12th Physics Study Material 2023-24 PDF

KVS ZIET CHANDIGARH 2023-24  INDEX No. NAME OF CHAPTER PAGE No. CH 1 ELECTRIC CHARGES AND FIELDS 8-20 CH 2 ELECTRIC POTENTIAL AND CAPACITANCE 21-35 CH 3 CURRENT ELECTRICITY...

KVS ZIET CHANDIGARH 2023-24  INDEX No. NAME OF CHAPTER PAGE No. CH 1 ELECTRIC CHARGES AND FIELDS 8-20 CH 2 ELECTRIC POTENTIAL AND CAPACITANCE 21-35 CH 3 CURRENT ELECTRICITY 36-46 CH 4 MOVING CHARGES AND MAGNETISM 47-57 CH 5 MAGNETISM AND MATTER 58-66 CH 6 ELECTROMAGNETIC INDUCTION 67-76 CH 7 ALTERNATING CURRENT 77-84 CH 8 ELECTROMAGNETIC WAVES 85-93 CH 9 RAY OPTICS AND OPTICAL INSTRUMENTS 94-108 CH 10 WAVE OPTICS 109-116 CH 11 DUAL NATURE OF RADIATION AND MATTER 117-126 CH 12 ATOMS 127-135 CH 13 NUCLEI 136-144 CH 14 SEMICONDUCTOR ELECTRONICS: MATERIAL DEVICES 145-153 ITS ALSO INCLUDES MIND MAP & SQP CHAPTERWISE KVS ZIET CHANDIGARH 1  PHYSICS Class XII (Code No.42) (2023-24) Senior Secondary stage of school education is a stage of transition from general education to discipline-based focus on curriculum. The present updated syllabus keeps in view the rigor and depth of disciplinary approach as well as the comprehension level of learners. Due care has also been taken that the syllabus is comparable to the international standards. Salient features of the syllabus include: Emphasis on basic conceptual understanding of the content. Emphasis on use of SI units, symbols, nomenclature of physical quantities and formulations as per international standards. Providing logical sequencing of units of the subject matter and proper placement of concepts with their linkage for better learning. Reducing the curriculum load by eliminating overlapping of concepts/content within the discipline and other disciplines. Promotion of process-skills, problem-solving abilities and applications of Physics concepts. Besides, the syllabus also attempts to ❖ Strengthen the concepts developed at the secondary stage to provide firm foundation for further learning in the subject. ❖ Expose the learners to different processes used in Physics-related industrial and technological applications. ❖ Develop process-skills and experimental, observational, manipulative, decision making and investigatory skills in the learners. ❖ Promote problem solving abilities and creative thinking in learners. ❖ Develop conceptual competence in the learners and make them realize and appreciate the interface of Physics with other disciplines KVS ZIET CHANDIGARH 2 KVS ZIET CHANDIGARH 3 Unit I: Electrostatics 26 Periods Chapter–1: Electric Charges and Fields Electric charges, Conservation of charge, Coulomb's law-force between two- point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric field. Electric flux, statement of Gauss's theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside). Chapter–2: Electrostatic Potential and Capacitance Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two-point charges and of electric dipole in an electrostatic field. Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarization, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor (no derivation, formulae only). Unit II: Current Electricity 18 Periods Chapter–3: Current Electricity Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm's law, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, temperature dependence of resistance, Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff's rules, Wheatstone bridge. Unit III: Magnetic Effects of Current and Magnetism 25 Periods Chapter–4: Moving Charges and Magnetism Concept of magnetic field, Oersted's experiment. Biot - Savart law and its application to current carrying circular loop. Ampere's law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields. Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; Current loop as a magnetic dipole and its magnetic dipole moment, moving coil galvanometer- its current sensitivity and conversion to ammeter and voltmeter. Chapter–5: Magnetism and Matter Bar magnet, bar magnet as an equivalent solenoid (qualitative treatment only), magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis (qualitative treatment only), torque on a magnetic dipole (bar magnet) in a uniform magnetic field (qualitative treatment only), magnetic field lines. KVS ZIET CHANDIGARH 4 Magnetic properties of materials- Para-, dia- and ferro - magnetic substances with examples, Magnetization of materials, effect of temperature on magnetic properties. Unit IV: Electromagnetic Induction and Alternating Currents 24 Periods Chapter–6: Electromagnetic Induction Electromagnetic induction; Faraday's laws, induced EMF and current; Lenz's Law, Self and mutual induction. Chapter–7: Alternating Current Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LCR series circuit (phasors only), resonance, power in AC circuits, power factor, wattless current. AC generator, Transformer Unit V: Electromagnetic waves 04 Periods Chapter–8: Electromagnetic Waves Basic idea of displacement current, Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses. Unit VI: Optics 30 Periods Chapter–9: Ray Optics and Optical Instruments Ray Optics: Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lens maker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism. Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers. Chapter–10: Wave Optics Wave optics: Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young's double slit experiment and expression for fringe width (No derivation final expression only), coherent sources and sustained interference of light, diffraction due to a single slit, width of central maxima (qualitative treatment only) Unit VII: Dual Nature of Radiation and Matter 08 Periods Chapter–11: Dual Nature of Radiation and Matter Dual nature of radiation, Photoelectric effect, Hertz and Lenard's observations; Einstein's photoelectric equation-particle nature of light. Experimental study of photoelectric effect Matter waves-wave nature of particles, de-Broglie relation. KVS ZIET CHANDIGARH 5 Unit VIII: Atoms and Nuclei 15 Periods Chapter–12: Atoms Alpha-particle scattering experiment; Rutherford's model of atom; Bohr model of hydrogen atom, Expression for radius of nth possible orbit, velocity and energy of electron in nth orbit, hydrogen line spectra (qualitative treatment only). Chapter–13: Nuclei Composition and size of nucleus, nuclear force Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion. Unit IX: Electronic Devices 10 Periods Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Intrinsic and extrinsic semiconductors- p and n type, p-n junction Semiconductor diode - I-V characteristics in forward and reverse bias, application of junction diode -diode as a rectifier KVS ZIET CHANDIGARH 6 The strategy to score 90+ in CBSE 12 Physics– Students who are weaker in Mathematics should try to read Modern Physics initially. The Semiconductor chapter is one of the most important chapters as its weight-age is around 8 marks. Atom and Nuclei together constitute 6 marks. The next set of chapters will consist of Dual Nature of Radiation which carries 4 marks. Having command over these chapters help students a great deal in scoring higher marks in Physics. Now you need to go for important and difficult chapters like Optics which carries 14 marks in the board exam. In Optics, you need to complete Ray optics first, which is easier to study. In Wave Optics, you should concentrate more on problems of Interference, Diffraction, and Young’s double slit experiment. Current and Electricity is also an easier chapter and it carries 7 marks. If one has not studied Electrostatics and Magnetism till date then it is better to leave these chapters because these two units consume more time than the others. However, these chapters are very important for competitive exams. EMI is another unit that you need to focus on to score well. It is an important unit both for the board exam and competitive exams. Practice the ray diagram and other important diagrams carefully. Drawing and studying will help you understand many topics faster and in an easier way. Try to solve objective problems to make yourself efficient in solving 1-mark questions in the exam; this will be a bonus for you. KVS ZIET CHANDIGARH 7 KVS ZIET CHANDIGARH 8  KEY FEATURES 1. Charge- Charge is the property associated with matter due to which it produces and experiences electric and magnetic effect. 2. Conductors and Insulators Those substances which readily allow the passage of electricity through them are called conductors, e.g. metals, the earth and those substances which offer high resistance to the passage of electricity are called insulators, e.g. plastic rod and nylon. 3. Transference of electrons is the cause of frictional electricity. 4. Additivity of Charges- Charges are scalars and they add up like real numbers. It means if a system consists of n charges q1, q2, q3 , … ,qn, then total charge of the system will be q1 +q2 + … +qn. 5. Conservation of Charge The total charge of an isolated system is always conserved, i.e. initial and final charge of the system will be same. 6. Quantisation of Charge - Charges exists in discrete amount rather than continuous value and hence, quantised. Mathematically, charge on an object, q=±ne where, n is an integer and e is electronic charge. When any physical quantity exists in discrete packets rather than in continuous amount, the quantity is said to be quantised. Hence, charge is quantised. 