Physics Support Material PDF - Kendriya Vidyalaya 2024-25

XII/Physics-Support Material/Bengaluru Region/2024-25 KENDRIYA VIDYALAYA SANGATHAN (BANGALURU REGION) STUDY MATERIAL...

XII/Physics-Support Material/Bengaluru Region/2024-25 KENDRIYA VIDYALAYA SANGATHAN (BANGALURU REGION) STUDY MATERIAL SESSION 2024-25 CLASS -XII PHYSICS (042) "Learning gives creativity, creativity leads to thinking, thinking provides knowledge, and knowledge makes you great." XII/Physics-Support Material/Bengaluru Region/2024-25 PREFACE This study material is the culmination of an extensive collaborative effort involving a dedicated team of educators and subject matter experts. With meticulous care, we have meticulously designed this resource to provide students with a succinct yet comprehensive tool for consolidating their knowledge. Under the diligent guidance of our esteemed subject experts and the unwavering enthusiasm of our team, we have incorporated the entire curriculum and an extensive collection of practice questions spanning all chapters. Our paramount objective has been to ensure perfect alignment with the latest curriculum and examination patterns as set forth by the CBSE. We firmly believe that this material will prove to be an invaluable resource, serving as a clear and concise repository of essential information for effective subject revision. It encompasses all the critical components necessary to assist in students' preparation and enhance their understanding of the subject matter. Our aspiration is that this study material will emerge as a dependable aid for swift and efficient revision, instilling confidence in students and ultimately contributing to their academic success. We strongly encourage you to actively engage with the content, pose questions, and fully utilize this resource in your educational journey. We extend our heartfelt best wishes for your studies and sincerely hope that this material becomes your trusted companion on the path to academic excellence. KENDRIYA VIDYALAYA SANGATHAN REGIONAL REGION K. KAMARAJA ROAD BENGALURU 560042 XII/Physics-Support Material/Bengaluru Region/2024-25 MESSAGE FROM THE DEPUTY COMMISSIONER Dear students and teachers! It is a matter of great pride and delight that KVS Bengaluru Region is putting forward the Students' Support Material (SSM) for class XIIth subject PHYSICS for the session 2024-25. I believe firmly that the subject experts have left no stone unturned to enable our students to add on more to their quality of performance by deep rooting more towards accessing required understating in the subject. Certainly, use of this SSM will help students in empowering themselves as one of the tools and will lead in bringing success. With devotion, dedication & persistent hard work the team of experts has crafted out this SSM meticulously to complement the classroom learning experience of the students as well as to cope up with the Competency Based Questions as per the new pattern of examinations aligned with NEP-2020 and NCFSE-2023. This SSM, being well-structured and presented in a manner which makes it to be comprehended easily, will definitely serve as a precious supplement for self-study of students. I am pleased to place on record my appreciation and commendation for the commitment and dedication of the team comprising of the subject experts in carving out such a useful edition of Students' Support Material for the students. Wishing all the best! XII/Physics-Support Material/Bengaluru Region/2024-25 OUR PATRON SHRI DHARMENDRA PATLE DEPUTY COMMISSIONER, RO BENGALURU SHRI P C RAJU SHRI R PRAMOD SMT HEMA K AC, RO BENGALURU AC, RO BENGALURU AC, RO BENGALURU XII/Physics-Support Material/Bengaluru Region/2024-25 MESSAGE FROM THE COORDINATOR Dear Students, I feel thrilled to commence on the study material for the Physics for class XII. My sincere appreciation and gratitude to the diligent team for the designing the practice material that caters to the revised pattern of CBSE. Congratulations to the committed team for their vital role in designing the practice material with inclusive competencies, analytical and critical reasoning questions summarizing all concepts. I feel greatly honoured to be associated as a coordinator of diligent team in bringing out the Support Material for class XII Physics for the year 2024-25 and truly convinced that it would definitely help in learning and scoring high in exams. Wishing success to all in the journey of learning. I/C Principal, PM SHRI K V Vijayapura, Karnataka SHRI BANSHI LAL I/C PRINCIPAL,PM SHRI KV VIJAYAPURA XII/Physics-Support Material/Bengaluru Region/2024-25 MATERIAL PREPARED BY- SLNO. NAME OF THE TEACHER NAME OF THE KV CHAPTER / UNIT 1. Sh. G. Gugan, PGT Physics KV RWF Yelahanka Reviewing committee members 2. Sh. Satish Kumar Sivalanka, PGT KV Belagavi No.2 Reviewing committee members Physics (Cantonment) 3. 1. Mrs Nisha M Mohan KV Hebbal 1. Electric Charges and Fields 2. Mr A Jose Herbert Raj 2. Electrostatic Potential and 3. Mrs. Neethu K Capacitance 4. 1. Mr. Rajendrnath U KV MEG & Centre 3.Current Electricity 2. Mrs. Rashmi K 5. 1. Mr. Birender Singh KV Yelahanka - AFS 4. Moving Charges and Magnetism 2. Mrs. Anupama Sharma 5. Magnetism and Matter 3. Mrs Lalithambika 6. 1. Mr. K Srihari KV DRDO 6. Electromagnetic Induction 2. Dr Lalit Upadhyay 7. Alternating Current 7. 1. M. Govindan KV IISc. 8. Electromagnetic Waves 2. Rajani C 9. Ray Optics and Optical 8. Bheemarao M KV Bidar - AFS 10. Wave Optics 9. 1. N Chandramouli KV Malleshwaram 11. Dual Nature of Radiation and 2. Mr. Boominathan B Matter 10. 1. Mrs. Susmitha.LK KV Jalahalli No.2 12. Atoms 2. Mr. S N V Harinath Babu 13. Nuclei 11. 