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      Class 12 PHYSICS – JEE

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      • Class 12
      • Class 12 PHYSICS – JEE
      CoursesClass 12PhysicsClass 12 PHYSICS – JEE
      • 1.Electrostatics (1)
        8
        • Lecture1.1
          Charge, Coulombs Law and Coulombs law in Vector form 41 min
        • Lecture1.2
          Electric Field; Electric Field Lines; Field lines due to multiple charges 42 min
        • Lecture1.3
          Charge Distribution; Finding Electric Field due to Different Object 01 hour
        • Lecture1.4
          Solid angle; Area Vector; Electric Flux; Flux of closed surface; Gauss Law 47 min
        • Lecture1.5
          Finding E Using Concept of Gauss law and Flux 01 hour
        • Lecture1.6
          Chapter Notes – Electrostatics (1)
        • Lecture1.7
          NCERT Solutions – Electrostatics
        • Lecture1.8
          Revision Notes Electrostatics
      • 2.Electrostatics (2)
        7
        • Lecture2.1
          Work done by Electrostatic Force; Work done by man in E-Field; Electrostatic Potential Energy 49 min
        • Lecture2.2
          Finding Electric Potential, Equipotential Surface and Motion in Electric Field 01 hour
        • Lecture2.3
          Electric Dipole and Dipole in Uniform and Non-uniform Electric field 01 hour
        • Lecture2.4
          Analysis of charge on conductors; Potential due to induced charge 58 min
        • Lecture2.5
          Conductors with cavity- Case 1: Empty cavity, Case 2: Charge Inside Cavity 41 min
        • Lecture2.6
          Connecting Two Conductors; Grounding of conductor; Electric field just outside conductor; Electrostatic pressure; Self potential Energy 54 min
        • Lecture2.7
          Chapter Notes – Electrostatics (2)
      • 3.Current Electricity (1)
        9
        • Lecture3.1
          Current, Motion of Electrons in Conductor; Temp. Dependence of Resistor 26 min
        • Lecture3.2
          Circuit Theory and Kirchoffs Laws 31 min
        • Lecture3.3
          Some Special Circuits- Series & Parallel Circuits, Open Circuit, Short Circuit 26 min
        • Lecture3.4
          Wheatstone Bridge, Current Antisymmetric 21 min
        • Lecture3.5
          Equivalent Resistance- Series and parallel, Equipotential Points, Wheatstone Bridge 25 min
        • Lecture3.6
          Current Antisymmetric, Infinite Ladder, Circuit Solving, 3D circuits 20 min
        • Lecture3.7
          Chapter Notes – Current Electricity
        • Lecture3.8
          NCERT Solutions – Current Electricity
        • Lecture3.9
          Revision Notes Current Electricity
      • 4.Current Electricity (2)
        4
        • Lecture4.1
          Heating Effect of Current; Rating of Bulb; Fuse 19 min
        • Lecture4.2
          Battery, Maximum power theorem; Ohmic and Non Ohmic Resistance; Superconductor 31 min
        • Lecture4.3
          Galvanometer; Ammeter & Voltmeter and Their Making 44 min
        • Lecture4.4
          Potentiometer and its applications ; Meter Bridge; Post Office Box; Colour Code of Resistors 32 min
      • 5.Capacitor
        6
        • Lecture5.1
          Capacitor and Capacitance; Energy in Capacitor 38 min
        • Lecture5.2
          Capacitive Circuits- Kirchoff’s Laws; Heat Production 01 hour
        • Lecture5.3
          Equivalent Capacitance; Charge on both sides of cap. Plate 52 min
        • Lecture5.4
          Dielectric Strength; Polar and Non-Polar Dielectric; Equivalent Cap. with Dielectric 01 hour
        • Lecture5.5
          Inserting and Removing Dielectric- Work (Fringing Effect), Force; Force between plates of capacitor 38 min
        • Lecture5.6
          Revision Notes Capacitor
      • 6.RC Circuits
        3
        • Lecture6.1
          Maths Needed for RC Circuits, RC circuits-Charging Circuit 19 min
        • Lecture6.2
          RC circuits-Discharging Circuit, Initial & Steady State, Final (Steady) State, Internal Resistance of Capacitor 44 min
        • Lecture6.3
          Revision Notes RC Circuits
      • 7.Magnetism and Moving Charge
        16
        • Lecture7.1
          Introduction, Vector Product, Force Applied by Magnetic Field, Lorentz Force, Velocity Selector 40 min
        • Lecture7.2
          Motion of Charged Particles in Uniform Magnetic Field 40 min
        • Lecture7.3
          Cases of Motion of Charged Particles in Uniform Magnetic Field 56 min