7. Units of Charge (i) SI unit coulomb (C) (ii) CGS system (a) electrostatic unit, esu of charge or stat-coulomb (stat-C) (b) electromagnetic unit, emu of charge or ab-C (ab-coulomb) 1 ab-C = 10 C, 1 C = 3 x 109 stat-C 8. Coulomb’s Law It states that the electrostatic force of interaction or repulsion acting between two stationary point charges is given by 9. Electrostatic forces (Coulombian forces) are conservative forces. KVS ZIET CHANDIGARH 9 10. Principle of Superposition of Electrostatic Forces This principle states that the net electric force experienced by a given charge particle q0 due to a system of charged particles is equal to the vector sum of the forces exerted on it due to all the other charged particles of the system. 11. Electrostatic Force due to Continuous Charge Distribution The region in which charges are closely spaced is said to have continuous distribution of charge. It is of three types given as below: KVS ZIET CHANDIGARH 10 12. Electric Field Intensity The electric field intensity at any point due to source charge is defined as the force experienced per unit positive test charge placed at that point without disturbing the source charge. It is expressed as 13. Electric Field Intensity (EFI) due to a Point Charge 14. Electric Field due to a System of Charges Same as the case of electrostatic force, here we will apply principle of superposition, i.e. 15. Electric Field Lines Electric field lines are a way of pictorially mapping the electric field around a configuration of charge(s). These lines start on positive charge and end on negative charge. The tangent on these lines at any point gives the direction of field at that point. 16. Electric field lines due to positive and negative charge and their combinations are shown as below: 17. Electric Dipole Two-point charges of same magnitude and opposite nature separated by a small distance altogether form an electric dipole. KVS ZIET CHANDIGARH 11 18. Electric Dipole Moment The strength of an electric dipole is measured by a vector quantity known as electric dipole moment (p) which is the product of the charge (q) and separation between the charges (2l). 19. Electric Field due to a Dipole Electric field of an electric dipole is the space around the dipole in which the electric effect of the dipole can be experienced. 21. Torque on an electric dipole placed in a uniform electric field (E) is given by 24. Dipole is in stable equilibrium in uniform electric field when angle between p and E is 0° and in unstable equilibrium when angle θ= 180°. 25. Net force on electric dipole placed in a uniform electric field is zero. 26. There exists a net force and torque on electric dipole when placed in non-uniform electric field. 27. Work done in rotating the electric dipole from θ1 to θ2 is W = pE (cos θ1 – cos θ2) 28. Potential energy of electric dipole when it rotates from θ1 = 90° to θ2 =0 U = pE (cos 90° – cos θ) = -pE cos θ = – p.E KVS ZIET CHANDIGARH 12 29. Work done in rotating the dipole from the position of stable equilibrium to unstable equilibrium, i.e. when θ1 = 0° and θ2 = π. W = 2 pE 30. Work done in rotating the dipole from the position of stable equilibrium to the position in which dipole experiences maximum torque, i.e. when θ1 = 0° and θ2 = 90°. W = pE 31. Electric flux. The electric flux through a small surface ts defined as the electric lines of force passing through that are when held normally to the lines of force. Mathematically-- φ= ⃗𝑬⃗. ⃗⃗⃗⃗⃗ ∆𝑺 where E is the electric field and AS is the area vector representing the elementary surface area. Unit. In SI, unit of electric flux is newton metre2 coulomb-2 (N m2 C-2 ). 32. Gauss’ theorem. It states that the total outward electric flux through a closed surface is 1 times the charge enclosed |by the closed surface. ∈𝑜 𝒒 𝑬. ⃗⃗⃗⃗⃗ Mathematically: ∮ ⃗⃗⃗ 𝒅𝑺 = ∈ 𝑶 where q is charge enclosed by the closed surface. 33. Gaussian surface. Any closed surface around the charge distribution (may be a point charge, a line charge, a surface charge or a volume charge) so that Gauss’ theorem can be conveniently applied to find electrical field due to it is called the gaussian surface. 34.Electric field due to infinitely long straight wire of linear charge density λ 𝛌 E = 𝟐𝝅∈ 𝑶 𝒓 where r is perpendicular distance of the observation point from the wire. 35. Electric field due to an infinite plane sheet of charge of surface charge density σ Electric field between two infinite plane parallel sheets of charge of surface charge density 0 and -o: 𝛔 E = 𝟐∈ 𝑶 36. Electric field due to spherical shell of surface charge density o and radius R: 𝟏 𝒒 E= 𝟒𝝅∈ for r>R (outside the shell) 𝑶 𝒓𝟐 E=0, for rR (outside the sphere) 𝟒𝝅∈𝑶 𝒓𝟐 𝟏 𝒒𝒓 E= 𝟒𝝅∈ 𝟑 for r + q then in between 1 the charges the electric field is zero at a point (a) closer to + Q (b) exactly at the mid-point of line segment joining + Q and + q. (c) closer to + q (d) nowhere on the line segment joining + Q and + q. 2 Assertion: A metallic shield in form of a hollow shell may be built to block an electric field. 1 Reason: In a hollow spherical shield, the electric field inside it is zero at every point. a- Both assertion and reason are correct and the reason is the correct explanation of assertion. b- Both assertion and reason are correct and reason is not a correct explanation of assertion. c- Assertion is correct but the reason is incorrect d- Assertion is incorrect but the reason is correct. 3 Electric lines of force about a negative point charge are 1 (a) circular anticlockwise (b) circular clockwise (c) radial, inwards (d) radial, outwards 4 The electric field at a point on equatorial line of a dipole and direction of the dipole moment 1 (a) will be parallel (b) will be in opposite direction (c) will be perpendicular (d) are not related 5 Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge 2 ‘q’ and other an equal negative charge. Are their masses after charging equal? 6 An electric dipole free to move is placed in an electric field. What is the action on it, when it is 2 placed in (a) a uniform electric field (b) a non-uniform electric field? 7 Derive a relation for the intensity of electric field at an equatorial point of an electric dipole. 3 Case study-based questions (questions no 8- 10) 4 In a uniform electric field of strength E, the net electric force is zero; but a torque equal to pE sin θ acts on the dipole (where θ is the angle between directions of dipole moment p and electric field E). This torque tends to align the dipole along the direction of electric field. Torque in vector form 𝜏 = 𝑝 𝑥 𝐸⃗ 8. When is the torque applied is maximum? 1 9. What is the direction of torque applied 1 10. What is net force and net when an electric dipole is placed in uniform electric field? 2 OR 10. What is net force and net when an electric dipole is placed in non-uniform electric field? 2 11 (a) A point charge (+Q) is kept in the vicinity of uncharged conducting plate. 5 Sketch electric field lines between the charge and the plate. (b) Two infinitely large plane thin parallel sheets having surface charge densities σ1 and σ2 (σ1> σ2) are shown in the figure. Write the magnitudes and directions of the net fields in the regions marked II and III. KVS ZIET CHANDIGARH 20 KVS ZIET CHANDIGARH 21  KEY FEATURES 1. Electric potential difference. The electric potential difference between two points in an electric field is defined as the amount of work done per unit positive test charge in moving the test charge from one point to the other against the electrostatic force due to the field. Mathematically - If W is work done in moving a small positive test charge q, from point A to B in the electrostatic field of charge q, then potential difference between points B and A, 𝑊𝐴𝐵 𝑞 1 1 𝑉𝐵 − 𝑉𝐴 = == (𝑟 − ) 𝑞𝑂 4𝜋∈𝑂 𝐵 𝑟𝐴 Here, 𝑟𝐴 and 𝑟𝐵 are distances of points A and B from the source charge q. Unit. Its unit in SI is volt (V) 1 volt (V) = 1 joule coulomb! (J C1) 2. Electric potential. The electric potential at a point in an electric field is defined as the amount of work done per unit positive test charge in moving the test charge from infinity to that point against the electrostatic force due to the field. Mathematically - If W is work done in moving a small positive test charge from infinity to point A in the electrostatic field of charge q, then potential at point A, 𝑊𝐴𝐵 1 𝑞 V= = 𝑞𝑂 4𝜋∈𝑂 𝑟 Here, r is the distance of the point A from the source charge q. Unit. Its unit is also volt. 1V=1J/s 3. Electric potential due to group of charges. The electric potential at a point due to a group of charges is equal to the algebraic sum of the electric potentials due to individual charges at that point. It is because of the reason that electric potential is a scalar quantity. 1 𝑞 𝑞2 𝑞3 𝑞𝑛 V= ( 𝑟1 + + + ⋯……+ ) 4𝜋∈𝑂 1 𝑟2 𝑟3 𝑟𝑛 4. Potential gradient. The rate of change of potential with distance at a point is called potential gradient at that point. The electric field at a point is equal to the negative potential gradient at that point. 𝑑𝑉 Mathematically - E= - 𝑑𝑟 Unit. Its unit in SI is volt /metre (V/m). 5. Equipotential surface. The surface at every point of which, the electric potential is same, is called equipotential surface. Two equipotential surfaces can never intersect each other. 6. Electrostatic potential energy of a system of charges. It is defined as the work done to put the charges constituting the system at their respective locations after having been removed to infinity. Mathematically- A. Potential energy of the system of two charges q1 and q2 1 𝑞1 𝑞2 U = 4𝜋∈𝑂 𝑟 KVS ZIET CHANDIGARH 22 B. Potential energy of the system of three charges q1, q2 and q3 1 𝑞1 𝑞2 𝑞2 𝑞3 𝑞3 𝑞1 U= ( + + ) 4𝜋∈𝑂 𝑟12 𝑟23 𝑟31 Unit. Its unit in SI is joule (J) or electron volt (eV). 