1. Sarithan Swamy KV Yelahanka- CRPF 14. Semiconductor Electronics: Materials, Devices and Simple Circuits 12. Mr. Birender Singh KV Yelahanka - AFS SAMPLE PAPER WITH SOLUTION 13. Mrs Lalithambika KV Yelahanka - AFS PRACTICE PAPER -01 14. Mr. K. VEERABHADRAPPA K V Bellari PRACTICE PAPER -02 XII/Physics-Support Material/Bengaluru Region/2024-25 INDEX S. No Content 1. CHAPTER- 1. ELECTRIC CHARGES AND FIELDS 2. CHAPTER- 2. ELECTROSTATIC POTENTIAL AND CAPACITANCE 3. CHAPTER- 3. CURRENT ELECTRICITY 4. CHAPTER- 4. MOVING CHARGES & MAGNETISM 5. CHAPTER- 5. MAGNETISM & MATTER 6. CHAPTER- 6. ELECTRO MAGNETIC INDUCTION 7. CHAPTER- 7. ALTERNATING CURRENTS & ELECTRICAL MACHINES 8. CHAPTER- 8. ELECTRO MAGNETIC WAVES 9. CHAPTER- 9. RAY OPTICS & OPTICAL INSTRUMENTS 10. CHAPTER- 10. WAVE OPTICS 11. CHAPTER- 11. DUAL NATURE OF RADIATION & MATTER 12. CHAPTER- 12. ATOMS 13. CHAPTER- 13. NUCLEI 14. CHAPTER- 14. SEMICONDUCTOR ELECTRONICS & DEVICES 15. CBSE SAMPLE PAPER 2024-25 16. CBSE PRACTICE PAPER -01 17. CBSE PRACTICE PAPER -02 XII/Physics-Support Material/Bengaluru Region/2024-25 1. ELECTRIC CHARGES AND FIELDS SYLLABUS 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). GIST/FORMULAS/IMPORTANT DEFINITIONS/DIAGRAMS ELECTRIC CHARGE It is a scalar. SI unit is C. METHODS OF CHARGING By friction By Induction By conduction BASIC PROPERTIES OF ELECTRIC CHARGE Additivity of charges Charge is conserved Quantisation of charge q =ne COULOMB’S LAW F = kq1q2/r2 k = 1/4πϵ0, ϵ0 = 8.854 × 10–12 C2 N–1m–2 FORCE BETWEEN MULTIPLE CHARGES XII/Physics-Support Material/Bengaluru Region/2024-25 SUPERPOSITION PRINCIPLE CONTINUOUS CHARGE DISTRIBUTION Linear charge density , λ = q/l Surface charge density , σ =q/A Volume charge density, ρ= q/V ELECTRIC FIELD E = F /q0 SI unit = N/C, it is a vector. ELECTRIC FIELD DUE TO A POINT CHARGE The direction of E is on a radial line from q, pointing outward if q is positive and inward if q is negative. ELECTRIC FIELD LINES A line of force is a curve drawn in a field such that the tangent at each point on the curve gives the direction of the field at that point. PROPERTIES i) Field lines start from positive charges and end at negative charges. If there is a single charge, they may start or end at infinity. (ii) In a charge-free region, electric field lines can be taken to be continuous curves without any breaks. (iii) Two field lines can never cross each other. (iv) Electrostatic field lines do not form any closed loops.This follows from the conservative nature of electric field. XII/Physics-Support Material/Bengaluru Region/2024-25 FIELD LINES DUE TO SOME SIMPLE CHARGE CONFIGURATIONS XII/Physics-Support Material/Bengaluru Region/2024-25 ELECTRIC DIPOLE An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a. ELECTRIC DIPOLE MOMENT (p) p = q x 2a It is a vector whose direction is from –q to +q. SI unit is Cm. ELECTRIC FIELD DUE TO A DIPOLE FOR A POINT ON THE AXIS FOR POINTS ON THE EQUITORIAL PLANE XII/Physics-Support Material/Bengaluru Region/2024-25 TORQUE ON A DIPOLE IN A UNIFORM ELECTRIC FIELD ELECTRIC FLUX (φ) The flux Δφ of electric field through a small area element ΔS is given by Δφ = E. ΔS It is a scalar. SI unit is Nm2C-1 GAUSS’S LAW Gauss’s law: The flux of electric field through any closed surface S is 1/ε0 times the total charge enclosed by surface. Φ = q / ε0 XII/Physics-Support Material/Bengaluru Region/2024-25 APPLICATIONS OF GAUSS’S LAW1). FIELD DUE TO AN INFINITELY LONG STRAIGHT UNIFORMLY CHARGED WIRE where nˆ is the radial unit vector in the plane normal to the wire passing through the point. E is directed outward if λ is positive and inward if λ is negative. (ii)FIELD DUE TO A UNIFORMLY CHARGED INFINITE PLANE SHEET where nˆ is a unit vector normal to the plane and going away from it. FIELD DUE TO A UNIFORMLY CHARGED THIN SPHERICAL SHELL. Field outside the shell XII/Physics-Support Material/Bengaluru Region/2024-25 Field inside the shell. XII/Physics-Support Material/Bengaluru Region/2024-25 MIND MAP XII/Physics-Support Material/Bengaluru Region/2024-25 COMPETENCY BASED QUESTIONS MCQ (1mark) 1. Two positive ions each carrying a charge q are separated by a distance d. If F is the force of repulsion between the ions , the number of electrons missing from each ion 2. The total electric flux emanating from an alpha particle is (a) 2e/𝞮0 ( b)e/ 𝞮0 ( c ) 4e/ 𝞮0 (d) e2 / 𝞮0 3. A charge Q is placed at each of the opposite corners of a sphere. A charge q is placed at each of the other corners.If the net electrical force on Q is zero then Q/q is equal to (a) -2√2 (b) -1 (c) 1 (d) -1/√2 4. A cylinder of radius r and length l is placed in a uniform electric field parallel to the axis of the cylinder. The total flux for the surface of the cylinder is given by (a) zero (b) π r2 (c) π E r2 (d) 2E(π r2 ) 5. 𝑞1, 𝑞2, 𝑞3 and q4are point charges located at points as shown in the figure and S is a spherical Gaussian surface of radius R. Which of the following is true according to the Gauss’s law 6. Seven charges of equal magnitude q are placed at the corners of a cube of side b. The force experienced by another charge Q placed at the center of the cube is (a) Zero (b) KQq/ 3b (c) 7KQq/3b (d) 2KQq/3b 7. Electric charge is uniformly distributed along a long straight wire of radius 1mm. The charge per cm of the wire is Q coulomb. Another cylindrical surface of length L meter encloses the wire symmetrically, The total flux through the surface is (a) Q/𝞮0 (b) LQ/ 𝞮0 (c) QL/ 10-3𝞮0 (d) Q/L 10-3𝞮0 8 A hemisphere is uniformly charged positively. The electric field at a point on the diameter away from the centre is directed (a) perpendicular to the diameter (b) parallel to the diameter. (c) at an angle tilted towards the diameter (d) zero XII/Physics-Support Material/Bengaluru Region/2024-25 ANSWERS 1.(a) 2. (a) 3. (a) 4. (a) 5.(b) 6. (d) 7. (b) 8. (a) COMPETENCY BASED QUESTIONS 2 MARKS QUESTIONS 1. Calculate the electric field strength required to just support a water drop of mass 10-7 kg and having a charge 1.6x 10-19 C. Ans. Force on water drop due to electric field = Weight of water drop. qE =mg E = 6.125x1012N/C 2. A spherical rubber ballon carries some charge distributed uniformly over its surface. The balloon is blown up to increase in its size. How does the total electric flux coming out of the surface change? Ans. The total electric flux coming out of the surface of ballon remains unchanged. This is because total charge enclosed by the balloon is independent of size of the ballon. 3. Two dipoles, made from charges and have equal dipole moments. Give the (i) Ratio between the separations of these two pairs of charges (ii) Angle between the dipole axes of these two dipoles Ans. As the two dipoles have equal dipole moments (i) Qx 2a’ =q X 2a, 2a’/2a=q/Q () = 4. The electrostatic force (F) acts between two point charges in a vacuum. If a brass plate is placed between the two charges. What would be the value of the electrostatic force? Ans. For any metal , =∞, Fbrass=Fvac/ K=F/∞=0 5. A uniformly charged conducting sphere of diameter 24m has a surface charge density of 80μc/m2. Find the charge on the sphere and the total electric flux leaving the sphere. 6. Two-point charges Q and -3Q are placed at some distance apart at some distance apart. If electric field at location of Q is E Find the field in the location of -3Q Ans. 7. Why do the electrostatic field lines not form closed loops? XII/Physics-Support Material/Bengaluru Region/2024-25 Ans. Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field. lines do not form closed loops. COMPETENCY BASED QUESTIONS 3 MARKS QUESTIONS 1. Five point charges, each of value +q are placed on five vertices of a regular hexagon of side Lm. What is the magnitude of the force on a point charge of value -q coulomb placed at the centre of the hexagon? Ans. If there had been a sixth charge +q at the remaining vertex of hexagon due to all the six charges on - q at O will be zero (as the forces due to individual charges will balance each other). The resultant force , F = ( 1/4π ε0 ) q/ L2 2. Given is a line of charge of uniform linear density. A charge +q is distributed uniformly between y = 0 and y = a and charge –q is distributed uniformly between y = 0 and y = -a. Explain how the direction of the resultant electric field at point P can be obtained. Represent using a vector diagram Ans. The x-components of E1 and E2, due to two equidistant points on either side of O, cancel each other.The resultant electric field is due to the superposition of the y-components of E1 and E2.The direction of the net electric field is along the negative y-axis. This is true for all pairs of equidistant points on either side of O. XII/Physics-Support Material/Bengaluru Region/2024-25 3. A positively charged ball A hangs from a string. A non-conducting ball B is brought near ball A. Ball A is seen to be attracted to ball B. (a) Give reason why it is NOT possible to determine whether ball B is negatively charged or neutral for sure from the above experiment alone. (b) Suggest any ONE additional experiment with ball B required to determine whether ball B is negatively charged or neutral for sure. Ans. (a) The attraction between A and B could be due to the following reasons: - B is negatively charged and hence A and B attract each other. - B is neutral. The two balls attract each other due to the polarization of molecules in neutral ball B. It is not possible to determine for sure that ball B is negative or neutral from this experiment alone. (b) Possible additional experiments: - A known neutral ball can be brought near ball B (without ball A nearby). If the neutral ball is attracted to ball B, then ball B is negatively charged for sure. If there is no interaction between the two balls, then ball B is neutral for sure. 4. A spherical Gaussian surface encloses a positive charge q. Explain with a reason what happens to the net electric flux through the Gaussian surface if: (a) the charge is tripled (b) the volume of the sphere is tripled (c) the shape of the Gaussian surface is changed into a cuboid. XII/Physics-Support Material/Bengaluru Region/2024-25 Ans. (a) The net flux is also tripled because as per Gauss law the net flux is proportional to the net charge enclosed. (b) Regardless of the volume of the enclosed surface, if the net charge enclosed is the same, the net flux remains the same as per Gauss law. (c) No change in the net flux as it doesn’t depend upon the shape of the closed surface. 5. Two small identical electric dipoles AB and CD each of dipole moment P are kept at an angle 120 0 to each other in an electric field E pointing along the X axis. Find the dipole moment of the arrangement and the magnitude and direction of torque acting on it. Ans. Dipole moment of AB is P j Dipole moment of CD = pcos30 i – pcos60 j= p√3/2 i -p/2 j COMPETENCY BASED QUESTIONS 5 MARKS QUESTIONS 1. a). Define electric flux. Is it a scalar or vector quantity? b).A point charge Q is placed at a distance a/2 above the centre of the square surface of edge a as shown in the figure. What is the electric flux through the square surface? c. If the point charge is now moved to a distance ‘d’ from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected. Ans. (a) No. of electric lines of force passing normal to the surface. It is a scalar quantity. (b) Draw an imaginary square enclosing that point charge at the centre. XII/Physics-Support Material/Bengaluru Region/2024-25 Then from gauss law, ϕ=Q/ϵo Electric flux through one face=1/6 (Q/ϵo) No change 2. a). Derive an expression for electric field at a point outside and inside for a uniformly charged spherical shell b).An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negertive charge up to a radius R. The atom as a whole is neutral. For this model, what is the eletric field at a distance r from the nucleus? Ans. XII/Physics-Support Material/Bengaluru Region/2024-25 According to Gauss’s law, 3. (a) Find the electric field due to an electric dipole on a point in an equatorial plane. (b) Two similar balls, each of mass m and charge q, are hung from a common point by two slik threads, each of length l. Prove that separation between the ball is if θ is small. Find the rate dq/dt with which the charge should leak off each sphere if thevelocity of approach varies as v=a/√x, where a is a constant. XII/Physics-Support Material/Bengaluru Region/2024-25 Ans. XII/Physics-Support Material/Bengaluru Region/2024-25 1 MARK QUESTIONS 1. Two charges q1 and q2 are placed at the centres of two spherical conducting shells of radius r1 and r2 respectively. The shells are arranged such that their centres are d [> (r1+r2) ] distance apart. The force on q2 due to q1 is (a) (1/4Πε0) q1q2/d2 (b) (1/4Πε0) q1q2/(d –r1)2 ( c) Zero (d) (1/4Πε0) q1q2/[d –(r1-r2)]2 2. When a negative charge (-Q) is brought near one face of a metal cube, the (a) cube becomes positively charged (b) cube becomes negatively charged. (c) face near the charge becomes positively charged and the opposite face becomes negatively charged. (d)face near the charge becomes negatively charged and the opposite face becomes positively charged. 3. Let F1 be the magnitude of the force between two small spheres, charged to a constant potential in free space and F2 be the magnitude of the force between them in a medium of dielectric constant K, Then F1/F2 is (a) 1/K (b) K (c ) K2 (d) 1/ K2 4. A charge Q is placed at the centre of the line joining two charges q and q. The system of the three charges will be in equilibrium if Q is (a) q/3 (b) – q/3 (c ) q/4 (d) – q/4 5. A point charge situated at a distance r from a short electric dipole on its axis, experience a force F. If the distance of the charge is 2r, the force on the charge will be (a) F/16 (b) F/8 (c ) F/4 (d) F/2 6. The magnitude of the electric field due to a point charge, object at a distance of 4m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of (a) 1m (b) 2m (c ) 3m (d) 6m 7. An isolated point charge particle produces an electric field E at a point 3m away from it. The distance of the point at which the field is E/4 (a)2m (b) 3m (c ) 4m (d) 6m 8. Which one of the following is not a scalar quantity ? (a)Electric field (b) Voltage (c ) Resistivity (d) Power 9. An electric dipole of length 2cm is placed at an angle of 30˚ with an electric field of 2x105 N/C. If the dipole experiences a torque of 8x10-3 Nm, the magnitude of either charge of the dipole , is (a) 4 µC (b) 7 μC ( c ) 8 mC (d) 2mC 10. An electric dipole placed in a non-uniform electric field will experience XII/Physics-Support Material/Bengaluru Region/2024-25 (a) Only a force (b) only a torque (c ) both force and torque (d) neither force nor torque ANSWER KEY (1) C ( 2) C (3) B (4) D (5) B (6) C (7) D (8) A (9) A (10) C ASSERSION- REASON QUESTIONS Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below A) Both A and R are true and R is the correct explanation of A B) Both A and R are true and R is NOT the correct explanation of A C) A is true but R is false D) A is false and R is also false. 