\u221232]\u00d7N<br \/>\n\u21d2B\u2248\u03bc0IR22\u00d7(5R24)\u221232[1\u22126d5R+1+6d5R]\u21d2B\u2248\u03bc0IR22\u00d7(5R24)\u221232[1\u22126d5R+1+6d5R]<br \/>\n\u21d2B\u2248(45)32\u03bc0INR=0.72(\u03bc0INR)\u21d2B\u2248(45)32\u03bc0INR=0.72(\u03bc0INR)<br \/>\nClearly, we proved that the field along the axis around the mid-point between the coils is uniform.<\/h3>\n<h2>17. A Toroid Has a Core (non-Ferromagnetic) of Inner Radius 25 Cm and Outer Radius 26 Cm, Around Which 3500 Turns of a Wire Are Wound. If the Current in the Wire is 11 A, What is the Magnetic Field<br \/>\na) Outside the Toroid<\/h2>\n<h3>Ans: We are given,<br \/>\nInner radius of the toroid, r1=25cm=0.25mr1=25cm=0.25m<br \/>\nOuter radius of the toroid, r2=26cm=0.26mr2=26cm=0.26m<br \/>\nNumber of turns on the coil, N=3500N=3500<br \/>\nCurrent in the coil, I=11AI=11A<br \/>\nMagnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.<\/h3>\n<h2>b) Inside the Core of the Toroid.<\/h2>\n<h3>Ans: Magnetic field inside the core of a toroid is given by the relation, B=\u03bc0NIlB=\u03bc0NIl<br \/>\nWhere, Permeability of free space \u03bc0=4\u03c0\u00d710\u22127TmA\u22121\u03bc0=4\u03c0\u00d710\u22127TmA\u22121<br \/>\nll is the length of toroid, given by<br \/>\nl=2\u03c0(r1+r22)=\u03c0(0.25+0.26)=0.51\u03c0l=2\u03c0(r1+r22)=\u03c0(0.25+0.26)=0.51\u03c0<br \/>\n\u21d2B=4\u03c0\u00d710\u22127\u00d73500\u00d7110.51\u03c0\u22483.0\u00d710\u22122T\u21d2B=4\u03c0\u00d710\u22127\u00d73500\u00d7110.51\u03c0\u22483.0\u00d710\u22122T<br \/>\nThus, magnetic field inside the core of the toroid is 3.0\u00d710\u22122T3.0\u00d710\u22122T.<\/h3>\n<h2>c) In the Empty Space Surrounded by the Toroid.<\/h2>\n<h3>Ans: The empty space that is surrounding the toroid has a magnetic field equal to zero.<\/h3>\n<h2>18. Answer the Following Questions:<br \/>\na) A Magnetic Field That Varies in Magnitude from Point to Point but Has a Constant Direction (east to West) Is Set up in a Chamber. A Charged Particle Enters the Chamber and Travels Undeflected Along a Straight Path With Constant Speed. What Can You Say About the Initial Velocity of the Particle?<\/h2>\n<h3>Ans: The initial velocity of the particle could either be parallel or be anti-parallel to the magnetic field. So, it travels along a straight path without undergoing any deflection in the field.<\/h3>\n<h2>b) A Charged Particle Enters an Environment of a Strong and Non-Uniform Magnetic Field Varying from Point to Point Both in Magnitude and Direction, and Comes Out of it Following a Complicated Trajectory. Would Its Final Speed Equal the Initial Speed If it Suffered No Collisions With the Environment?<\/h2>\n<h3>Ans: Yes, the final speed of the charged particle would be equal to its initial speed as the magnetic force has the potential to change the direction of velocity even though not its magnitude.<\/h3>\n<h2>c) An Electron Travelling West to East Enters a Chamber Having a Uniform Electrostatic Field in the North to South Direction. Specify the Direction in Which a Uniform Magnetic Field Should Be Set up to Prevent the Electron from Deflecting from Its Straight-Line Path.<\/h2>\n<h3>Ans: An electron travelling from West to East enters a chamber having a uniform electrostatic field along the North-South direction.