1 eV =1.6 x 10-19 J 7. Potential energy of an electric dipole in a uniform electric field. 1. If the electric dipole is rotated from initial orientation making angle 6, with the electric field to the final orientation making angle 𝝑 , with the field, then U = pE (cos𝝑𝟐 — cos𝝑𝟏 ) 2. If the electric dipole is rotated from its initial orientation perpendicular to the field to the final orientation so as to make an angle 𝝑 with the field, then ⃗ ⃗⃗⃗⃗ U = -pEcos 𝝑 = - 𝒑.𝑬 8. Behaviour of a charged conductor A. Charges reside only at the surface of the charged conductor. B. The electric potential is constant at the surface and inside the conductor. C. The electric field is zero inside the conductor and just outside it, the electric field is normal to the surface. 9. Electrical capacitance. The ability of a conductor to store charge is called its electrical capacitance. 𝑞 Mathematically- C = 𝑉 Unit. Its unit in SI is farad (F). 1 farad (F) = 1 coulomb /volt (C/ V) Capacitance of a spherical conductor. C = 4𝜋 ∈𝑂 r , r is radius (in metre) of the spherical conductor. 10. Capacitor. It is an arrangement for storing a very large amount of charge. 11. Principle. The capacitance of a conductor gets increased greatly, when an earth connected conductor is placed near it. ∈𝑂 𝐴 12. Parallel plate capacitor C= 𝑑 (when air is between the plates) 𝐾 ∈𝑂 𝐴 C= 𝑑 (when dielectric is between the plates) Here, A is area of each plate and d is separation between the two plates. 13. Energy stored in a capacitor. Work done in charging a capacitor gets stored in the capacitor in the form of its electrostatic potential energy. 1 1 1 𝑄2 Mathematically: U = 𝐶𝑉 2 = qV = = 2 2 2 𝐶 14. Dielectric constant. The ratio of the strength of the applied electric field to the strength of reduced value of electric field on inserting the dielectric slab between the plates of a capacitor is called the dielectric constant of the slab. 15. Dielectric strength. The maximum value of electric field (or potential gradient) that can be applied to the dielectric without its electric breakdown is called dielectric strength of the dielectric. Unit. Its unit in V/m (same as that of electric field). 16. Effect of dielectric slab on the capacitance of a parallel plate capacitor. 1. When a dielectric slab of dielectric constant K and thickness t (t < d) is introduced between the plates, then ∈𝑂 𝐴 C= 𝑡 𝑑−𝑡(1− ) 𝐾 ∈𝑂 𝐴 2. When a conducting slab of thickness t (f < d) is introduced, then C= 𝑡 𝑑(1− ) 𝐾 KVS ZIET CHANDIGARH 23  QUESTIONS WITH ANSWERS Q.1 Name the physical quantity whose SI unit is JC–1. Is it a scalar or a vector quantity? Ans. Electric potential. It is a scalar quantity. Q.2 In the given figure, charge +Q is placed at the centre of a dotted circle. Work done in taking another charge +q from A to B is W1 and from B to C is W2. Which one of the following is correct: W1 > W2, W1=W2 and W1 < W2? Ans- The points A and C are at same distance from the charge +Q at the centre, so VA = VC Therefore, VA – VB = VC – VB Hence, the magnitude of work done in taking charge +q from A to B or from B to C will be the same i.e., W1 = W2 Q.3 The field lines of a negative point charge are as shown in the figure. Does the kinetic energy of a small negative charge increase or decrease in going from B to A? Ans-The kinetic energy of a negative charge decreases while going from point B to point A, against the movement of force of repulsion. Q. 4 A point charge Q is placed at point ‘O’ as shown in figure. Is the potential at point A, i.e., VA, greater, smaller or equal to potential, VB, at point B, when Q is (i) positive, and (ii) negative charge? Ans- Q.5 Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction Ans- The equipotential surfaces are the equidistant planes normal to the z-axis, i.e., planes parallel to the X–Y plane. Q.6 Why do the equipotential surfaces due to a uniform electric field not intersect each other? Ans. This is because at the point of intersection there will be two values of electric potential, which is not possible. Q.7 Why is there no work done in moving a charge from one point to another on an equipotential surface? Ans. The potential difference between any two points of equipotential surface is zero. 𝑊 We have 𝑉1 − 𝑉2 = =0⇒ 𝑊 =0 𝑞 KVS ZIET CHANDIGARH 24 Q.8 The figure shows the field lines of a positive point charge. What will be the sign of the potential energy difference of a small negative charge between the points Q and P? Justify your answer. Ans-The sign of the potential energy difference of a small negative charge will be positive. This is because negative charge moves from a point at a lower potential energy to a point at a higher potential energy. Q.9 Two uniformly large parallel thin plates having charge densities +σ and – σ are kept in the X-Z plane at a distance ‘d’ apart. Sketch an equipotential surface due to electric field between the plates. If a particle of mass m and charge ‘–q’ remains stationary between the plates, what is the magnitude and direction of this field? Ans- The equipotential surface is at a distance d/2 from either plate in X-Z plane. For a particle of charge (–q) at rest between the plates, then (i)weight mg acts vertically downward (ii) electric force qE acts vertically upward. So, mg = qE 𝑚𝑔 E= vertically downward, i.e., along (–)Y-axis. 𝑞 Q.10 (a) A parallel plate capacitor (C1) having charge Q is connected, to an identical uncharged capacitor C2 in series. What would be the charge accumulated on the capacitor C2? (b) Three identical capacitors each of capacitance 3 µF are connected, in turn, in series and in parallel combination to the common source of V volt. Find out the ratio of the energies stored in two configurations. Ans- Q.11 A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere? Ans. Potential at centre of sphere = 10 V. Potential at all points inside the hollow metal sphere (or any surface) is always equal to the potential at its surface. KVS ZIET CHANDIGARH 25 Q.12 Net capacitance of three identical capacitors in series is 1 µF. What will be their net capacitance if connected in parallel? Find the ratio of energy stored in the two configurations if they are both connected to the same source. Ans- Q.13 Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 µF. When the ends X and Y are connected to a 6 V battery, find out (i) the charge and (ii) the energy stored in the network. Ans- KVS ZIET CHANDIGARH 26 Q.14 Four charges +q, – q, + q and – q are to be arranged respectively at the four corners of a square ABCD of side ‘a’. (a) Find the work required to put together this arrangement. (b) A charge q0 is brought to the centre of the square, the four charges being held fixed. How much extra work is needed to do this? Ans- Q.15 The two graphs are drawn below, show the variations of electrostatic potential (V) with 1/r (r being the distance of field point from the point charge) for two-point charges q1 and q2. (i) What are the signs of the two charges? (ii) Which of the two charges has the larger magnitude and why? Ans- (i)The potential due to positive charge is positive and due to negative charge, it is negative, so, q1 is positive and q2 is negative. 1 𝑞 (ii) V = 4𝜋∈𝑂 𝑟 𝑞 The graph between V and 1/r is a straight line passing through the origin with slope. 4𝜋∈𝑂 As the magnitude of slope of the line due to charge q2 is greater than that due to q1, q2 has larger magnitude. Q.16 Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination. Ans- KVS ZIET CHANDIGARH 27 Q.17 A parallel plate capacitor each with plate area A and separation ‘d’ is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any, will take place in (i) charge on the plates, (ii) electric field intensity between the plates, (iii) capacitance of the capacitor? Justify your answer in each case. Ans- Q. 18 A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of KVS ZIET CHANDIGARH 28 (i) capacitance (ii) potential difference between the plates (iii) electric field between the plates, and (iv) the energy stored in the capacitor. Ans- Q.19 A parallel plate is charged by a battery. When the battery remains connected, a dielectric slab is inserted in the space between the plates. Explain what changes if any, occur in the values of (i) potential difference between the plates (ii) electric field strength between the plates (iii) capacitance (iv) charge on the plates (v) energy stored in the capacitor Ans- (i) When battery remains connected, the potential difference remains the same. 𝑽 (ii) As electric field, E =𝒅 , V = constant and d = constant; therefore, electric field strength remains the same. (iii) The capacitance of capacitor increases as K > 1. (iv) The charge Q = CV, V = same, C = increases; therefore, charge on plates increases. 