1. Assertion: The charge on anybody can be increased or decreased in terms of e. Reason: Quantization of charge means that the charge on a body is the integral multiple of e. 2. Assertion : A Charge, which is less than charge of one electron is not possible Reason : Charge is quantized. 3. Assertion: The properties that the force with which two charges attract or repel each other are not affected by the presence of a third charge. Reason: Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to other charges, taken one at a time. 4. Assertion(A) : A metallic shield in the form of a hollow shell, can be built to block an electric field. Reason(R): In a hollow spherical metallic shell, electric field inside is zero at every point. 5. Assertion: Coulomb force is the dominating force in the universe. Reason: Coulomb force is weaker than the gravitational force. 6. Assertion (A): The range of gravitational force and coulomb force is infinity. Reason(R): The Coulomb force is stronger than the gravitational force. 7. Assertion (A): The range of gravitational force and coulomb force is infinity. Reason(R): The Coulomb force is stronger than the gravitational force. 8. Assertion (A): Positive electric flux indicates that electric lines of force are directed outwards Reason (R): Positive electric flux is due to a positive charge. 9. Assertion (A): Electric field inside a metallic charged conductor is always zero whatever of amount of charge. Reason (R): Electric field lines are always perpendicular to surface of the metal. 10.. Assertion (A): Charge on a body is 2.3x10-19C is not possible. Reason (R): Electric charge on a body is quantized and integral multiple of charge of an electron. ANSWER KEY XII/Physics-Support Material/Bengaluru Region/2024-25 (1) A ( 2) B (3) B (4) A (5) D (6) C (7) C (8) B (9) B (10) A 2 MARKS QUESTIONS 1. Derive the expression for the torque acting on an electric dipole, when it is held in a uniform electric field Ans. Force on +q charge = + q E along the direction of E Force on –q charge = -q E opposite to the direction of E Net force on the dipole is F Net = +qE- qE =0 But the two force act at different point on the dipole. They form a couple and exerts torque. Torque = Force X perpendicular distance between lines of action of forces. τ = qE X 2a Sin θ τ = pE sin θ 2. Use Gauss’s law to obtain the expression for electric field due to an infinitely long straight uniformly charged wire Ans. The surface area of the curved part =2πrl Flux through the Gaussian surface= 𝜑 = 𝐸 × 2𝜋𝑟𝑙 XII/Physics-Support Material/Bengaluru Region/2024-25 3. Use Gauss’s law to obtain the expression for the elctric field due to a uniformly charged infinite plane thin sheet. Ans. 4. Ans. Let 𝑥 be the distance of 𝑄 from either charge. Let us assume the following figure: 5. XII/Physics-Support Material/Bengaluru Region/2024-25 Ans. 6. Ans. 7. Ans. 8. Ans. (i) Electric flux , φ = q/ε 0 Φ = 8.85 x 10-8 / 8.85 x 10-12 Φ = 104 Nm2 /C (ii) Electric flux remains the same as the charge enclosed is same. XII/Physics-Support Material/Bengaluru Region/2024-25 9. Ans. 10. What is the electric flux through one face of a cube enclosing a dipole of dipole moment 2x10-7 Cm? Justify your answer. Ans. Net charge of dipole is zero. Hence flux is zero. 3 MARKS QUESTIONS 1. Using Gauss’s law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius R and charge density σ C/m2. Draw the graph of electric field E versus distance r. Draw the field lines when the charge density of the sphere is (i)positive (ii) negative Ans. Electric field outside the shell: XII/Physics-Support Material/Bengaluru Region/2024-25 Consider a point P outside the shell with the radial vector r. Let P be the point on the Gaussian surface to be a sphere of radius R with the centre O passing through P. The electric field at each point of Gaussian surface has the same magnitude E and is along the radius vector at each point. Thus E and ΔS at every point are parallel. Flux = E ΔS. The total flux over the Gaussian surface= E * 4π r2 Charge q = σ X 4 πR2 ,From Gauss Law, Φ = q / ε0 E x 4 Π r 2 = σ x 4 Π R2 E = σ R2 / ε0 r 2 But charge q = σ 4π R2, The direction of the field is radially out ward, if the charge on the shell is +ve. field lines when the charge density of the sphere is (i) Positive (ii) Negative XII/Physics-Support Material/Bengaluru Region/2024-25 2. Derive the expression for electric field at a point on the equatorial plane of an electric dipole. Ans. 3. XII/Physics-Support Material/Bengaluru Region/2024-25 Ans. CASE STUDY QUESTIONS ( 4 MARKS EACH) 1. Observe the figure, read the data given below and answer the following questions XII/Physics-Support Material/Bengaluru Region/2024-25 Ans. (i) d (ii) b (iii) b (iv) d 2. When a charged particle is placed in an electric field, it experiences an electrical force. If this is the only force on the particle, it must be the net force. The net force will cause the particle to accelerate according to Newton’s second law. So, F=qE= ma. If E is uniform, then a is constant and a= q E /m. If the particle has a positive charge, its acceleration is in the direction of the field. If the particle has a negative charge, its acceleration is in the direction opposite to the electric field. Since the acceleration is constant, the kinematic equations can be used. (i) An electron of mass m, charge e falls through a distance h metre in a uniform electric field E. then time of fall, (ii) The electric flux through a closed surface area S enclosing charge Q is f. If the surface area is doubled, then the flux is (iii)A Gaussian surface encloses a dipole. The electric flux through this surface is (iv) In an electric field directed upwards, an electron will experience a force directed (a) Downward force of magnitude e E (b)Upward force of magnitude e E (c )Downward force of magnitude e/E (d)Upward force of magnitude e/E Ans. (i) a (ii) d (iii) d (iv) a XII/Physics-Support Material/Bengaluru Region/2024-25 3. Electric field strength is proportional to the density of lines of force i.e., electric field strength at a point is proportional to the number of lines of force cutting a unit area element placed normal to the field at that point. As illustrated in the given figure, the electric field at P is stronger that at Q. (i) Electric lines of force about a positive point charge are (a) radially outwards (b) circular clockwise (c) radially inwards (d) parallel straight lines. (d) parallel straight lines. (ii) Which of the following is false for electric lines of force? (a) They always start from positive charges and terminate on negative charges. (b) They are always perpendicular to the surface of a charged conductor. (c) They always form closed loops. (d) They are parallel and equally spaced in a region of uniform electric field. (iii) Which one of the following patterns of electric line of force in not possible in filed due to stationary charges? (iv)The figure below shows the electric field lines due to two positive charges. The magnitudes EA, EB and EC of the electric fields at points A, B and C respectively are related as (a) EA > EB > EC (b) EB > EA > EC (c) EA = EB > EC (d) EA > EB = EC Ans. (i) a (ii) c (iii) c (iv) a XII/Physics-Support Material/Bengaluru Region/2024-25 5 MARKS QUESTIONS 1. Ans. XII/Physics-Support Material/Bengaluru Region/2024-25 2. Ans. XII/Physics-Support Material/Bengaluru Region/2024-25 3. Ans. XII/Physics-Support Material/Bengaluru Region/2024-25 iii. p 1= q X 2 along OA p2= q X 2 along OD ASSIGNMENTS 2 MARKS QUESTIONS 1 An attractive force of 5N is acting between two charges of +2.0 μC & -2.0 μC placed at some distance. If the charges are mutually touched and placed again at the same distance, what will be the new force between them? 2. A spherical balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason 3. Two point charges placed at a distance r in air exert a force F on each other. At what distance will these charges experience the same force F in a medium of dielectric constant k? 4. A force F is acting between two charges placed some distance apart in vacuum. If a brass rod is placed between these charges, how does the force change? 5. Define electric lines of force and give its two important properties 3 MARKS QUESTIONS 1. An electric dipole is held in a uniform electric field. (i)Using suitable diagram, show that it does not undergo any translator motion. (ii)Derive an expression for the torque acting on it and specify its direction. 2. A charge is distributed uniformly over a ring of radius a. Obtain an expression for the electric intensity E a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge. 3. A long charged cylinder of linear charge density λ1 is surrounded by a hollow co-axial conducting cylinder of linear charge density – λ2. Use Gauss’s law to obtain expressions for the electric field at a point (i) In the space between the cylinders. XII/Physics-Support Material/Bengaluru Region/2024-25 (ii) Outside the larger cylinder. 4. (a) State Gauss’s law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density λ. (b)A wire AB of length L has linear charge density λ = kx where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for the electric flux through the surface. 5. Two large parallel thin metallic plates are placed close to each other. The plates have surface charge densities of opposite signs and of magnitude 20 x 10 -12 C/ m2. Calculate the electric field intensity (i) in the outer region of the plates (ii) in the interior region between the plates 5 MARKS QUESTIONS 1. (a) Derive an expression for the electric field at any point on the axial line of an electric dipole. (b)Two identical point charge q each are kept 2m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. 2. (a) Derive an expression for the electric field E due to a dipole of length ‘2a’at a point distant r from the centre of the dipole on the axial line (b) Draw a graph of E versus r for r>>a. (c) If this dipole were kept in a uniform external electric field Eo diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases. 3..(i) Use Gauss’s law to obtain the expression for the electric field due to an infinitely long thin straight wire with uniform linear charge density λ. (ii)An infinitely long positively charged straight wire has a linear charge density λ. An electron is revolving in a circle with a constant speed v such that the wire passes through the centre, and is perpendicular to the plane, of the circle. Find the kinetic energy of the electron in terms of magnitudes of its charge and linear charge density λ on the wire. (iii) Draw a graph of kinetic energy as a function of linear charge density λ. **************************** XII/Physics-Support Material/Bengaluru Region/2024-25 2. ELECTROSTATIC POTENTIAL AND CAPACITANCE SYLLABUS: 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). GIST 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. Unit. Its SI unit is volt (V) 1 volt (V) = 1 joule /coulomb 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, V= 𝑊𝐴𝐵/𝑞 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. 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 of potential gradient at that point. Mathematically E= - 𝑑𝑉/𝑑𝑟 Unit. Its unit in SI is volt /metre (V/m). Equipotential surface. The surface inan electric field having same potential at all points is called equipotential surface. Two equipotential surfaces can never intersect each other. Electrostatic potential energy of a system of charges. The electric potential energy of a system of point charges is the work needed to bring the charges from an infinite separation to their final positions. Its SIunit is joule (J) or electron volt (eV). 