<br \/>\nThis moving electron stays undeflected when the electric force acting on it is equal and opposite of the magnetic field.<br \/>\nThe magnetic force would stay directed towards the South. Also, according to Fleming\u2019s left-hand rule, the magnetic field must be applied in a vertically downward direction.<\/h3>\n<h2>19. An Electron Emitted by a Heated Cathode and Accelerated Through a Potential Difference of 2.0kV2.0kV, Enters a Region With Uniform Magnetic Field of 0.15T0.15T. Determine the Trajectory of the Electron If the Field<br \/>\na) Is Transverse to Its Initial Velocity.<\/h2>\n<h3>Ans: We are given,<br \/>\nMagnetic field strength, B=0.15TB=0.15T<br \/>\nCharge on the electron, e=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C<br \/>\nMass of the electron, m=9.1\u00d710\u221231kgm=9.1\u00d710\u221231kg<br \/>\nPotential difference, V=2.0kV=2\u00d7103VV=2.0kV=2\u00d7103V<br \/>\nNow we have the kinetic energy of the electron given by,<br \/>\nK.E=eVK.E=eV<br \/>\nSubstituting the given values we get,<br \/>\neV=12mv2eV=12mv2<br \/>\n\u21d2v=2eVm\u2212\u2212\u2212\u221a\u21d2v=2eVm\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (1)<br \/>\nWhere, vvis the velocity of the electron<br \/>\nSince the magnetic force on the electron provides the required centripetal force of the electron, the electron traces a circular path of radius rr.<br \/>\nNow, the magnetic force on the electron is given by the relation,<br \/>\nF=BevF=Bev<br \/>\nCentripetal force,<br \/>\nFC=mv2rFC=mv2r<br \/>\n\u21d2Bev=mv2r\u21d2Bev=mv2r<br \/>\n\u21d2r=mvBe\u21d2r=mvBe\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (2)<br \/>\nFrom the equations (1) and (2), we get,<br \/>\nr=mBe[2eVm]12r=mBe[2eVm]12<br \/>\nSubstituting the given values,<br \/>\n\u21d2r=9.1\u00d710\u2212310.15\u00d71.6\u00d710\u221219(2\u00d71.6\u00d710\u221219\u00d72\u00d71039.1\u00d710\u221231)12\u21d2r=9.1\u00d710\u2212310.15\u00d71.6\u00d710\u221219(2\u00d71.6\u00d710\u221219\u00d72\u00d71039.1\u00d710\u221231)12<br \/>\n\u21d2r=100.55\u00d710\u22125\u21d2r=100.55\u00d710\u22125<br \/>\n\u21d2r=1mm\u21d2r=1mm<br \/>\nThus, we found that the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.<\/h3>\n<h2>b) Makes an Angle of 30\u221830\u2218 with the Initial Velocity.<\/h2>\n<h3>Ans: When the field makes an angle \u03b8\u03b8 of 30\u221830\u2218with initial velocity, the initial velocity will be,<br \/>\nv1=vsin\u03b8v1=vsin\u2061\u03b8<br \/>\nFrom equation (2), we can write the following expression:<br \/>\nr1=mv1Ber1=mv1Be<br \/>\n\u21d2r1=mvsin\u03b8Be\u21d2r1=mvsin\u2061\u03b8Be<br \/>\n\u21d2r1=9.1\u00d710\u2212310.15\u00d71.6\u00d710\u221219[2\u00d71.6\u00d710\u221219\u00d72\u00d71039\u00d710\u221231]sin30\u2218\u21d2r1=9.1\u00d710\u2212310.15\u00d71.6\u00d710\u221219[2\u00d71.6\u00d710\u221219\u00d72\u00d71039\u00d710\u221231]sin\u206130\u2218<br \/>\n\u21d2r=0.5\u00d710\u22123m=0.5mm\u21d2r=0.5\u00d710\u22123m=0.5mm<br \/>\nClearly, we found that the electron has a helical trajectory of radius0.5mm0.5mm, with axis of the solenoid along the magnetic field direction.