1 (v) Energy stored by capacitor U = 2 𝐶𝑉 2 , also increases. Q.20 Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium K = 4. (i) Calculate the capacitance of each capacitor if equivalent capacitance of the combination is 4 µF. (ii) Calculate the potential difference between the plates of X and Y. (iii) Estimate the ratio of electrostatic energy stored in X and Y. Ans- KVS ZIET CHANDIGARH 29 Q.21 Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B, calculate the charge drawn from the battery by the circuit. Ans- Hence, charge drawn from battery (Q) = CV = 10 × 10 mC = 100 mC = 10–4 C KVS ZIET CHANDIGARH 30 Q.22 Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also find the charge drawn from the battery in each case. Ans- Q.23 Two identical parallel plate (air) capacitors C1 and C2 have capacitances C each. The space between their plates is now filled with dielectrics as shown. If the two capacitors still have equal capacitance, obtain the relation between dielectric constants K, K1 and K2. Ans- KVS ZIET CHANDIGARH 31 Q.24 If N drops of same size each having the same charge, coalesce to form a bigger drop. How will the following vary with respect to single small drop? (i) Total charge on bigger drop (ii) Potential on the bigger drop (iii) Capacitance. Ans- Q. 25 You are given an air-filled parallel plate capacitor C1. The space between its plates is now filled with slabs of dielectric constants K1 and K2 as shown in C2. Find the capacitances of the capacitor C2. if area of the plates is A and distance between the plates is d. Ans- KVS ZIET CHANDIGARH 32 KVS ZIET CHANDIGARH 33  REVISION PAPER UNIT- II –ELECTROSTATIC-POTENTIAL AND CAPACITANCE Note: Q. No. 1-4 is of 01 mark each, Q. 5-6 is of 02 marks each, Q.No.7 is of 03 marks, Q. No. 8 is a case study based and is of 04 marks, Q. No. 11 is of 5 marks. S Question Ma N rks 1 A positively charged particle is released from rest in a uniform electric field. The electric 1 potential energy of the charge (a) remains a constant because the electric field is uniform. (b) increases because the charge moves along the electric field. (c) decreases because the charge moves along the electric field. (d) decreases because the charge moves opposite to the electric field. 2 Assertion: When two conductors charged to different potentials are connected to each other, the 1 negative charge always flows from lower potential to higher potential. Reason: In the charging process, there is always a flow of electrons only. e- Both assertion and reason are correct and the reason is the correct explanation of assertion. f- Both assertion and reason are correct and reason is not a correct explanation of assertion. g- Assertion is correct but the reason is incorrect h- Assertion is incorrect but the reason is correct. 3 A capacitor is charged by a battery. The battery is removed and another identical uncharged 1 capacitor is connected in parallel. The total electrostatic energy of resulting system (a) increases by a factor of 4. (b) decreases by a factor of 2. (c) remains the same. (d) increases by a factor of 2. 4 A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting 1 the charging battery, the distance between the plates of the capacitor is increased using an insulating handle. As a result, the potential difference between the plates (a) increases (b) decreases (c) does not change (d) becomes zero 5 Can electrostatic potential at a point be zero, while electric field at that point is not zero? 2 6 If a dielectric slab is introduced between the plates of a parallel plate capacitor after the battery is 2 disconnected. How do the following quantities change? (i) Charge (ii) Potential difference (iii) Capacitance (iv) Energy. 7 Define an equipotential surface. Draw equipotential surfaces. 3 (i) in the case of a single point charge and (ii) in a constant electric field in Z-direction. Why the equipotential surfaces about a single charge are not equidistant? (iii) Can electric field exist tangential to an equipotential surface? Give reason. KVS ZIET CHANDIGARH 34  Case study-based questions (questions no 8- 10) Capacitor and Capacitance 4 A capacitor contains two oppositely charged metallic conductors at a finite separation. It is a device by which capacity of storing charge may be varied simply by changing separation and/or medium between the conductors. The capacitance of a capacitor is defined as the ratio of 𝑄 magnitude of charge (Q) on either plate and potential difference (V) across the plate, i.e., C = 𝑉 The unit of capacitance is coulomb/volt or farad (F) 8. What is a capacitor? 1 9. What is main purpose of using a capacitor? 1 10. Can we increase the capacitance by increasing potential applied across it? 2 OR 10. What will be the effect on capacitance by inserting a dielectric in between the plates? 2 11 (a) Derive an expression for the energy stored in a parallel plate capacitor C, charged to a 5 potential difference V. Hence derive an expression for the energy density of a capacitor. 3 (b) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1:2 so that the energy stored in the two cases becomes the same 2 KVS ZIET CHANDIGARH 35 KVS ZIET CHANDIGARH 36  KEY FEATURES Electric current. If is the rate of flow of electric charge through a conductor. Mathematically - I= a Unit. In SI, the unit of electric current is ampere (A). 1 ampere (A) = 1 coulomb second™! (C s~}) Ohm’s law. It states that physical conditions remaining unchanged, the current flowing through a conductor is always directly proportional to the potential difference across its two ends. Mathematically - Val or V=RI Here, R is called resistance of the conductor. Unit. The unit of resistance is ohm (Ω) 1 ohm (Ω) = 1 volt/ampere (V/A) Resistance of a conductor. The resistance of a conductor of length / and area of cross-section A is given by 𝑙 R=ρ 𝐴 Here, ρ is resistivity of the material of the conductor. Resistivity. The resistivity of the material of a conductor is the resistance offered by a wire of this material of unit length and unit area of cross-section. It is also known as specific resistance of the material of the conductor. Unit. The SI unit of resistivity is ohm metre (Ω m) Conductance. The reciprocal of the resistance of a conductor is called its conductance (G). Thus, 1 G= 𝑅 Unit. The SI unit of conductance is ohm=! (Ω ) or siemen (S). ohm-1 is also written as mho. -1 Conductivity. The reciprocal of the resistivity of the material of a conductor is called its conductivity. Thus, 1 σ= 𝜌 Unit. The SI unit of conductivity is ohm/ metre (Ω/m) or siemen /metre (S/m). ohm-1 metre-1 is also written as mho metre’. Drift velocity. It is the velocity with which a free electron in the conductor gets drifted under the influence of the applied external electric field. 𝒆𝑬 𝑰 Mathematically - 𝒗𝒅 = 𝒎 𝝉 = 𝒏𝒆𝑨 Here, 𝜏 t is average relaxation time and n is number of free electrons per unit volume in the conductor. The other symbols have their usual meanings. Temperature coefficient of resistance. It is defined as the change in resistance per unit resistance per degree rise in temperature. If resistance increases linearly up to temperature θ, then temperature coefficient, 𝑅𝑡 − 𝑅𝑂 α= 𝑅𝑂 𝜃 Unit. The unit of temperature coefficient is °C-1. KVS ZIET CHANDIGARH 37 E.M.F. The work done per unit charge by the source in taking the charge from its one terminal to the other is called the electromotive force or e.m.f. of the source. It is equal to the potential difference between the two terminals of the source, when no current is drawn from it. Internal resistance. The resistance offered by the electrolyte of the cell, when the electric current flows through it, is known as internal resistance of the cell. If V is potential difference across the two terminals of a cell, when a current I is drawn from it, then 𝑟 V = E – Ir = E (1 − ) 𝑅+𝑟 Here, E is e.m.f. of the cell and R, the external resistance in the circuit. 𝐸 Mathematically- r = R ( − 1) 𝑉 Kirchhoff’s Laws. These laws are used to analyse electric circuits. First law, It states that the algebraic sum of the currents meeting at a point in an electrical circuit is always zero. Second law. It states that in any closed part of an electrical circuit, the algebraic sum of the e.m.fs. is equal to the algebraic sum of the products of the resistances and the currents flowing through them. Wheatstone bridge. It is an arrangement of four resistances used to determine an unknown resistance. In a balanced Wheatstone bridge, 𝑃 𝑅 = 𝑄 𝑆 where P, Q, R and S are resistances in the four arms of the Wheatstone bridge. QUESTIONS WITH ANSWERS Q.1 How does the mobility of electrons in a conductor change, if the potential difference applied across the conductor is doubled, keeping the length and temperature of the conductor constant? Ans- Mobility is defined as the magnitude of drift velocity per unit electric field. 𝑣 𝑒𝐸𝜏 𝑒𝜏 µ = 𝐸𝑑 = 𝑚𝐸 = 𝑚 µ𝛼𝜏 1 At constant temperature and length, there is no change in relaxation time i.e., t ∝ 𝑇. Also it does not depend on potential difference. Hence, on changing the potential difference, there is no change in mobility of electrons. Q.2 Plot a graph showing variation of current versus voltage for the material GaAs Ans- The variation of electric current with applied voltage for GaAs is as shown KVS ZIET CHANDIGARH 38 Q.3 Two wires, one of copper and the other of manganin, have same resistance and equal thickness. Which wire is longer? Justify your answer Ans- Copper Reason: Let l1 and l2 be lengths of copper and manganin wires having same resistance R and thickness i.e., area of cross-section (A). Q.4 Nichrome and copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more? Justify your answer Ans- Q.5 Why are alloys used for making standard resistance coils? Ans. Alloys have (i)low value of temperature coefficient and the resistance of the alloy does not vary much with rise in temperature. (ii) high resistivity, so even a smaller length of the material is sufficient to design high standard resistance. Q.6 Define the terms (i) drift velocity, (ii) relaxation time. Ans. (i) Drift Velocity: The average velocity acquired by the free electrons of a conductor in a direction opposite to the externally applied electric field is called drift velocity. The drift velocity will remain the same with lattice ions/atoms. (ii) Relaxation