        • Lecture7.4
          Force on a Current Carrying Wire on Uniform B and its Cases, Questions and Solutions 59 min
        • Lecture7.5
          Magnetic Field on Axis of Circular Loop, Magnetic field due to Moving Charge, Magnetic Field due to Current 52 min
        • Lecture7.6
          Magnetic Field due to Straight Wire, Different Methods 40 min
        • Lecture7.7
          Magnetic Field due to Rotating Ring and Spiral 41 min
        • Lecture7.8
          Force between Two Current Carrying Wires 36 min
        • Lecture7.9
          Force between Two Current Carrying Wires 58 min
        • Lecture7.10
          Miscellaneous Questions 55 min
        • Lecture7.11
          Solenoid, Toroid, Magnetic Dipole, Magnetic Dipole Momentum, Magnetic Field of Dipole 54 min
        • Lecture7.12
          Magnetic Dipole in Uniform Magnetic Field, Moving Coil Galvanometer, Torsional Pendulum 01 hour
        • Lecture7.13
          Advanced Questions, Magnetic Dipole and Angular Momentum 56 min
        • Lecture7.14
          Chapter Notes – Magnetism and Moving Charge
        • Lecture7.15
          NCERT Solutions – Magnetism and Moving Charge
        • Lecture7.16
          Revision Notes Magnetism and Moving Charge
      • 8.Magnetism and Matter
        10
        • Lecture8.1
          Magnetic Dipole, Magnetic Properties of Matter, Diamagnetism; Domain Theory of Ferro 47 min
        • Lecture8.2
          Magnetic Properties of Matter in Detail 39 min
        • Lecture8.3
          Magnetization and Magnetic Intensity, Meissner Effect, Variation of Magnetization with Temperature 55 min
        • Lecture8.4
          Hysteresis, Permanent Magnet, Properties of Ferro for Permanent Magnet, Electromagnet 31 min
        • Lecture8.5
          Magnetic Compass, Earth’s Magnetic Field 20 min
        • Lecture8.6
          Bar Magnet, Bar Magnet in Uniform Field 49 min
        • Lecture8.7
          Magnetic Poles, Magnetic Field Lines, Magnetism and Gauss’s Law 32 min
        • Lecture8.8
          Chapter Notes – Magnetism and Matter
        • Lecture8.9
          NCERT Solutions – Magnetism and Matter
        • Lecture8.10
          Revision Notes Magnetism and Matter
      • 9.Electromagnetic Induction
        14
        • Lecture9.1
          Introduction, Magnetic Flux, Motional EMF 01 min
        • Lecture9.2
          Induced Electric Field, Faraday’s Law, Comparison between Electrostatic Electric Field and Induced Electric Field 43 min
        • Lecture9.3
          Induced Current; Faraday’s Law ; Lenz’s Law 56 min
        • Lecture9.4
          Faraday’s Law and its Cases 50 min
        • Lecture9.5
          Advanced Questions on Faraday’s Law 37 min
        • Lecture9.6
          Cases of Current Electricity 59 min
        • Lecture9.7
          Lenz’s Law and Conservation of Energy, Eddy Current, AC Generator, Motor 01 hour
        • Lecture9.8
          Mutual Induction 53 min
        • Lecture9.9
          Self Inductance, Energy in an Inductor 34 min
        • Lecture9.10
          LR Circuit, Decay Circuit 01 hour
        • Lecture9.11
          Initial and Final Analysis of LR Circuit 38 min
        • Lecture9.12
          Chapter Notes – Electromagnetic Induction
        • Lecture9.13
          NCERT Solutions – Electromagnetic Induction
        • Lecture9.14
          Revision Notes Electromagnetic Induction
      • 10.Alternating Current Circuit
        8
        • Lecture10.1
          Introduction, AC/DC Sources, Basic AC Circuits, Average & RMS Value 46 min
        • Lecture10.2
          Phasor Method, Rotating Vector, Adding Phasors, RC Circuit 35 min
        • Lecture10.3
          Examples and Solutions 21 min
        • Lecture10.4
          Power in AC Circuit, Resonance Frequency, Bandwidth and Quality Factor, Transformer 51 min
        • Lecture10.5
          LC Oscillator, Question and Solutions of LC Oscillator, Damped LC Oscillator 53 min
        • Lecture10.6
          Chapter Notes – Alternating Current Circuit
        • Lecture10.7
          NCERT Solutions – Alternating Current Circuit
        • Lecture10.8
          Revision Notes Alternating Current Circuit
      • 11.Electromagnetic Waves
        4
        • Lecture11.1
          Displacement Current; Ampere Maxwell Law 14 min
        • Lecture11.2
          EM Waves; EM Spectrum; Green House Effect; Ozone Layer 36 min
        • Lecture11.3
          Chapter Notes – Electromagnetic Waves