1 eV =1.6 x 10-19J Potential energy of an electric dipole in a uniform electric field. If the electric dipole is rotated from initial orientation making angle Ɵ1, with the electric field to the final orientation making angle Ɵ2 , with the field, then XII/Physics-Support Material/Bengaluru Region/2024-25 U = pE (cos Ɵ2 — cos Ɵ1) Behaviour of a charged conductor Charges reside only at the surface of the charged conductor. The electric potential is constant at the surface and inside the conductor.. The electric field is zero inside the conductor and just outside it, the electric field is normal to the surface. Capacitor :It is an arrangement for storing a very large amount of charge. Electrical capacitance. The ability of a conductor to store charge is called its electrical capacitance. Mathematically C = Q/ V Its SI unit 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. Principle. The capacitance of a conductor gets increased greatly, when an earth connected conductor is placed near it. Parallel plate capacitor It consists of two flat, parallel metal plates separated by a small distance. The space between the plates may have vacuum or some other insulating material such as mica, glass or paper. Energy stored in a capacitor. Work done in charging a capacitor gets stored in the capacitor in the form of its electrostatic potential energy. 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. 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. Its unit is V/m (same as that of electric field). Effect of dielectric slab on the capacitance of a parallel plate capacitor. When a dielectric slab of dielectric constant K is introduced between the plates, then , 𝐾𝜖0 𝐴 𝐶′ = 𝐶 ′ = 𝐾𝐶0. 𝑑 XII/Physics-Support Material/Bengaluru Region/2024-25 MIND MAP Electric Potential Electric potential at a point in an electric field is equal to the work done in bringing a unit positive charge (against electric field) from infinity to that point along any path. 𝑊 𝑞 𝑉 = 𝑞 = 4𝜋∈ 0 0𝑟 Unit- Volt (JC-1).Scalar quantity. ELECTRIC POTENTIAL DUE TO A SINGLE POINT CHARGE Consider a point charge +q placed at point O in free space/air as shown in Fig. It is desired to find electric potential at P due to charge +q. Let r be the distance of point P from O i.e,.OP = r. XII/Physics-Support Material/Bengaluru Region/2024-25 At point A at a distance x from charge +q, electric field intensity is Small amount of work done/C in moving positive test charge from A to B (where AB = dx) is dW= – E dx The negative sign is taken because dx is measured along the negative direction of x. Total amount of work done/C in bringing a small positive test charge from infinity to r is 𝑞 Therefore, 𝑉 = 4𝜋∈ 0𝑟 ELECTRIC FIELD INTENSITY AND POTENTIAL DIFFERENCE, EQUIPOTENTIAL SURFACE Any surface over which the potential is constant is called an equipotential surface. Few equipotential surfaces Equipotential surface due to a uniform field Equipotential surface due to point charge Equipotential surface due to equal charges Equipotential surface due to a dipole XII/Physics-Support Material/Bengaluru Region/2024-25 Properties of equipotential surfaces. (i) Work done in moving a charge over an equipotential surface is zero. The work done in moving a test charge +q0 from A to B is given by; WAB= (VB– VA) q0 Since VB– VA= 0, WAB= 0 Hence, no work is done in taking a charge from one point to another over an equipotential surface. (ii) The electric field is always perpendicular to an equipotential surface. we have, 𝑉𝐵 − 𝑉𝐴 = −𝐸⃗. ⃗⃗⃗⃗ 𝑑𝑟 But 𝑉𝐵 − 𝑉𝐴 = 0 Therefore, −𝐸⃗. ⃗⃗⃗⃗ 𝑑𝑟 = 0 It implies that, E perpendicular to r. (iii) Two equipotential surfaces can never intersect. If two equipotential surfaces could intersect, then at the point of intersection there would be two values of electric potential which is not possible. (iv) The spacing between equipotential surfaces enables us to identify regions of strong and weak fields. ELECTRIC POTENTIAL ENERGY The electric potential energy of a system of point charges is the work needed to bring the charges from an infinite separation to their final positions. POTENTIAL DUE TO AN ELECTRIC DIPOLE Consider an electric dipole (±𝑞, 2𝑎) having dipole moment |𝑝| = 2𝑞𝑎, from –q to +q. Let P be a point at(r,θ)from the centreO of the dipole (i.e., OP = r). It is desired to find the Electric Potential at P due to the dipole as shown in fig. XII/Physics-Support Material/Bengaluru Region/2024-25 XII/Physics-Support Material/Bengaluru Region/2024-25 POTENTIAL ENERGY OF TWO-CHARGE SYSTEM (i) First assume that the two charges +q1 and +q2 are infinite apart at rest. If we bring charge +q1 from infinity to its original position A, no work is done because no electrostatic force acts on it due to any other charge. (ii) When we bring charge +q2 from infinity (where V = 0) to its original position B, work will have to be done due to the repulsive force of +q1. This work done is equal to potential difference between B and ∞multiplied by charge +q2, i.e. By definition, this is the electric potential energy (U) of the two-charge system. Potential energy of a system of two charges in an external field Important electrostatic properties of a conductor 1. Inside a conductor, electrostatic field is zero. 2. At the surface of a charged conductor, electrostatic field must be normal to the surface at every point because surface of the conductor is an equipotential surface. 3. The interior of a conductor can have no excess charge in the static situation because E=0 and therefore ∑Q=0(from Gauss theorem) 4. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface because work done is zero from surface to inside as at surface E is perpendicular and inside E=0. 5. Electric field at the surface of a charged conductor 𝜎 𝐸=𝜖 0 ELECTROSTATIC SHIELDING: The phenomenon of protecting a certain region of space from external electric field is called electrostatic shielding. XII/Physics-Support Material/Bengaluru Region/2024-25 Since, Electric field inside a conductor is zero therefore if we want to protect delicate instruments from external electric field, we enclose them in hollow conductors. Dielectric Strength:The maximum electric field that a dielectric medium can withstand without its electrical break-down. e.g. for air it is about 3 × 106 Vm–1. Capacitor It consists of two metallic conductors electrically insulated from each other as well as their surroundings. It is used to store electrical energy in the form of electric field lines. The total charge of the capacitor is zero. Principle of Capacitor:The charge storing capacity of a conductor can be increased considerably by bringing an uncharged earthed conductor near it. Capacitance(C):The ratio of charge of capacitor to the potential difference across its ends. 