Time: The average time of free travel of free electrons between two successive collisions is called the relaxation time. KVS ZIET CHANDIGARH 39 Q.7 Plot a graph showing variation of voltage Vs the current drawn from the cell. How can one get information from this plot about the emf of the cell and its internal resistance? Ans- Q.8 Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires. Ans- Q.9 A conductor of length ‘l’ is connected to a dc source of potential ‘V’. If the length of the conductor is tripled by gradually stretching it, keeping ‘V’ constant, how will (i) drift speed of electrons and (ii) resistance of the conductor be affected? Justify your answer. Ans- Q.10 A potential difference V is applied across the ends of copper wire of length l and diameter D. What is the effect on drift velocity of electrons if (i) V is halved? (ii) l is doubled? (iii) D is halved? Ans- KVS ZIET CHANDIGARH 40 Q.11 In the circuit shown in the figure, find the total resistance of the circuit and the current in the arm CD. Ans- Q. 12 Find the magnitude and direction of current in 1Ω resistor in the given circuit Ans- Q.13 Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points B and E? Ans KVS ZIET CHANDIGARH 41 Q.14 Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E. Ans- Q.15 Calculate the steady current through the 2 Ω resistor in the circuit shown below Ans- In steady state there is no current in capacitor branch, so equivalent circuit is shown in fig. Net resistance of circuit, KVS ZIET CHANDIGARH 42 Q.16 Show, on a plot, variation of resistivity of (i) a conductor, and (ii) a typical semiconductor as a function of temperature. Using the expression for the resistivity in terms of number density and relaxation time between the collisions, explain how resistivity in the case of a conductor increases while it decreases in a semiconductor, with the rise of temperature. 𝒎 Ans- We know that, ρ = 𝒏𝒆𝟐 𝝉 Where m is mass of electron t = charge density, τ = relaxation time e = charge on the electron. (i) In case of conductors with increase in temperature, relaxation time decreases, so resistivity increases. (ii) In case of semiconductors with increase in temperature number density (n) of free electrons increases, hence resistivity decreases Q.17 State Kirchhoff’s rules. Use these rules to write the expressions for the currents I1, I2 and I3 in the circuit diagram shown. Ans- KVS ZIET CHANDIGARH 43 KVS ZIET CHANDIGARH 44  REVISION PAPER 3 UNIT- III – CURRENT ELECTRICITY Note: Q. No. 1-4 is of 01 mark each, Q. 5-6 is of 02 marks each, Q.No.7 is of 03 marks, Q. No. 8 is a case study based and is of 04 marks, Q. No. 11 is of 5 marks. S Question Ma N rks 1 The relaxation time in conductors 1 (a) increases with the increases of temperature (b) decreases with the increases of temperature (c) it does not depend on temperature (d) all of sudden changes at 400 K 2 Assertion: For a conductor resistivity increases with increase in temperature. 1 𝑚 Reason: Since ρ =𝑛𝑒 2 𝜏, when temperature increases the random motion of free electrons increases and vibration of ions increases which decreases. i- Both assertion and reason are correct and the reason is the correct explanation of assertion. j- Both assertion and reason are correct and reason is not a correct explanation of assertion. k- Assertion is correct but the reason is incorrect l- Assertion is incorrect but the reason is correct. 3 With increase in temperature the conductivity of 1 (a) metals increases and of semiconductor decreases. (b) semiconductors increases and metals decreases. (c) in both metals and semiconductors increases. (d) in both metal and semiconductor decreases. 4 In the series combination of two or more than two resistances 1 (a) the current through each resistance is same (b) the voltage through each resistance is same (c) neither current nor voltage through each resistance is same (d) both current and voltage through each resistance are same. 5 Define the terms (i) drift velocity, (ii) relaxation time. 2 6 Two conducting wires X and Y of same diameter but different materials are joined in series across a 2 battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires. 7 Using the concept of free electrons in a conductor, derive the expression for the conductivity of a wire in 3 terms of number density and relaxation time. Hence obtain the relation between current density and the applied electric field E. Case study-based questions (questions no 8- 10) Emf of a cell 4 Emf of a cell is the maximum potential difference between two electrodes of the cell when no current is drawn from the cell. Internal resistance is the resistance offered by the electrolyte of a cell when the electric current flows through it. The internal resistance of a cell depends upon the following factors; (i) distance between the electrodes (ii) nature and temperature of the electrolyte (iii) nature of electrodes A (iv) area of electrodes. KVS ZIET CHANDIGARH 45  8. What is EMF of a cell? 1 9. Define internal resistance of a cell 1 10. List the factors on which EMF of a cell depends 2 OR 10..Explain the effect of temperature on internal resistance of a cell 2 11 (i) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of 5 number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend? 3 (ii) Why alloys like constantan and manganin are used for making standard resistors? 2 KVS ZIET CHANDIGARH 46 KVS ZIET CHANDIGARH 47  KEY FEATURES 1. Magnetic Effect of Current: A magnetic field is associated with an electric current flowing through a metallic wire. This is called magnetic effect of current. On the other hand, a stationary electron produces electric field only. 2. Source and Units of Magnetic Field Oersted’s Experiment: A Danish physicist, Hans Christian Oersted, in 1820, demonstrated that a magnetic needle is deflected by a current carrying wire. He concluded that the magnetic field is caused by current elements (or moving charges). The unit of magnetic field strength in SI system is tesla (T) or weber/metre2 (Wb/m2) or newton/ampere-metre (N A– 1 m–1). In CGS system, the unit of magnetic field is gauss (G). 1T=104 G 3. Biot-Savart Law It states that the magnetic field strength dB produced due to a current element (of current I and length dl) at a point having position vector r relative to current element is µ𝟎 𝒊𝒅𝒍 𝒔𝒊𝒏𝜽 dB = 𝟒𝝅 , 𝒓𝟐 where µ0 is permeability of free space. Its value is µ0 = 4π ×10–7 Wb/A-m. 𝜇𝑜 𝑖 4, Magnetic field at the center of a circular loop is given by B= 2 𝑟 5. Magnetic field due to a straight conductor of finite length carrying current I at a point at perpendicular distance a from it is given by 𝝁 𝒊 B = 𝟒𝝅𝒐 𝒓 [ sin∅𝟏 + 𝐬𝐢𝐧∅𝟐 ] where ∅1 and ∅2 are angles, which the lines joining the two ends of the conductor to the observation point make with the perpendicular from the observation point to the conductor. 6. Magnetic field due to a straight conductor of infinite length carrying current I at a point at 𝝁 𝟐𝒊 perpendicular distance a from it is given by B =𝟒𝝅𝒐 𝒓 7. Magnetic field due toa straight conductor of infinite length carrying current I at a point near its one 𝝁 𝒊 end at a perpendicular distance a from it is given by patio. (by =0° and > =90°) B =𝟒𝝅𝒐 𝒓 8. The magnetic field due to a current carrying straight conductor of infinite length varies inversely as the distance of the observation points from the conductor. 9. For a given distance from the current carrying straight conductor, field is maximum, when the observation point lies along a direction perpendicular to it. 10. Ampere's Circuital Law It states that the line integral of magnetic field B " along a closed path is equal to µ0-times the current (I) passing through the closed path. ⃗⃗⃗ = 𝜇𝑂 I ⃗. 𝑑𝑙 ∮𝐵 KVS ZIET CHANDIGARH 48 11. Force on a Moving Charged Particle in Magnetic Field The force on a charged particle moving with velocity v in a uniform magnetic field B is given by ⃗ ) = qvBsin𝜃 𝐹𝑚 = 𝑞(𝑣 𝑋 𝐵 This is known as Lorentz force. The direction of this force is determined by using Fleming’s lef- hand rule. ⃗ , The direction of this force is perpendicular to both 𝑣 𝑎𝑛𝑑 𝐵 ⃗ , then 𝐹𝑚 =0 When 𝑣 is parallel to 𝐵 ⃗ , then 𝐹𝑚 is maximum, i.e., 𝐹𝑚 = qvB. When 𝑣 is perpendicular to 𝐵 12. Path of Charged Particle in a Uniform Magnetic Field KVS ZIET CHANDIGARH 49  ⃗ ) =ilBsinθ 13. Magnetic Force on a Current Carrying Conductor of Length l " is given by 𝐹𝑚 = 𝑖(𝑙 𝑋 𝐵 Magnitude of force is Fm = IlB sin θ Direction of force Fm " is normal to l " and B " given by Fleming’s Left Hand Rule. If θ = 0 (.i e l , is parallel to B), then magnetic force is zero 14. Force between Parallel Current Carrying Conductors Two parallel current carrying conductors attract while antiparallel current carrying conductors repel. The magnetic force per unit length on either current carrying conductor at separation ‘r’ is given by 𝐹 𝜇𝑜 𝐼1 𝐼2 = 𝑙 2𝜋𝑟 Its unit is newton/metre abbreviated as N/m. 15. Torque Experienced by a Current Loop (of Area A) Carrying Current I in a Uniform Magnetic Field B is given by ⃗ ) = (𝑀 𝜏 = 𝑁𝑖(𝐴 𝑋 𝐵 ⃗⃗ 𝑋 𝐵 ⃗) where M = Ni A is magnetic moment of loop. The unit of magnetic moment in SI system is ampere × metre2 (Am2) 16. Potential energy of a current loop in a magnetic field When a current loop of magnetic moment M is placed in a magnetic field, then potential energy of magnetic dipole is U = - 𝑀 ⃗⃗. 