        • Lecture11.4
          Revision Notes Electromagnetic Waves
      • 12.Photoelectric Effect
        5
        • Lecture12.1
          Recalling Basics; Photoelectric Effect 50 min
        • Lecture12.2
          Photo-electric Cell 35 min
        • Lecture12.3
          Photon Flux; Photon Density; Momentum of Photon; Radiation Pressure- Full Absorption, Full Reflection; Dual nature 52 min
        • Lecture12.4
          Chapter Notes – Photoelectric Effect
        • Lecture12.5
          Revision Notes Photoelectric Effect
      • 13.Ray Optics (Part 1)
        12
        • Lecture13.1
          Rays and Beam of Light, Reflection of Light, Angle of Deviation, Image Formation by Plane Mirror 01 hour
        • Lecture13.2
          Field of View, Numerical on Field of Line, Size of Mirror 42 min
        • Lecture13.3
          Curved Mirrors, Terms Related to Curved Mirror, Reflection of Light by Curved Mirror 40 min
        • Lecture13.4
          Image Formation by Concave Mirror, Magnification or Lateral or Transverse Magnification 01 hour
        • Lecture13.5
          Ray Diagrams for Concave Mirror 45 min
        • Lecture13.6
          Image Formation by Convex Mirror; Derivations of Various Formulae used in Concave Mirror and Convex Mirror 01 hour
        • Lecture13.7
          Advanced Optical Systems, Formation of Images with more than one Mirror 24 min
        • Lecture13.8
          Concept of Virtual Object, Formation of Image when Incident ray are Converging, Image Characteristics for Virtual Object, 55 min
        • Lecture13.9
          Newton’s Formula, Longitudinal Magnification 23 min
        • Lecture13.10
          Formation of Image when Two Plane Mirrors kept at an angle and parallel; Formation of Image by two Parallel Mirrors. 43 min
        • Lecture13.11
          Chapter Notes – Ray Optics
        • Lecture13.12
          NCERT Solutions – Ray Optics
      • 14.Ray Optics (Part 2)
        13
        • Lecture14.1
          Refractive Index, Opaque, Transparent, Speed of Light, Relative Refractive Index, Refraction and Snell’s Law, Refraction in Denser and Rarer Medium 42 min
        • Lecture14.2
          Image Formation due to Refraction; Derivation; Refraction and Image formation in Glass Slab 57 min
        • Lecture14.3
          Total Internal Reflection, Critical Angle, Principle of Reversibility 01 hour
        • Lecture14.4
          Application of Total Internal Reflection 45 min
        • Lecture14.5
          Refraction at Curved Surface, Image Formation by Curved Surface, Derivation 56 min
        • Lecture14.6
          Image Formation by Curved Surface, Snell’s Law in Vector Form 01 hour
        • Lecture14.7
          Lens, Various types of Lens, Differentiating between various Lenses; Optical Centre, Derivation of Lens Maker Formula 01 hour
        • Lecture14.8
          Lens Formula, Questions and Answers 39 min
        • Lecture14.9
          Property of Image by Convex and Concave Lens; Lens Location, Minimum Distance Between Real Image and Object 01 hour
        • Lecture14.10
          Power of Lens, Combination of Lens, Autocollimation 35 min
        • Lecture14.11
          Silvering of Lens 44 min
        • Lecture14.12
          Cutting of Lens and Mirror, Vertical Cutting, Horizontal Cutting 49 min
        • Lecture14.13
          Newton’s Law for Lens and Virtual Object 01 hour
      • 15.Ray Optics (Part 3)
        6
        • Lecture15.1
          Prism, Angle of Prism, Reversibility in Prism 51 min
        • Lecture15.2
          Deviation in Prism, Minimum and Maximum Deviation, Asymmetric, Thin Prism, Proof for formula of Thin Prism 59 min
        • Lecture15.3
          Dispersion of Light, Refractive Index, Composition of Light, Dispersion through Prism 01 hour
        • Lecture15.4
          Rainbow Formation, Scattering of Light, Tyndall Effect, Defects of Image, Spherical Defect, Chromatic Defect, Achromatism. 57 min
        • Lecture15.5
          Optical Instruments, The Human Eye, Defects of Eye and its Corrections 01 hour
        • Lecture15.6
          Microscope & Telescope 02 hour
      • 16.Wave Optics
        21
        • Lecture16.1
          Introduction to Wave Optics 11 min
        • Lecture16.2
          Huygens Wave Theory 14 min
        • Lecture16.3
          Huygens Theory of Secondary Wavelets 10 min
        • Lecture16.4
          Law of Reflection by Huygens Theory 10 min
        • Lecture16.5