𝑄 𝐶= 𝑉 SI unit of capacitance is 1 farad (=1 coulomb volt-1) or 1 F = 1 C V–1. It does not depend on charge and potential. It depends only on the dimension of the capacitor. Graph between Q and V is a straight line. PARALLEL PLATE CAPACITOR :It consists of two flat, parallel metal plates separated by a small distance as shown in Fig. The space between the plates may have vacuum or some other insulating material such as mica, glass or paper. CAPACITANCE OF PARALLEL PLATE CAPACITOR Consider a parallel plate capacitor having air/vacuum in the space between the plates. Let A = area of each plate, d = distance between the plates V = p.d. across the plates, q = charge on each plate 𝜎 = surface charge density on either plate = q/A The electric field between the plates is uniform and its magnitude is given by XII/Physics-Support Material/Bengaluru Region/2024-25 CAPACITANCE OF PARALLEL PLATE CAPACITOR WITH A DIELECTRIC SLAB Consider a parallel plate capacitor having dielectric in the space between the plates. Let A = area of each plate d = distance between the plates V = p.d. across the plates q = charge on each plate 𝜎 = surface charge density on either plate = q/A In the absence of dielectric, field between the plates 𝜎 𝑞 𝐸0 = = 𝜖0 𝐴𝜖0 The electric field due to the charged plates induces a net dipole moment in the dielectric. This effect, called polarization ,gives rise to a field in the opposite direction. The net electric field inside the dielectric is thus reduced and given by 𝐸0 𝑞 𝐸′ = = 𝐾 𝐾𝐴𝜖0 𝑞 Also, 𝑉 ′ = 𝐸 ′ 𝑑 = 𝐾𝐴𝜖 0 𝑞 𝐾𝜖0 𝐴 = 𝑉′ 𝑑 𝐾𝜖0 𝐴 But q/v’ is the capacitance,therefore, 𝐶′ = ⇒𝐶 ′ = 𝐾𝐶0 𝑑 RELATIVE PERMITTIVITY or DIELECTRIC CONSTANT Effect of introducing a dielectric of constant K between the plates of parallel plate capacitor Physical Quantity When the capacitor connected When the capacitor disconnected Capacitance K times K times Charge K times Constant Potential Difference constant 1 times 𝐾 Electric Field 1 1 times times 𝐾 𝐾 Potential Energy K times 1 times 𝐾 XII/Physics-Support Material/Bengaluru Region/2024-25 Energy stored in a Capacitor During charging a capacitor, transfer of electrons take place from one plate of the capacitor to the other. This work done by the battery in moving the electron is stored in the form of electric potential energy in the 𝑄2 1 1 electric field between the plates. 𝑈 = = 2 𝐶𝑉 2 = 2 𝑄𝑉 2𝐶 ENERGY DENSITY Consider a charged parallel plate capacitor of plate area A and plate separation das shown in Fig. When two capacitors are connected then 𝐶1 𝑉1+ 𝐶2 𝑉2 (i) Common Potential, 𝑉𝐶 = 𝐶1 +𝐶2 (ii) New charge on each capacitor, 𝑞1 = 𝐶1 𝑉𝐶 , 𝑞2 = 𝐶2 𝑉𝐶 1 𝐶 𝐶 (iii) Energy loss, ∆𝐸 = 2 𝐶 1+𝐶2 (𝑉1 − 𝑉2 )2 1 2 (iv) This energy is lost in the form of heat and electromagnetic radiation. (v) For uncharged capacitor take V=0 COMPETENCY BASED QUESTIONS 1 Consider a uniform electric field in the z-direction. The potential is a constant (a) for any x for a given z (b) for any y for a given z (c) on the x-y plane for a given z (d) all of these ANS d 2 In a parallel plate capacitor, the capacity increases if (a) area of the plate is decreased. (b) distance between the plates increases. (c) area of the plate is increased. (d) dielectric constant decreases ANS c XII/Physics-Support Material/Bengaluru Region/2024-25 3 When two identical charges approaches each other then electric potential energy of the system (a) Increases (b) decreases ( c) may increase or decrease (d) remains constant ANS a 4 Two charges + 3 µC and – 3 µC are placed at points A(0,0,4cm) and B( 0,0,-4cm.) respectively. Work done to move a charge of 2 µC from point P (0,1cm,0) to point Q ( 2cm,0,0) will be (a) 2 mJ (b) Zero (c) 1 mJ (d) 4 mJ ANS b 5 Three capacitors each of capacitance 3 µF are first connected in series and then in parallel. Ratio of effective capacitance in series Cs and in parallel CP is (a) 9 ( b) 1/9 (c) 2/9 (d) 9/2 ANS b 6 Two capacitors 3 µF and 6 µF are connected in series and connected to 100 V d.c. source. What is the ratio of energies stored in them (a) 1:2 ( b) 1:3 (c) 2:1 (d) 3:1 ANS c 7 SI unit of line integral of electric field is (a) J (b) N /C (c) N c -1 m2 (d) J / C ANS d 8 Two parallel plate air capacitors A and B have plate separation in ratio 1:3 and plate area in ratio 2:3. Ratio of their capacitances (a) 1:2 ( b) 1:3 (c) 2:1 (d) 3:1 ANS c 9 To protect a given region from the effect of electric fields and charges (a) it should be enclosed in a non magnetic material (b) it should be enclosed in a metallic conductor (c) it should be enclosed in a magnetic material (d) it should be enclosed in an insulating material ANS b 10 Four capacitor each of capacitance 16μF areare given. Equivalent capacitance will be 4μF when (a) all are connected in parallel (b) three are connected in parallel and one in series with them (c) all are connected in series (d) three are connected in series and one in parallel with them ANS c XII/Physics-Support Material/Bengaluru Region/2024-25 ONE MARK QUESTIONS 1 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 The electric field inside the shell is zero. This implies that potential is constant inside the shell (as no work is done in moving a charge inside the shell) and, therefore, equals its value at the surface, which is 10 V. 2 What is the work done in moving a test charge q through a distance of 1 cm along the equatorial axis of an electric dipole? ANS Zero 3 Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface? ANS 4 A point charge Q is placed at point O as shown. Is the potential difference ( VA-VB) positive, negative or zero if Q is positive ANS If q is positive charge, VA – VB = negative 5 What will happen to the Capacitance when a dielectric slab is placed between the plates of the capacitor? ANS Increases 2 MARKS QUESTIONS 1 Find the equivalence capacitance between X and Y. ANS 9 µ F 2 The given graph shows variation of charge ‘q’ versus potential difference ‘V’ for two capacitors C1 and C2. Both the capacitors have same plate seperation but plate area of C2 is XII/Physics-Support Material/Bengaluru Region/2024-25 greater than that of C1. Which line (A or B) corresponds to C1 and why? ANS Line B corresponds to C1 Reason: Since slope (qv) of ‘B’ is less than that of ‘A’ 3. Draw 3 equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains constant along Z-direction. How are these surfaces different from that of a constant electric field along Z-direction? ANS d2 < d1 for increasing field and d2 = d1 for uniform field. 