𝐵 ⃗ = -MBcosθ (i) When θ=0, U=–MB (minimum or stable equilibrium position) (ii) When θ=π, U=+MB (maximum or unstable equilibrium position) 𝜋 (iii) When θ = 2 potential energy is zero 17. Conversion of Galvanometer into Ammeter A galvanometer may be converted into ammeter by using very small resistance in parallel with the galvanometer coil. The small resistance connected in parallel is called a shunt. If G is resistance of galvanometer, Ig is current in galvanometer for full scale deflection, then for conversion of galvanometer into ammeter of range I ampere, the shunt is given by 𝐺𝐼𝑔 S= 𝐼− 𝐼𝑔 18. Conversion of Galvanometer into Voltmeter A galvanometer may be converted into voltmeter by connecting high resistance (R) in series with the coil of galvanometer. If V volt is the range of voltmeter 𝑉 formed, then series resistance is given by R = 𝐼 - G 𝑔 QUESTIONS WITH ANSWERS Q. 1 Is the source of magnetic field analogue to the source of electric field? Ans. No. It is because, the source of magnetic field is not a magnetic charge. In case of electric field, the source of electric field is electric charge. Q. 2 Does a current carrying circular coil produce uniform magnetic field? KVS ZIET CHANDIGARH 50 Ans. No, magnetic field produced due to a current carrying circular coil is not uniform. However, it may be considered as uniform at the centre of the circular coil. Q. 3 What is the effect of increasing the number of turns on magnetic field produced due to a circular coil? Ans. The magnetic field produced by a coil of m turns is n times the magnetic field produced by a 𝜇𝑂 𝑁𝑖 coil of single turn. B= 2𝑟 Q. 4 Looking at a circular coil, the current is found to be flowing in anticlockwise direction. Predict the direction of magnetic field produced at a point on the axis of the coil on the same side as the observer. Ans. The direction of magnetic field is perpendicular to the plane of the coil and directed towards the observer. Q. 5 What kind of magnetic field is produced by an infinitely long current carrying conductor? Ans. Magnetic field lines are concentric circular loops in a plane perpendicular to the straight conductor. The centres of the circular magnetic field lines lie on the conductor. Q. 6. In what respect does a wire carrying a current differ from a wire, which carries no current? Ans. A current carrying wire produces magnetic field. It is because, when current flows through a wire, electrons move inside it along a definite direction. On the other hand, in a wire which carries no current, electrons are in motion in random direction. Such a wire does not produce anymagnetic field. Q.7 An electric charge enters in electric field at right angles to the direction of electric field. What is the nature of the path followed? Ans. The electric charge will move along a parabolic path. Q.8 What is the magnitude of transverse acceleration produced in the motion of the electric charge, when it passes through the electric field? Ans. If a charge q having mass m passes transversely through an electric field E, then acceleration, 𝑞𝐸 a= 𝑚 Q. 8 Under what condition is the force acting on a charge moving through a uniform magnetic field minimum? Ans. A charge moving through a magnetic field, experiences no force (minimum), when it moves along the direction of magnetic field. Q. 9 An electron is projected in the direction of magnetic field. How will its motion be affected by the action of magnetic field? Ans. No force acts on the electron due to the magnetic field, when it is projected in the direction of magnetic field. Hence, its motion will not be affected. Q. 10. What will be the path of a charged particle moving perpendicular to the direction of a uniform magnetic field? Ans. When the charged particle moves perpendicular to the direction of a uniform magnetic field, it experiences a force perpendicular to its direction of motion. As such, it moves alone a circular path. KVS ZIET CHANDIGARH 51 Q. 11. Does a stationary charge experience a force in an electric field? Ans. The force due to electric field does not depend, whether the charge is at rest or is in motion. A stationary charge experiences force in an electric field, which is given by F = qE Q. 12 When is the force on a moving charge due to a magnetic field maximum and when is it minimum? Ans. We know, Fm =Bqvsin𝜃 For force to be maximum, sin𝜃 = 1 ie. 𝜃 = 90° ie. when the charged particle moves perpendicular to the direction of magnetic field. For force to be minimum, sin 𝜃 = 0 i.e. 𝜃 = 0° 1.e. when the charged particle moves along the direction of magnetic field. Q. 13 Why does a charged particle moving at right angle to the direction of a magnetic field follow a circular path? Ans. When a charged particle moves at right angle to the direction of a magnetic field, it experiences force which always acts perpendicular to the velocity. Hence, the magnitude of its velocity remains constant and only the direction of the velocity of the charged particle changes. In other words, the force on the charged particle acts as centripetal force and it follows a circular path. Q.14. Why does not a charged particle moving at right angle to the direction of a magnetic field undergo any change in kinetic energy? Or The energy of a charged particle moving in a uniform magnetic field does not change. Why? Explain. Ans. The force on a charged particle moving in a uniform magnetic field always acts in a direction perpendicular to the direction of motion of the charge. As work done by the magnetic field on the charge is zero, the energy of the charged particle does not change. Q. 15 What is the nature of force, when the two parallel conductors carry currents in the (i) same direction (ii) opposite direction? Ans. (i) Force is attractive. (ii) Force is repulsive. Q.16. Does the torque on a planar current loop 1n magnetic field change, when its shape is changed without changing its geometrical area? Ans. The torque on a planar current loop in a magnetic field does not change, when its shape is changed without changing the area of the loop. Q.17 A current carrying loop free to turn is placed in a uniform magnetic field B. What will be its orientation relative to B in the equilibrium state? Ans. In equilibrium state, the current carrying loop will orient itself, such that B is perpendicular to the plane of the coil. It is because of the fact that in this orientation, the torque on the current loop becomes zero. Q. 18. Under what circumstances, will a current carrying loop not rotate in the magnetic field? Ans. If the current carrying loop is placed in a magnetic field, with its plane perpendicular to the field, then it will not rotate. KVS ZIET CHANDIGARH 52  Q. 19 Is the resistance of an ammeter greater than or less than that of the galvanometer of which it is formed? Ans. The resistance of an ammeter is always less than that of the galvanometer, of which it is formed. Q. 20 Why should an ammeter have a low resistance? Ans. For measuring current in a circuit, an ammeter is connected in series. So that the current in the circuit remains practically unchanged on connecting the ammeter, the resistance of the ammeter should be low. Q. 21 How is an ammeter connected in an electric circuit? Ans. An ammeter is connected in series in an electric circuit. Q.22. How can a galvanometer be converted into voltmeter? Ans. A galvanometer can be converted into a voltmeter by connecting a suitable high resistance in series to its coil. Q. 23 Is the resistance of a voltmeter greater than or less than that of the galvanometer of which it is formed? Ans. The resistance of a voltmeter is always greater than that of the galvanometer, of which it is formed. Q. 24 What is the resistance of an ideal voltmeter and an ammeter? Ans. The resistance of an ideal voltmeter is infinite and that of an ammeter is zero. Q.25 How is a voltmeter connected in an electric circuit? Ans. A voltmeter is connected in parallel in an electric circuit. Q.26 Give two differences between a voltmeter and an ammeter. Ans. (i) An ammeter is a low resistance instrument and is used to measure current in an electrical circuit. (ii) A voltmeter is a high resistance instrument and is used to measure potential difference in an electrical circuit. Q.27 The wire shown in the diagram carries a current of 10 A. Determine the magnitude of magnetic field induction at the centre O. Given that radius of the bent coil is 3 cm. Ans- The magnetic field induction at the centre of a current carrying circular wire of radius a is given by KVS ZIET CHANDIGARH 53  Q.28 A wire loop is formed by joining two semi-circular wires of radii r, and r, as shown in diagram. If the loop carries a current I, find the magnetic field at the centre O. Ans. The magnetic field at the point O due to the semi-circular part ABC, Q. 29 Two parallel straight wires X and Y separated by a distance 5 cm in air carry current of 10 A and 5 A respectively in opposite direction as shown in diagram. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y. Ans- Force on a unit length of the wire Y due to the wire X, Q.30 A galvanometer has a resistance of 60 Ω and a full-scale deflection is produced by 1.0 mA. How will you convert it in to (a) an ammeter to read 1A (full scale) and (b) voltmeter to read 3 V (full scale)? Ans- Here G = 60 Ω , I= 1A , V= 3V, Ig 1.0mA KVS ZIET CHANDIGARH 54 KVS ZIET CHANDIGARH 55  REVISION PAPER UNIT- IV–MOVING CHARGES AND MAGNETISM Note: Q. No. 1-4 is of 01 mark each, Q. 5-6 is of 02 marks each, Q.No.7 is of 03 marks, Q. No. 8 is a case study based and is of 04 marks, Q. No. 11 is of 5 marks. S Question Ma N rks 1 The strength of magnetic field at the centre of circular coil is 1 2 Assertion (A): The coils of a spring come close to each other, when current is passed through it. 1 Reason (R): It is because, the coils of a spring carry current in the same direction and hence attract each other. m- Both assertion and reason are correct and the reason is the correct explanation of assertion. n- Both assertion and reason are correct and reason is not a correct explanation of assertion. o- Assertion is correct but the reason is incorrect p- Assertion is incorrect but the reason is correct. 