<\/h3>\n<h2>20. A Magnetic Field Set up Using Helmholtz Coils (Described in Exercise 4.164.16) is Uniform in a Small Region and Has a Magnitude of 0.75T0.75T. In the Same Region, a Uniform Electrostatic Field Is Maintained in a Direction Normal to the Common Axis of the Coils. A Narrow Beam of (single Species) Charged Particles All Accelerated Through 15kV15kV Enters This Region in a Direction Perpendicular to Both the Axis of the Coils and the Electrostatic Field. If the Beam Remains Undeflected When the Electrostatic Field Is 9.0\u00d710\u22125Vm\u221219.0\u00d710\u22125Vm\u22121 Make a Simple Guess as to What the Beam Contains. Why Is the Answer Not Unique?<\/h2>\n<h3>Ans: We are given,<br \/>\nMagnetic field, B=0.75TB=0.75T<br \/>\nAccelerating voltage, V=15kV=15\u00d7103VV=15kV=15\u00d7103V<br \/>\nElectrostatic field, E=9\u00d7105Vm\u22121E=9\u00d7105Vm\u22121<br \/>\nMass of the electron=m=m<br \/>\nCharge of the electron =e=e<br \/>\nVelocity of the electron =v=v<br \/>\nKinetic energy of the electron =eV=eV<br \/>\nThus,<br \/>\n12mv2=eV12mv2=eV<br \/>\n\u21d2em=v22V\u21d2em=v22V\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (1)<br \/>\nSince the particle remains undeflected by electric and magnetic fields, we could infer that the electric field is balancing the magnetic field.<br \/>\neE=evBeE=evB<br \/>\n\u21d2v=EB\u21d2v=EB\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (2)<br \/>\nNow we could substitute equation (2) in equation (1) to get,<br \/>\nem=12(EB)2V=E22VB2em=12(EB)2V=E22VB2<br \/>\n\u21d2em=(9.0\u00d7105)22\u00d715000\u00d7(0.75)2=4.8\u00d7107C\/kg\u21d2em=(9.0\u00d7105)22\u00d715000\u00d7(0.75)2=4.8\u00d7107C\/kg<br \/>\nThis value of specific charge (em)(em) is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He++He++, Li+++Li+++<\/h3>\n<h2>21. A Straight Horizontal Conducting Rod of Length 0.45m0.45m and Mass 60g60g is Suspended by Two Vertical Wires at Its Ends. A Current of 5.0A5.0A is Set up in the Rod Through the Wires.<br \/>\na) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?<\/h2>\n<h3>Ans: We are given,<br \/>\nLength of the rod, l=0.45ml=0.45m<br \/>\nMass suspended by the wires, m=60g=60\u00d710\u22123kgm=60g=60\u00d710\u22123kg<br \/>\nAcceleration due to gravity, g=9.8ms\u22122g=9.8ms\u22122<br \/>\nCurrent in the rod flowing through the wire, I=5AI=5A<br \/>\nWe could say that magnetic field (B) is equal and opposite to the weight of the wire i.e.,<br \/>\nBIl=mgBIl=mg<br \/>\n\u21d2B=mgIl=60\u00d710\u22123\u00d79.85\u00d70.45\u21d2B=mgIl=60\u00d710\u22123\u00d79.85\u00d70.45<br \/>\n\u21d2B=0.26T\u21d2B=0.26T<br \/>\nClearly, a horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up.<\/h3>\n<h2>b) What will be the Total Tension in the Wires If the Direction of Current is Reversed Keeping the Magnetic Field Same as Before? (Ignore the Mass of the Wires.) g=9.8ms\u22122g=9.8ms\u22122<\/h2>\n<h3>Ans: When the direction of the current is reversed, BIlBIl and mgmg will act downwards. Clearly, the effective tension in the wires is found to be,<br \/>\nT=0.26\u00d75\u00d70.45+(60\u00d710\u22123)\u00d79.8T=0.26\u00d75\u00d70.45+(60\u00d710\u22123)\u00d79.8<br \/>\n\u21d2T=1.176N\u21d2T=1.176N<\/h3>\n<h2>22. The Wires Which Connect the Battery of an Automobile to Its Starting Motor Carry a Current of 300A300A(for a short time). What Is the Force Per Unit Length Between the Wires if they are 70cm70cm Long and 1.5cm1.5cm Apart? Is the Force Attractive or Repulsive?