          Deriving Laws of Refraction by Huygens Wave Theory 10 min
        • Lecture16.6
          Multiple Answer type question on Huygens Theory 41 min
        • Lecture16.7
          Conditions of Constructive and Destructive Interference 22 min
        • Lecture16.8
          Conditions of Constructive and Destructive Interference 06 min
        • Lecture16.9
          Conditions of Constructive and Destructive Interference 23 min
        • Lecture16.10
          Incoherent Sources of Light 38 min
        • Lecture16.11
          Youngs Double Slit Experiment 12 min
        • Lecture16.12
          Fringe Width Positions of Bright and Dark Fringes 15 min
        • Lecture16.13
          Numerical problems on Youngs Double Slit Experiment 11 min
        • Lecture16.14
          Numerical problems on Youngs Double Slit Experiment 19 min
        • Lecture16.15
          Displacement of Interference Pattern 19 min
        • Lecture16.16
          Numerical problems on Displacement of Interference Pattern 28 min
        • Lecture16.17
          Shapes of Fringes 37 min
        • Lecture16.18
          Colour of Thin Films 59 min
        • Lecture16.19
          Interference with White Light 32 min
        • Lecture16.20
          Chapter Notes – Wave Optics
        • Lecture16.21
          NCERT Solutions – Wave Optics
      • 17.Atomic Structure
        6
        • Lecture17.1
          Thomson and Rutherford Model of Atom; Trajectory of Alpha particle; Bohr’s Model ; Hydrogen Like Atom; Energy Levels 58 min
        • Lecture17.2
          Emission Spectra, Absorption Spectra; De Broglie Explanation of Bohr’s 2nd Postulate; Limitations of Bohr’s Model 37 min
        • Lecture17.3
          Momentum Conservation in Photon Emission, Motion of Nucleus, Atomic Collision 58 min
        • Lecture17.4
          Chapter Notes – Atomic Structure
        • Lecture17.5
          NCERT Solutions – Atomic Structure
        • Lecture17.6
          Revision Notes Atomic Structure
      • 18.Nucleus
        6
        • Lecture18.1
          Basics- Size of Nucleus, Nuclear Force, Binding Energy, Mass Defect; Radioactive Decay 01 hour
        • Lecture18.2
          Laws of Radioactive Decay 36 min
        • Lecture18.3
          Nuclear Fission; Nuclear Reactor; Nuclear Fusion- Reaction Inside Sun 30 min
        • Lecture18.4
          Chapter Notes – Nucleus
        • Lecture18.5
          NCERT Solutions – Nucleus
        • Lecture18.6
          Revision Notes Nucleus
      • 19.X-Ray
        4
        • Lecture19.1
          Electromagnetic Spectrum, Thermionic Emission; Coolidge Tube – Process 1 22 min
        • Lecture19.2
          Coolidge Tube – Process 2; Moseley’s Law; Absorption of X-rays in Heavy Metal 39 min
        • Lecture19.3
          Chapter Notes – X-Ray
        • Lecture19.4
          Revision Notes X-Ray
      • 20.Error and Measurement
        2
        • Lecture20.1
          Least Count of Instruments; Mathematical Operation on Data with Random Error 18 min
        • Lecture20.2
          Significant Digits; Significant Digits and Mathematical Operations 30 min
      • 21.Semiconductors
        9
        • Lecture21.1
          Conductor, Semiconductors and Insulators Basics Difference, Energy Band Theory, Si element 21 min
        • Lecture21.2
          Doping and PN Junction 01 hour
        • Lecture21.3
          Diode and Diode as Rectifier 01 hour
        • Lecture21.4
          Voltage Regulator and Zener Diode and Optoelectronic Jn. Devices 01 hour
        • Lecture21.5
          Transistor, pnp, npn, Modes of operation, Input and Output Characteristics, , Current Amplification Factor 01 hour
        • Lecture21.6
          Transistor as Amplifier, Transistor as Switch, Transistor as Oscillator, Digital Gates 01 hour
        • Lecture21.7
          Chapter Notes – Semiconductors
        • Lecture21.8
          NCERT Solutions – Semiconductors
        • Lecture21.9
          Revision Notes Semiconductors
      • 22.Communication Systems
        5
        • Lecture22.1
          Basic working and terms; Antenna; Modulation and Types of Modulation 32 min
        • Lecture22.2
          Amplification Modulation, Transmitter, Receiver, Modulation index 40 min
        • Lecture22.3
          Chapter Notes – Communication Systems
        • Lecture22.4
          NCERT Solutions – Communication Systems
        • Lecture22.5
          Revision Notes Communication Systems