3 What is the net force on the rectangular coil? 1 (a) 25 × 10-7 N towards wire. (b) 25 × 10-7 N away from wire. (c) 35 × 10-7 N towards wire. (d) 35 × 10-7 N away from wire. 4 A positive charge enters in a magnetic field and travels parallel to but opposite the field. If 1 experiences (a) an upward force. (b) a downward force. (c) an accelerated force. (d) no force. 5 An α–particle and a proton are moving in the plane of paper in a region where there is a uniform 2 magnetic field B " directed normal to the plane of the paper. If the particles have equal linear momenta, what would be the ratio of the radii of their trajectories in the field? 6 State two reasons why a galvanometer cannot be used as such to measure current in a given 2 circuit. 7 Write any two important points of similarities and differences each between Coulomb’s law for 3 the electrostatic field and Biot-Savart’s law for the magnetic field. Case study-based questions (questions no 8- 10) Conversion of Galvanometer into Ammeter 4 A galvanometer may be converted into ammeter by using very small resistance in parallel with the galvanometer coil. The small resistance connected in parallel is called a shunt. If G is resistance of galvanometer, Ig is current in galvanometer for full scale deflection, then for 𝑰𝒈 conversion of galvanometer into ammeter of range I ampere, the shunt is given by S = 𝑰 −𝑰 G 𝒈 8. What is a shunt? 1 9. Can we increase or decrease the range of an ammeter? 1 10. What is the net resistance of an ammeter? 2 OR KVS ZIET CHANDIGARH 56  10. A galvanometer has a resistance of 15 Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 A? 2 11 (i) State Biot-Savart Law. Using this law, find an expression for the magnetic field at the centre of a 5 circular coil of N-turns, radius R, carrying current I. 3 (ii) Sketch the magnetic field for a circular current loop, clearly indicating the direction of the field. 2 KVS ZIET CHANDIGARH 57 KVS ZIET CHANDIGARH 58  KEY FEATURES Magnetic dipole. An arrangement of two unlike poles of equal strength and separated by a small distance ts called magnetic dipole. In SI, the unit of magnetic pole strength is ampere metre (A m). The distance between the two magnetic poles is called the magnetic length of the magnetic dipole. It is denoted by 2 | , a vector from south to north pole of the magnetic dipole. Magnetic dipole moment. The product of the pole strength of the either magnetic pole and the magnetic ⃗⃗ length of the magnetic dipole is called its magnetic dipole moment. It is denoted by 𝑀 ⃗⃗⃗ ) ⃗⃗ =m (2𝑙 Mathematically - 𝑀 Here, m is pole strength of the magnetic dipole. The SI unit of magnetic dipole moment is ampere/metre2 (A m2). Current loop and magnetic dipole. A current loop of area A carrying current I behave as a magnetic dipole having magnetic dipole moment, M=IA Torque on a magnetic dipole in a magnetic field. When a magnetic dipole of magnetic dipole moment M is placed in a uniform magnetic field of strength B_- making an angle 4 with the direction of magnetic field, it experiences a torque, which is given by |𝝉 ⃗⃗⃗ 𝑿 ⃗𝑩 ⃗ | = |𝑴 ⃗ | = MB sinθ Potential energy stored in a magnetic dipole on rotating inside a magnetic field. The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside the magnetic dipole in the form of its potential energy. When the magnetic dipole is rotated from its initial position θ1, to the final position θ2, then the potential energy stored is given by U=MB (cosθ2 - cosθ1) Magnetic intensity. It is defined as the ratio of magnetic induction in vacuum to the absolute magnetic permeability of free space. It is given by 𝐵𝑂 H= 𝜇𝑂 where 𝜇𝑂 = 4π x 10-7 tesla metre/ampere is absolute permeability of vacuum. Magnetic intensity is also known as H-field or magnetising field strength. The unit of magnetic intensity i.e. A/ m is also equivalent to N /m2 T or N/Wb or J/m-3 T Intensity of magnetisation. It is defined as the magnetic dipole moment developed per unit volume or the pole strength developed per unit area of cross-section of the specimen. It is given by 𝑀 𝑚 I= 𝑉 = 𝑎 Here, V is volume and A is area of cross-section of the specimen. KVS ZIET CHANDIGARH 59 In SI, the unit of intensity of magnetisation is ampere/metre (A/m}). Magnetic induction. It is defined as the number of magnetic lines of induction (magnetic field lines inside the material) crossing per unit area normally through the magnetic material. If H is the strength of the magnetising field, then magnetic induction is given by B= 𝜇𝑂 (H +I) In SI, the unit of the strength of magnetising field is ampere/ metre (A/ m) and that of magnetic induction is tesla (T) or weber /metre2 (Wb/ m2 ) Magnetic susceptibility. The magnetic susceptibility of a material is defined as the ratio of the intensity of magnetisation (1) and the strength of magnetising field (H). It is given by 𝐼 𝜒𝑚 = 𝐻 The magnetic susceptibility has no units. Magnetic permeability. The magnetic permeability of a material is defined as the ratio of the magnetic induction (B) of the material to the strength of magnetising field (H). It is given by 𝐵 𝜇𝑚 = 𝐻 In SI, the unit of magnetic permeability is tesla metre/ ampere (T m/ A). Diamagnetic substances. Those substances, which when placed in a magnetic field are feebly magnetised in a direction opposite to that of the magnetising field. Paramagnetic substances. Those substances, which when placed in a magnetic field are feebly magnetised in the direction of the magnetising field. Ferromagnetic substances. Those substances, which when placed in a magnetic field are strongly magnetised in the direction of the magnetising field. QUESTIONS WITH ANSWERS Q.1 What do you mean by directive property of a magnetic dipole? Ans. A freely suspended magnet always aligns itself along the N-S line. Q.2 A bar magnet is cut into two equal pieces trans-verse to its length. What happens to its dipole moment? Ans. Let m and 2l be the pole strength and the length of the given bar magnet. When the magnet is cut into two equal pieces transverse to its length, each piece will be a magnet having pole strength m (unchanged) and length l. Therefore, the magnetic moment of each piece will be ml i.e. one half of that of the original magnet. Q.3 Why ordinarily a piece of iron does not behave as a magnet? Ans. In an ordinary piece of iron, the molecular magnets are randomly oriented and form closed chains. Since the molecular magnets cancel the effect of each other, the ordinary iron piece does not behave as a magnet. Q.4 What is the source of magnetic field (magnetism)? Ans. Magnetism is of electrical origin. The electrons revolving in an atom behave as tiny current loops and these current loops give rise to magnetism. Q.5 Does an isolate magnetic pole exists like an isolate electric charge? Ans. No, an isolate magnetic pole does not exist. KVS ZIET CHANDIGARH 60 Q.6 What is the unit of magnetic pole strength? Ans. Unit of magnetic pole strength, ampere metre (A m). Q.7 What is the unit and direction of magnetic dipole moment? Ans. The unit of magnetic dipole moment is A m2 and its direction is from S-pole to N-pole of the magnetic dipole. Q.8 Can a current loop be treated as magnetic dipole? Ans. A current loop can be treated as a magnetic dipole. If the current loop has an area A and carries a current I, then its magnetic dipole moment is given by M=IA Q.9 Define Bohr magneton and write its value. Ans. Bohr magneton ts defined as the magnetic dipole moment associated with an atom due to orbital motion of an electron in the first orbit of hydrogen atom. Bohr magneton, 𝜇𝐵 = 9.27 x 10-24 A/m2 Q.10 Does a bar magnet exert a torque on itself due to its own field? Does one element of a current- carrying wire exert a force on another element of the same wire? Ans. No, a bar magnet does not exert a force or torque on itself due to its own field. But an element of a current carrying conductor experiences force due to another element of the conductor. Q.11 When does a magnetic dipole possess maximum potential energy inside a magnetic field? Ans. A magnetic dipole possesses maximum potential energy, when its magnetic moment M and the magnetic field B are antiparallel. Q.12 When does a magnetic dipole possess minimum potential energy inside a magnetic field? Ans. A magnetic dipole possesses maximum potential energy, when its magnetic moment M and the magnetic field B are parallel. Q.13 Compare the magnetic fields due to a straight solenoid and a bar magnet. Ans. The magnetic field of a bar magnet and a straight solenoid are identical. The two ends of the straight solenoid behave as the north and south poles as in case of a bar magnet. Q.14 What is the basic difference between magnetic lines of force and electric lines of force? Ans. The electric lines of force originate from positive charge and end at negative charge and are thus discontinuous curves. But as the isolated magnetic poles do not exist, the magnetic field lines are closed loops. Q.15 Why two magnetic lines of force do not cross each other? Ans. Two magnetic field lines cannot intersect each other. It is because, if they do so, then at the point of intersection, the