<\/h2>\n<h3>Ans: We are given,<br \/>\nCurrent in both wires, I=300AI=300A<br \/>\nDistance between the wires, r=1.5cm=0.015mr=1.5cm=0.015m<br \/>\nLength of the two wires, l=70cm=0.7ml=70cm=0.7m<br \/>\nWe know that, Force between the two wires is given by the relation,<br \/>\nF=\u03bc0I22\u03c0rF=\u03bc0I22\u03c0r<br \/>\nWhere, Permeability of free space\u03bc0=4\u03c0\u00d710TmA\u22121\u03bc0=4\u03c0\u00d710TmA\u22121<br \/>\nAs the direction of the current in the wires is found to be opposite, a repulsive force exists between them.<\/h3>\n<h2>23. A Uniform Magnetic Field of 1.5T1.5T Exists in a Cylindrical Region of Radius 10.0cm10.0cm, its Direction Parallel to the Axis Along East to West. A Wire Carrying Current of 7.0A7.0A in the North to South Direction Passes Through This Region. What is the Magnitude and Direction of the Force on the Wire If,<br \/>\na) The Wire Intersects the Axis,<\/h2>\n<h3>Ans: We are given,<br \/>\nMagnetic field strength, B=1.5TB=1.5T<br \/>\nRadius of the cylindrical region, r=10cm=0.1mr=10cm=0.1m<br \/>\nCurrent in the wire passing through the cylindrical region, I=7AI=7A<br \/>\nIf the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus, l=2r=0.2ml=2r=0.2m<br \/>\nAngle between magnetic field and current, \u03b8=90\u2218\u03b8=90\u2218<br \/>\nWe know that, Magnetic force acting on the wire is given by the relation,<br \/>\nF=BIlsin\u03b8F=BIlsin\u2061\u03b8<br \/>\n\u21d2F=1.5\u00d77\u00d70.2\u00d7sin90\u2218\u21d2F=1.5\u00d77\u00d70.2\u00d7sin\u206190\u2218<br \/>\n\u21d2F=2.1N\u21d2F=2.1N<br \/>\nClearly, a force of 2.1 N acts on the wire in a vertically downward direction.<\/h3>\n<h2>b) The Wire is Turned from N-S to northeast-northwest Direction,<\/h2>\n<h3>Ans: The new length of the wire after turning it to the northeast-northwest direction can be given as:<br \/>\nl1=lsin\u03b8l1=lsin\u2061\u03b8<br \/>\nAngle between magnetic field and current, \u03b8=45\u2218\u03b8=45\u2218<br \/>\nForce on the wire,<br \/>\nF=BIl1sin\u03b8=BIl=1.5\u00d77\u00d70.2F=BIl1sin\u2061\u03b8=BIl=1.5\u00d77\u00d70.2<br \/>\n\u21d2F=2.1N\u21d2F=2.1N<br \/>\nThus, a force of 2.1 N acts vertically downward on the wire. This is independent of angle \u03b8\u03b8 as lsin\u03b8lsin\u2061\u03b8 is fixed.<\/h3>\n<h2>c) The Wire in the N-S Direction is Lowered from the Axis by a Distance of 6.0cm6.0cm?<\/h2>\n<h3>Ans: The wire is lowered from the axis by distance, d=6.0cmd=6.0cm<br \/>\nLet l2l2be the new length of the wire,<br \/>\n(l22)2=4(d+r)=4(10+6)=4\u00d716(l22)2=4(d+r)=4(10+6)=4\u00d716<br \/>\n\u21d2l2=8\u00d72=16cm=0.16m\u21d2l2=8\u00d72=16cm=0.16m<br \/>\nMagnetic force that is exerted on the wire is,<br \/>\nF2=BIl2=1.5\u00d77\u00d70.16F2=BIl2=1.5\u00d77\u00d70.16<br \/>\n\u21d2F=1.68N\u21d2F=1.68N<br \/>\nClearly, a force of 1.68N1.68Nacts in a vertically downward direction on the wire.