        NCERT Solutions – Magnetism and Matter

        5.1. Answer the following questions regarding earth’s magnetism:

        (a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.

        (b) The angle of dip at a location in southern India is about 18º.

        Would you expect a greater or smaller dip angle in Britain?

        (c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

        (d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

        (e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T−1 located at its centre. Check the order of magnitude of this number in some way.

        (f ) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

        ANSWER:

        (a) The three independent quantities conventionally used for specifying earth’s magnetic field are:

        (i) Magnetic declination,

        (ii) Angle of dip, and

        (iii) Horizontal component of earth’s magnetic field

        (b)The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.

        (c)It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.

        Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

        (d)If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

        (e)Magnetic moment, M = 8 × 1022 J T−1

        Radius of earth, r = 6.4 × 106 m

        Magnetic field strength, 

        Where,

         = Permeability of free space = 

        This quantity is of the order of magnitude of the observed field on earth.

        (f)Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

        5.2. Answer the following questions:

        (a) The earth’s magnetic field varies from point to point in space.

        Does it also change with time? If so, on what time scale does it change appreciably?

        (b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

        (c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?

        (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?

        (e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

        (f ) Interstellar space has an extremely weak magnetic field of the order of 10−12 T. Can such a weak field be of any significant consequence? Explain.

        [Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]

        ANSWER:

        (a) Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected.

        (b)Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth’s magnetism.

        (c)Theradioactivity in earth’s interior is the source of energy that sustains the currents in the outer conducting regions of earth’s core. These charged currents are considered to be responsible for earth’s magnetism.

        (d)Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.

        (e)Earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.

        (f)An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.

        5.3. A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10−2 J. What is the magnitude of magnetic moment of the magnet?

        ANSWER:

        Magnetic field strength, B = 0.25 T

        Torque on the bar magnet, T = 4.5 × 10−2 J

        Angle between the bar magnet and the external magnetic field,θ = 30°

        Torque is related to magnetic moment (M) as:

        T = MB sin θ

        Hence, the magnetic moment of the magnet is 0.36 J T−1.

        5.4. A short bar magnet of magnetic moment m = 0.32 J T−1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

        ANSWER:

        Moment of the bar magnet, M = 0.32 J T−1

        External magnetic field, B = 0.15 T

        (a)The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle θ, between the bar magnet and the magnetic field is 0°.

        Potential energy of the system

        (b)The bar magnet is oriented 180° to the magnetic field. Hence, it is in unstable equilibrium.

        θ = 180°

        Potential energy = − MB cos θ

         

        5.5. A closely wound solenoid of 800 turns and area of cross section 2.5 × 10−4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

        ANSWER:

        Number of turns in the solenoid, n = 800

        Area of cross-section, A = 2.5 × 10−4 m2

        Current in the solenoid, I = 3.0 A

        A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.