magnetic field will have two directions along the tangents to the two field lines. Q.16 An iron bar is magnetised with the help of another magnet or by subjecting it toa magnetising field. The magnetism acquired by the magnet is assumed due to the alignment of molecular magnets. Does the length of the iron bar undergo a change during the magnetisation process? Ans. Yes, the length of the iron bar increases in the direction of magnetisation. This effect is called magneto- striction and is used for producing ultrasonic waves. Q.17 The poles of a magnet cannot be separated. How does this statement derive support from the magnetic dipole behaviour of a current loop? Ans. A current loop behaves as a magnetic dipole. It’s one face behaves as -pole, while the other as s-pole. As the two faces of the current loop cannot be separated from each other, it follows that the magnetic poles developed on the two faces also cannot be separated from each other. KVS ZIET CHANDIGARH 61 Q.18 A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nails near a bar magnet, however, experiences a force of attraction in addition to a torque, why? Ans. The force and torque act on the nail due to the induced magnetic moment acquired by it. The iron needles will also not experience any force, if magnetic field is uniform. The magnetic field due to a bar magnet is not uniform. Therefore, an iron nail experiences both a force and torque, when placed near a bar magnet. It may be pointed that the nail experiences a net attractive force. It is because the attractive force on the nearer end (unlike induced pole) of the nail is greater than the repulsive force on its farther end (like induced pole). Q.19 Why does a magnetic dipole possess potential energy, when placed at some inclination with the direction of the field? Ans. In equilibrium, a magnetic dipole always aligns itself along the direction of the magnetic field. When the magnetic dipole is displaced from the equilibrium position, a restoring torque acts on the dipole to bring it back. Therefore, to place the dipole at some inclination with the field, work has to be done against the restoring torque. This work done is stored in the dipole as its potential energy. Q.20 What do you mean by magnetic lines of force? Why two such lines do not cross each other/? Ans. The magnetic field line is the path along which an isolated north pole will tend to move, if it is free to do so. Two magnetic field lines cannot intersect each other. It is because, if they do so, then at the point of intersection, the magnetic field will have two directions along the tangents to the two field lines. Q.21 Magnetic field arises due to charges in motion. Can a system have magnetic moment, even though its net charge is zero? Ans. A system can have a magnetic moment even though its net charge is zero. It is because, the average charge of a system may be zero, but it is not necessary that magnetic-moments due to various current loops will also be zero. For example, a neutron has zero charge, but possesses non zero magnetic moment. Q.22 Draw the magnetic field lines for a current carrying solenoid when a rod made of (a) copper, (b) aluminium and (c) iron are inserted within the solenoid as shown. Ans- (a) When a bar of diamagnetic material (copper) is placed in an external magnetic field, the field lines are repelled or expelled and the field inside the material is reduced. (b) When a bar of paramagnetic material (Aluminium) is placed in an external field, the field lines gets concentrated inside the material and the field inside is enhanced. (c) When a ferromagnetic material (Iron) is placed in an internal magnetic field, the field lines are highly concentrated inside the material Q.23 Explain the following: (i) Why do magnetic field lines form continuous closed loops? (ii) Why are the field lines repelled (expelled) when a diamagnetic material is placed in an external uniform magnetic field? KVS ZIET CHANDIGARH 62 Ans. (i) Magnetic lines of force form continuous closed loops because a magnet is always a dipole and as a result, the net magnetic flux of a magnet is always zero. (ii) When a diamagnetic substance is placed in an external magnetic field, a feeble magnetism is induced in opposite direction. So, magnetic lines of force are repelled Q.24 Write three points of differences between para-, dia- and ferro- magnetic materials, giving one example for each. Ans- Examples: Diamagnetic materials: Bi, Cu, Pb, Si, water, NaCl, Nitrogen (at STP) Paramagnetic materials: Al, Na, Ca, Oxygen (at STP), Copper chloride Ferromagnetic materials: Fe, Ni, Co, Alnico Q.25 A bar magnet of magnetic moment 1.5 JT–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? and (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? Ans- KVS ZIET CHANDIGARH 63 KVS ZIET CHANDIGARH 64  REVISION PAPER UNIT- V–MAGNETISM & MATTER Note: Q. No. 1-4 is of 01 mark each, Q. 5-6 is of 02 marks each, Q.No.7 is of 03 marks, Q. No. 8 is a case study based and is of 04 marks, Q. No. 11 is of 5 marks. S Question Ma N rks 1 Magnetism in substances is caused by 1 (a) orbital motion of electrons only (b) spin motion of electrons only (c) due to spin and orbital motions of electrons both (d) hidden magnets 2 Assertion (A): 1 Reason (R): q- Both assertion and reason are correct and the reason is the correct explanation of assertion. r- Both assertion and reason are correct and reason is not a correct explanation of assertion. s- Assertion is correct but the reason is incorrect t- Assertion is incorrect but the reason is correct. 3 A uniform magnetic field exists in space in the plane of paper and is initially directed from left to 1 right. When a bar of soft iron is placed in the field parallel to it, the lines of force passing through it will be represented by 4 Electro-magnets are made of soft iron because soft iron has 1 (a) small susceptibility and small retentivity (b) large susceptibility and small retentivity (c) large permeability and large retentivity (d) small permeability and large retentivity 5 Write two properties of a material suitable for making 2 (a) a permanent magnet, and (b) an electromagnet 6 Write three points of differences between para-, dia- and ferro- magnetic materials, giving one 2 example for each. 7 Explain the following: 3 (i) Why do magnetic field lines form continuous closed loops? (ii) Why are the field lines repelled (expelled) when a diamagnetic material is placed in an external uniform magnetic field? Case study-based questions (questions no 8- 10) DOMAIN THEORY 4 The atom of a ferromagnetic material also possesses non-zero magnetic moment as in case of a paramagnetic substance. However, due to a quantum mechanical effect, called exchange interaction, an unpaired electron in one atom interacts strongly with the unpaired electron in the neighbouring atom in such a way that they spontaneously align themselves in a common direction over a small volume of the material. KVS ZIET CHANDIGARH 65  These small volumes of uniform magnetisation are called domains. Although domains are extremely small in size (~ 10-18 m3 in volume), yet each domain contains a large number of atoms (~ 1011 atomic magnetic dipoles). 8.What are domains? 1 9. What is the volume of a domain? 1 10. What is the effect on orientation domains when external magnetic field applied? 2 OR 10. What is the effect on orientation domains when external magnetic field removed? 2 11 (a) Draw the magnetic field lines due to a circular loop of area A carrying current I. Show that it 5 acts as a bar magnet of magnetic moment m=IA (b) Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius ‘a’ having ‘n’ number of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distant ‘r’ from the centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar magnet of magnetic moment ‘m’? KVS ZIET CHANDIGARH 66 KVS ZIET CHANDIGARH 67  KEY FEATURES 1. Magnetic Flux. The number of magnetic field lines crossing a surface normally is called magnetic flux (∅𝐵 ) linked with the surface. Mathematically- ⃗⃗. 𝑨 φ=𝑩 ⃗ = BAcosθ where B is the magnetic field, A is the area of the surface and θ is the angle, which the direction of the magnetic field makes with normal to the surface. Unit. In SI, unit of magnetic flux is weber (Wb) 1 weber = 108 maxwell 2. Electromagnetic induction. It is the phenomenon of production of e.m.f. in a coil, when the magnetic flux linked with the coil is changed. The e.m.f. so produced is called induced e.m.f. and the resulting current is called induced current. 3. Faraday’s laws of electromagnetic induction- 1. Whenever magnetic flux linked with a circuit (a loop of wire or a coil or an electric circuit in general) changes, induced e.m.f. is produced. 2. The induced e.m.f. lasts as long as the change in the magnetic flux continues. 3. The magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux. 𝒅∅ ∅𝟐 − ∅𝟏 Mathematically: Induced e.m.f., e =- = − 𝒅𝒕 𝒕 4. Lenz’s law. It states that the induced current produced in a circuit always flows in such a direction that it opposes the change or the cause that produces it. Lenz’s law can be used to find the direction of the induced current. 5. Motional E.M.F. When a conductor of length / moves with a velocity v in a magnetic field B, so that magnetic field is perpendicular to both the length of the conductor and its direction of motion, the magnetic Lorentz force on the conductor gives rise to e.m.f. across the two ends of the conductor. Mathematically: e=Blv 6. Eddy currents. The currents induced in the body of a conductor, when the m