<\/h3>\n<h2>24. A Uniform Magnetic Field of 3000G3000G is Established Along the Positive Z-Direction. A Rectangular Loop of Sides 10cm10cm and 5cm5cm Carries a Current of 12A12A. What is the Torque on the Loop in the Different Cases Shown in Figure? What Is the Force on Each Case? Which Case Corresponds to Stable Equilibrium?<\/h2>\n<h2>a)<\/h2>\n<h3>Ans: We are given,<br \/>\nMagnetic field strength, B=3000G=3000\u00d710\u22124T=0.3TB=3000G=3000\u00d710\u22124T=0.3T<br \/>\nLength of the rectangular loop, l=10cml=10cm<br \/>\nWidth of the rectangular loop, b=5cmb=5cm<br \/>\nArea of the loop, A=l\u00d7b=(10\u00d75)cm2=50\u00d710\u22124m2A=l\u00d7b=(10\u00d75)cm2=50\u00d710\u22124m2<br \/>\nCurrent in the loop, I=12AI=12A<br \/>\nNow, we could take the anti-clockwise direction of the current as positive and vice-versa,<br \/>\nWe have the expression for torque given as,<br \/>\n\u03c4\u20d7=IA\u20d7 \u00d7B\u20d7 \u03c4\u2192=IA\u2192\u00d7B\u2192<br \/>\nWe could see from the given figure that A is normal to the y-z plane and B is directed along the z-axis. Substituting the given values,<br \/>\n\u03c4=12\u00d7(50\u00d710\u22124)i^\u00d70.3k^\u03c4=12\u00d7(50\u00d710\u22124)i^\u00d70.3k^<br \/>\n\u21d2\u03c4=\u22121.8\u00d710\u22122j^Nm\u21d2\u03c4=\u22121.8\u00d710\u22122j^Nm<br \/>\nNow, the torque is found to be directed along negative y-direction. Since the external magnetic field is uniform, the force on the loop would be zero.<\/h3>\n<h2>b)<\/h2>\n<h3>Ans: This case is very similar to case (a), and hence, the answer here would be same as (a).<\/h3>\n<h2>c)<\/h2>\n<h3>Ans: Torque here would be,<br \/>\n\u03c4=IA\u20d7 \u00d7B\u20d7 \u03c4=IA\u2192\u00d7B\u2192<br \/>\n\u21d2\u03c4=\u221212(50\u00d710\u22124)j^\u00d70.3k^\u21d2\u03c4=\u221212(50\u00d710\u22124)j^\u00d70.3k^<br \/>\n\u21d2\u03c4=\u22121.8\u00d710\u22122i^Nm\u21d2\u03c4=\u22121.8\u00d710\u22122i^Nm<br \/>\nThe direction here is along the negative x direction and the force is zero.<\/h3>\n<h2>d)<\/h2>\n<h3>Ans: Torque here would be,<br \/>\n|\u03c4|=IAB|\u03c4|=IAB<br \/>\n\u21d2\u03c4=12\u00d7(50\u00d710\u22124)\u00d70.3\u21d2\u03c4=12\u00d7(50\u00d710\u22124)\u00d70.3<br \/>\n\u21d2|\u03c4|=1.8\u00d710\u22122Nm\u21d2|\u03c4|=1.8\u00d710\u22122Nm<br \/>\nHere, the direction is found to be at 240\u2218240\u2218with positive x-direction and the force is zero.<\/h3>\n<h2>e)<\/h2>\n<h3>Ans: Torque,<br \/>\n\u03c4=IA\u20d7 \u00d7B\u20d7 \u03c4=IA\u2192\u00d7B\u2192<br \/>\n\u21d2\u03c4=(50\u00d710\u22124\u00d712)k^\u00d70.3k^\u21d2\u03c4=(50\u00d710\u22124\u00d712)k^\u00d70.3k^<br \/>\n\u21d2\u03c4=0\u21d2\u03c4=0<br \/>\nHere, both torque and force are found to be zero.<\/h3>\n<h2>f)<\/h2>\n<h3>Ans: Torque is given by,<br \/>\n\u03c4=IA\u20d7 \u00d7B\u20d7 \u03c4=IA\u2192\u00d7B\u2192<br \/>\n\u21d2\u03c4=(50\u00d710\u22124\u00d712)k^\u00d70.3k^\u21d2\u03c4=(50\u00d710\u22124\u00d712)k^\u00d70.3k^<br \/>\n\u21d2\u03c4=0\u21d2\u03c4=0<br \/>\nHere also both torque and force are found to be zero.<br \/>\nFor the case (e) the direction IA\u20d7 IA\u2192and B\u20d7 B\u2192 is the same and the angle between them is zero. They would come back to equilibrium on being displaced and so the equilibrium is stable.<br \/>\nFor the case (f), the direction of IA\u20d7 IA\u2192and B\u20d7 B\u2192 are opposite and the angle between them is 180\u2218180\u2218. Here it doesn\u2019t come back to its original position on being disturbed and hence, the equilibrium is unstable.