        The magnetic moment associated with the given current-carrying solenoid is calculated as:

        M = n I A

        = 800 × 3 × 2.5 × 10−4

        = 0.6 J T−1

        5.6. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

        ANSWER:

        Magnetic field strength, B = 0.25 T

        Magnetic moment, M = 0.6 T−1

        The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.

        Therefore, the torque acting on the solenoid is given as:

        5.7. A bar magnet of magnetic moment 1.5 J T−1 lies aligned with the direction of a uniform magnetic field of 0.22 T.

        (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

        (b) What is the torque on the magnet in cases (i) and (ii)?

        ANSWER:

        (a)Magnetic moment, M = 1.5 J T−1

        Magnetic field strength, B = 0.22 T

        (i)Initial angle between the axis and the magnetic field, θ1 = 0°

        Final angle between the axis and the magnetic field, θ2 = 90°

        The work required to make the magnetic moment normal to the direction of magnetic field is given as:

        (ii) Initial angle between the axis and the magnetic field, θ1 = 0°

        Final angle between the axis and the magnetic field, θ2 = 180°

        The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

        (b)For case (i): 

        ∴Torque, 

        For case (ii): 

        ∴Torque, 

        5.8. A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10−4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

        (a) What is the magnetic moment associated with the solenoid?

        (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10−2 T is set up at an angle of 30º with the axis of the solenoid?

        ANSWER:

        Number of turns on the solenoid, n = 2000

        Area of cross-section of the solenoid, A = 1.6 × 10−4 m2

        Current in the solenoid, I = 4 A

        (a)The magnetic moment along the axis of the solenoid is calculated as:

        M = nAI

        = 2000 × 1.6 × 10−4 × 4

        = 1.28 Am2

        (b)Magnetic field, B = 7.5 × 10−2 T

        Angle between the magnetic field and the axis of the solenoid, θ = 30°

        Torque, 

        Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 

        5.9. A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10−2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s−1. What is the moment of inertia of the coil about its axis of rotation?

        ANSWER:

        Number of turns in the circular coil, N = 16

        Radius of the coil, r = 10 cm = 0.1 m

        Cross-section of the coil, A = πr2 = π × (0.1)2 m2

        Current in the coil, I = 0.75 A

        Magnetic field strength, B = 5.0 × 10−2 T

        Frequency of oscillations of the coil, v = 2.0 s−1

        ∴Magnetic moment, M = NIA

        = 16 × 0.75 × π × (0.1)2

        = 0.377 J T−1

        Where,

        I = Moment of inertia of the coil

        Hence, the moment of inertia of the coil about its axis of rotation is 

        5.10. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22º with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

        ANSWER:

        Horizontal component of earth’s magnetic field, BH = 0.35 G

        Angle made by the needle with the horizontal plane = Angle of dip = 

        Earth’s magnetic field strength = B

        We can relate B and BHas:

        Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

        5.11. At a certain location in Africa, a compass points 12º west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

        ANSWER:

        Angle of declination,θ = 12°

        Angle of dip, 

        Horizontal component of earth’s magnetic field, BH = 0.16 G

        Earth’s magnetic field at the given location = B

        We can relate B and BHas:

        Earth’s magnetic field lies in the vertical plane, 12° West of the geographic meridian, making an angle of 60° (upward) with the horizontal direction. Its magnitude is 0.32 G.

        5.12. A short bar magnet has a magnetic moment of 0.48 J T−1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

        ANSWER:

        Magnetic moment of the bar magnet, M = 0.48 J T−1

        (a) Distance, d = 10 cm = 0.1 m

        The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation:

        Where,

         = Permeability of free space = 

        The magnetic field is along the S − N direction.

        (b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:

        The magnetic field is along the N − S direction.

        5.13. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null−point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

        ANSWER:

        Earth’s magnetic field at the given place, H = 0.36 G

        The magnetic field at a distance d, on the axis of the magnet is given as:

        Where,

        = Permeability of free space

        M = Magnetic moment

        The magnetic field at the same distance d, on the equatorial line of the magnet is given as:

        Total magnetic field,

        Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.

        5.14. If the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?

        ANSWER:

        The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:

        Where,

        M = Magnetic moment

        = Permeability of free space

        H = Horizontal component of the magnetic field at d1

        If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.