<\/h3>\n<h2>25. A Circular Coil of 20 Turns and Radius 10 Cm is Placed in a Uniform Magnetic Field of 0.10 T Normal to the Plane of the Coil. If the Current in the Coil Is 5.0 A, What is the: (The Coil is Made of Copper Wire of Cross-Sectional Area 10\u22125m210\u22125m2, and the Free Electron Density in Copper Is Given to be about1029m\u221231029m\u22123).<br \/>\na) Total Torque on the Coil?<\/h2>\n<h3>Ans: We are given,<br \/>\nNumber of turns on the circular coil, n=20n=20<br \/>\nRadius of the coil, r=10cm=0.1mr=10cm=0.1m<br \/>\nMagnetic field strength, B=0.10TB=0.10T<br \/>\nCurrent in the coil, I=5.0AI=5.0A<br \/>\nAs the angle between force and the normal to the loop is zero, the total torque on the coil is zero.<br \/>\nSo, \u03c4=NIABsin\u03b8\u03c4=NIABsin\u2061\u03b8is zero.<\/h3>\n<h2>b) Total Force on the Coil,<\/h2>\n<h3>Ans: There is no total force on the coil because the field is uniform.<\/h3>\n<h2>C) Average Force on Each Electron in the Coil Due to the Magnetic Field?<\/h2>\n<h3>Ans: Given that,<br \/>\nCross-sectional area of copper coil, A=10\u22125m2A=10\u22125m2<br \/>\nNumber of free electrons per cubic meter in copper, N=1029\/m3N=1029\/m3<br \/>\nCharge on the electron would be, e=1.6\u00d710\u221219Ce=1.6\u00d710\u221219C<br \/>\nMagnetic force, F=BevdF=Bevd<br \/>\nWhere, vdvd is drift velocity of electrons given by INeAINeA<br \/>\n\u21d2F=BeINeA=0.10\u00d75.01029\u00d710\u22125=5\u00d710\u221225N\u21d2F=BeINeA=0.10\u00d75.01029\u00d710\u22125=5\u00d710\u221225N<br \/>\nClearly, the average force on each electron is 5\u00d710\u221225N5\u00d710\u221225N.<\/h3>\n<h2>26. A Solenoid 60cm60cm Long and Radius 4.0cm4.0cmhas 3 Layers of Windings of 300turns Each. A 2.0cm2.0cm Long Wire of Mass 2.5g2.5g Lies Inside the Solenoid (near Its Centre) Normal to Its Axis; Both the Wire and the Axis of the Solenoid Are in the Horizontal Plane. the Wire Is Connected Through Two Leads Parallel to the Axis of the Solenoid to an External Battery Which Supplies a Current of 6.0a6.0a in the Wire. What Value of Current (with Appropriate Sense of Circulation) in the Windings of the Solenoid Can Support the Weight of the Wire? g=9.8ms\u22122g=9.8ms\u22122<\/h2>\n<h3>Ans: We are given:<br \/>\nLength of the solenoid, L=60cm=0.6mL=60cm=0.6m<br \/>\nRadius of the solenoid, r=4.0cm=0.04mr=4.0cm=0.04m<br \/>\nIt is given that there are 3 layers of windings of 300 turns each.<br \/>\nTotal number of turns, n=3\u00d7300=900n=3\u00d7300=900<br \/>\nLength of the wire, l=2cm=0.02ml=2cm=0.02m<br \/>\nMass of the wire, m=2.5g=2.5\u00d710\u22123kgm=2.5g=2.5\u00d710\u22123kg<br \/>\nCurrent flowing through the wire, i=6Ai=6A<br \/>\nAcceleration due to gravity, g=9.8ms\u22122g=9.8ms\u22122<br \/>\nMagnetic field produced inside the solenoid, B=\u03bc0nILB=\u03bc0nIL<br \/>\nWhere, Permeability of free space \u03bc0=4\u03c0\u00d710\u22127TmA\u22121\u03bc0=4\u03c0\u00d710\u22127TmA\u22121<br \/>\nCurrent flowing through the windings of the solenoid, II<br \/>\nMagnetic force is given by the relation,<br \/>\nF=Bil=\u03bc0nILilF=Bil=\u03bc0nILil<br \/>\nNow, we have the force on the wire equal to the weight of the wire.<br \/>\nmg=\u03bc0nIilLmg=\u03bc0nIilL<br \/>\n\u21d2I=mgL\u03bc0nil=2.5\u00d710\u22123\u00d79.8\u00d70.64\u03c0\u00d710\u22127\u00d7900\u00d70.02\u00d76\u21d2I=mgL\u03bc0nil=2.5\u00d710\u22123\u00d79.8\u00d70.64\u03c0\u00d710\u22127\u00d7900\u00d70.02\u00d76<br \/>\n\u21d2I=108A\u21d2I=108A<br \/>\nClearly, the current flowing through the solenoid is 108 A.<\/h3>\n<h2>27. A Galvanometer Coil Has a Resistance of 12\u03a912\u03a9and the Metre Shows Full Scale Deflection for a Current of 3mA3mA. How will you Convert the Metre Into a Voltmeter of Range 00 to 18V18V?<\/h2>\n<h3>Ans: We are given,<br \/>\nResistance of the galvanometer coil, G=12\u03a9G=12\u03a9<br \/>\nCurrent for which there is full scale deflection, Ig=3mA=3\u00d710\u22123AIg=3mA=3\u00d710\u22123A<br \/>\nRange of the voltmeter needs to be converted to 18V18V.<br \/>\nLet a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance can be given as:<br \/>\nR=VIg\u2212GR=VIg\u2212G<br \/>\nSubstituting the given values we get,<br \/>\nR=183\u00d710\u22123\u221212=6000\u221212R=183\u00d710\u22123\u221212=6000\u221212<br \/>\n\u21d2R=5988\u03a9\u21d2R=5988\u03a9<br \/>\nClearly, we found that a resistor of resistance 5988\u03a95988\u03a9is to be connected in series with the given galvanometer.<\/h3>\n<h2>28. A Galvanometer Coil Has a Resistance of 15\u03a915\u03a9 and the Metre Shows Full Scale Deflection for a Current of 4mA4mA. How will you Convert the Metre Into an Ammeter of Range 00 to 6A6A?<\/h2>\n<h3>Ans: We are given,<br \/>\nResistance of the galvanometer coil, G=15\u03a9G=15\u03a9<br \/>\nCurrent for which the galvanometer shows full scale deflection, Ig=4mA=4\u00d710\u22123AIg=4mA=4\u00d710\u22123A<br \/>\nWe said that, Range of the ammeter needs to be 6A6A.<br \/>\nIn order to convert the given galvanometer into an ammeter, a shunt resistor of resistance S is to be connected in parallel with the galvanometer.<br \/>\nThe value of S could be given as:<br \/>\nS=IgGI\u2212IgS=IgGI\u2212Ig<br \/>\nSubstituting the given values we get,<br \/>\nS=4\u00d710\u22123\u00d7156\u22124\u00d710\u22123=0.065.996\u22480.01\u03a9S=4\u00d710\u22123\u00d7156\u22124\u00d710\u22123=0.065.996\u22480.01\u03a9<br \/>\n\u21d2S=10m\u03a9\u21d2S=10m\u03a9<br \/>\nClearly, we found that a<br \/>\n10m\u03a910m\u03a9<br \/>\nshunt resistor is to be connected in parallel with the galvanometer.<\/h3>\n","protected":false},"excerpt":{"rendered":"<p>Class 12 Physics NCERT book solutions for Chapter 4 &#8211; Moving Charges And Magnetism 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-118547","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\/118547","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=118547"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118547\/revisions"}],"predecessor-version":[{"id":118579,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118547\/revisions\/118579"}],"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=118547"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=118547"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=118547"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}