        Hence, the magnetic field at a distance d2, on the equatorial line of the magnet can be written as:

        Equating equations (1) and (2), we get:

        The new null points will be located 11.1 cm on the normal bisector.

         

        5.15. A short bar magnet of magnetic moment 5.25 × 10−2 J T−1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º with earth’s field on

        (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

        ANSWER:

        Magnetic moment of the bar magnet, M = 5.25 × 10−2 J T−1

        Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 × 10−4 T

        (a) The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:

        Where,

         = Permeability of free space = 4π × 10−7 Tm A−1

        When the resultant field is inclined at 45° with earth’s field, B = H

        (b) The magnetic field at a distanced from the centre of the magnet on its axis is given as:

        The resultant field is inclined at 45° with earth’s field.

        5.16. Answer the following questions:

        (a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?

        (b) Why is diamagnetism, in contrast, almost independent of temperature?

        (c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?

        (d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

        (e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?

        (f ) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?

        ANSWER:

        (a) Owing to the random thermal motion of the molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.

        (b) Each molecule of the diamagnetic material is not a magnetic dipole in itself. Hence, the random thermal motion of the molecules of the diamagnetic material (which is related to the temperature) does not affect the diamagnetism of the material.

        (c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly less than a toroid whose core is empty.

        (d) The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field.

        (e) The permeability of a ferromagnetic material is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point. The proof of this fact is based on the boundary conditions of the magnetic fields at the interface of two media.

        (f) Yes, the maximum possible magnetisation of a paramagnetic sample will be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.

        5.17. Answer the following questions:

        (a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.

        (b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?

        (c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

        (d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

        (e) A certain region of space is to be shielded from magnetic fields.

        Suggest a method.

        ANSWER:

        The hysteresis curve (B–H curve) of a ferromagnetic material is shown in the following figure.

        (a) It can be observed from the given curve that magnetisation persists even when the external field is removed. This reflects the irreversibility of a ferromagnet.

        (b)The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat energy.

        (c)The value of magnetisation is memory or record of hysteresis loop cycles of magnetisation. These bits of information correspond to the cycle of magnetisation. Hysteresis loops can be used for storing information.

        (d)Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.

        (e)A certain region of space can be shielded from magnetic fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

        5.18. A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of west to 10° north of east. The magnetic meridian of the place happens to be 10º west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

        ANSWER:

        Current in the wire, I = 2.5 A

        Angle of dip at the given location on earth,  = 0°

        Earth’s magnetic field, H = 0.33 G = 0.33 × 10−4 T

        The horizontal component of earth’s magnetic field is given as:

        HH = H cos 

        The magnetic field at the neutral point at a distance R from the cable is given by the relation:

        Where,

         = Permeability of free space = 

        Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.

        5.19. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35º. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

        ANSWER:

        Number of horizontal wires in the telephone cable, n = 4

        Current in each wire, I = 1.0 A

        Earth’s magnetic field at a location, H = 0.39 G = 0.39 × 10−4 T

        Angle of dip at the location, δ = 35°

        Angle of declination, θ ∼ 0°

        For a point 4 cm below the cable:

        Distance, r = 4 cm = 0.04 m

        The horizontal component of earth’s magnetic field can be written as:

        Hh = Hcosδ − B

        Where,

        B = Magnetic field at 4 cm due to current I in the four wires

         = Permeability of free space = 4π × 10−7 Tm A−1

        = 0.2 × 10−4 T = 0.2 G

        ∴ Hh = 0.39 cos 35° − 0.2

        = 0.39 × 0.819 − 0.2 ≈ 0.12 G

        The vertical component of earth’s magnetic field is given as:

        Hv = Hsinδ

        = 0.39 sin 35° = 0.22 G

        The angle made by the field with its horizontal component is given as:

        The resultant field at the point is given as:

        For a point 4 cm above the cable:

        Horizontal component of earth’s magnetic field:

        Hh = Hcosδ + B

        = 0.39 cos 35° + 0.2 = 0.52 G

        Vertical component of earth’s magnetic field:

        Hv = Hsinδ

        = 0.39 sin 35° = 0.22 G

        Angle, θ  = 22.9°

        And resultant field:

        5.20. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45º with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.

        (a) Determine the horizontal component of the earth’s magnetic field at the location.

        (b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90º in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

        ANSWER:

        Number of turns in the circular coil, N = 30

        Radius of the circular coil, r = 12 cm = 0.12 m

        Current in the coil, I = 0.35 A

        Angle of dip, δ = 45°

        (a) The magnetic field due to current I, at a distance r, is given as:

        Where,

         = Permeability of free space = 4π × 10−7 Tm A−1

        = 5.49 × 10−5 T

        The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as:

        BH = Bsin δ

        = 5.49 × 10−5 sin 45° = 3.88 × 10−5 T = 0.388 G

        (b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 º, the needle will reverse its original direction. In this case, the needle will point from East to West.

        5.21. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60º, and one of the fields has a magnitude of 1.2 × 10−2 T. If the dipole comes to stable equilibrium at an angle of 15º with this field, what is the magnitude of the other field?

        ANSWER:

        Magnitude of one of the magnetic fields, B1 = 1.2 × 10−2 T

        Magnitude of the other magnetic field = B2

        Angle between the two fields, θ = 60°

        At stable equilibrium, the angle between the dipole and field B1, θ1 = 15°

        Angle between the dipole and field B2, θ2 = θ − θ1 = 60° − 15° = 45°

        At rotational equilibrium, the torques between both the fields must balance each other.

        ∴Torque due to field B1 = Torque due to field B2

        MB1 sinθ1 = MB2 sinθ2

        Where,

        M = Magnetic moment of the dipole

        Hence, the magnitude of the other magnetic field is 4.39 × 10−3 T.

        5.22. A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me= 9.11 × 10−19 C). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

        ANSWER:

        Energy of an electron beam, E = 18 keV = 18 × 103 eV

        Charge on an electron, e = 1.6 × 10−19 C

        E = 18 × 103 × 1.6 × 10−19 J

        Magnetic field, B = 0.04 G

        Mass of an electron, me = 9.11 × 10−19 kg

        Distance up to which the electron beam travels, d = 30 cm = 0.3 m

        We can write the kinetic energy of the electron beam as:

        The electron beam deflects along a circular path of radius, r.

        The force due to the magnetic field balances the centripetal force of the path.

        Let the up and down deflection of the electron beam be 

        Where,

        θ = Angle of declination

        Therefore, the up and down deflection of the beam is 3.9 mm.

        5.23. A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10−23 J T−1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

        ANSWER:

        Number of atomic dipoles, n = 2.0 × 1024

        Dipole moment of each atomic dipole, M = 1.5 × 10−23 J T−1

        When the magnetic field, B1 = 0.64 T

        The sample is cooled to a temperature, T1 = 4.2°K

        Total dipole moment of the atomic dipole, Mtot = n × M

        = 2 × 1024 × 1.5 × 10−23

        = 30 J T−1

        Magnetic saturation is achieved at 15%.

        Hence, effective dipole moment, 

        When the magnetic field, B2 = 0.98 T

        Temperature, T2 = 2.8°K

        Its total dipole moment = M2

        According to Curie’s law, we have the ratio of two magnetic dipoles as:

        Therefore,  is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.

        5.24. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

        ANSWER:

        Mean radius of a Rowland ring, r = 15 cm = 0.15 m

        Number of turns on a ferromagnetic core, N = 3500

        Relative permeability of the core material, 

        Magnetising current, I = 1.2 A

        The magnetic field is given by the relation:

        B 

        Where,

        μ0 = Permeability of free space = 4π × 10−7 Tm A−1

        Therefore, the magnetic field in the core is 4.48 T.

        5.25. The magnetic moment vectors μsand μlassociated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by:

        μs= –(e/m) S,

        μl = –(e/2m)l

        Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

        ANSWER:

        The magnetic moment associated with the orbital angular momentum is valid with the classical mechanics.

        The magnetic moment associated with the orbital angular momentum is given as

        For current i and area of cross-section A, we have the relation:

        Magnetic moment

        ………………………….(1)

        Where,

        e= Charge of the electron

        r= Radius of the circular orbit

        T= Time taken to complete one rotation around the circular orbit of radius r

        Orbital angular momentum, l= mvr

        …………………………….(2)

        Where,

        m= Mass of the electron

        v= Velocity of the electron

        r= Radius of the circular orbit

        Dividing equation (1) by equation (2), we get:

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