4. i) Can two equipotential surfaces intersect each other? Give reasons. (ii) Two charges -q and + q are located at points A (0, 0, – a) and B (0, 0, +a) respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q (-3,0,0)? ANS i) No, if they intersect, there will be two different directions of electric field at that point which is not correct. If they intersect, then at the same point of intersection, there will be two values of potential. This is not possible and hence two equipotential surfaces cannot intersect. (ii) Since both the points P and Q are on the equatorial line of the dipole and V = 0 at every point on it, work done will be zero. Also the force on any charge is perpendicular to the equatorial line, so work done is zero 5 A charge ‘q’ is moved from a point A above a dipole of dipole movement ‘p’ to a point B below the dipole in equitorial plane without acceleration. Find the work done in the process. ANS No work is done [W = q VAB = q × 0 = 0, since potential remains constant] 3 MARKS QUESTIONS 1 Figure shows two identical capacitors C1 and C2, each of 2 µF capacitance, connected to a battery of 5 V. Initially switch ‘S’ is left open and dielectric slabs of dielectric constant K = 5 are inserted to fill completely the space between the plates of the two capacitors. How will the XII/Physics-Support Material/Bengaluru Region/2024-25 charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted? ANS (i) When switch S is open and dielectric is introduced, charge on each capacitor will be q1 = C1 V, q2 = C2V q1 = 5CV = 5 × 2 × 5 = 50 µC, q2 = 50 µC Charge on each capacitor will become 5 times (ii) P.d. across C1 is still 5V and across C2, q = (5C) V 2..A network of four capacitors, each of capacitance 30 pF, is connected across a battery of 60 V as shown in the figure.Find the net capacitance and the energy stored in each capacitor. XII/Physics-Support Material/Bengaluru Region/2024-25 ANS Answer: 3 Net capacitance of three identical capacitors in series is 1 pF. 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 4 Two point charges q1 and q2 are located at r1→ and r2→ respectively in an external electric field E→.Obtain the expression for the total work done in assembling this configuration ANS Work done in bringing the charge q1 from infinity to position r1 W1 = q1V(r1) Work done in bringing charge q2 to the position r2 Hence, total work done in assembling the two charges XII/Physics-Support Material/Bengaluru Region/2024-25 w = W1 + W2 5 An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of 4√3 Nm. Calculate the potential energy of the dipole, if it has charge ± 8 nC ANS 1st method Given : 2a = 4 cm = 4 × 10-2 m, θ = 60° τ = 4 × √3 Nm, q = ±8 nC = ±8 x 10-9 CP.E. = |p| |E| cos θ, τ = |p| |E| sin θ 5 Marks Questions 1 (a) Define the SI unit of capacitance. (b) Obtain the expression for the capacitance of a parallel plate capacitor. (c) Derive the expression for the affective capacitance of a series combination of n capacitors. ANS (a) When a charge of one coulomb produces a potential difference of one volt between the plates of capacitor, the capacitance is one farad. (b) Capacity of a parallel plate capacitor. A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance. We first take the intervening medium between the plates to be vaccum. Let A be the area of each plate and d the separation between them. The two plates have charges Q and – Q. Since d is XII/Physics-Support Material/Bengaluru Region/2024-25 much smaller than the linear dimension of the plates (d2 > R, radius of cross-section. In (i) E is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below (a) case (i) contradicts Gauss's law for electrostatic fields (b)case (ii) contradicts Gauss's aw for magnetic fields XII/Physics-Support Material/Bengaluru Region/2024-25 OR The surface integral of a magnetic field over a surface (a) is proportional to mass enclosed (b) is proportional to charge enclosed (c) is zero (d) equal to its magnetic flux through that surface. 2. When the atomic dipoles are aligned partially or fully, there is a net magnetic moment in the direction of the field in any small volume of the material. The actual magnetic field inside material placed in magnetic field is the sum of the applied magnetic field and the magnetic field due to magnetisation. This field is called magnetic intensity (H). where M is the magnetisation of the material, llo is the permittivity of vacuum and B is the total magnetic field. The measure that tells us how a magnetic material responds to an external field is given by a dimensionless quantity is appropriately called the magnetic susceptibility: for a certain class of magnetic materials, intensity of magnetisation is directly proportional to the magnetic intensity. (i) Magnetization of a sample is (a) volume of sample per unit magnetic moment (b) net magnetic moment per unit volume (c) ratio of magnetic moment and pole strength (d) ratio of pole strength to magnetic moment (ii) Identify the wrongly matched quantity and unit pair. (a) Pole strength Am (b) Magnetic susceptibility dimensionless number (c) Intensity of magnetisation A m-1 XII/Physics-Support Material/Bengaluru Region/2024-25 (d) Magnetic permeability Henry m (iii) A bar magnet has length- 3 cm, cross-sectional area 2 cm2 and magnetic moment 3 A m2. The intensity of magnetisation of bar magnet is (iv) A solenoid has core of a material with relative permeability 500 and its windings carry a current of 1 A. The number of turns of the solenoid is 500 per metre. The magnetization of the material is nearly OR The relative permeability of iron is 6000. Its magnetic susceptibility is (a) 5999 (b) 6001 (c) 6000 x 10-7 (d) 6000 x 107 3. Before the 19th century, scientists believed that magnetic properties were confined to a few materials like iron, cobalt and nickel. But in 1846, Curie and Faraday discovered that all the materials in the universe are magnetic to some extent. These magnetic substances are categorised in two groups. Weak magnetic materials are called diamagnetic and para magnetic materials. Strong magnetic materials are called ferromagnetic materials. According to the modern theory of magnetism, the magnetic response of any material is due to circulating electrons in the atoms. Each such electron has a magnetic moment in a direction perpendicular to the plane of circulation. In magnetic materials all these magnetic moments due to the orbit and spin motion of all the electrons in any atom vectorially add up to a resultant magnetic moment. The magnitude and direction of these resultant magnetic moment is responsible for the behaviour of the materials. For diamagnetic materials χ is small and negative and for paramagnetic materials χ is small and positive. Ferromagnetic materials have a large χ and are characterised by non–linear relation between 𝐵̅and H (i) The universal (or inherent) property among all substance is (a) Diamagnetism (b) Para magnetism (c) Ferromagnetism (d) Both (a) and (b) (ii) When a bar is placed near a strong magnetic field and it is repelled, then the material of the bar is XII/Physics-Support Material/Bengaluru Region/2024-25 (a) Diamagnetic (b) Ferromagnetic (c) Paramagnetic (d) Anti-ferromagnetic (iii) Magnetic susceptibility of a diamagnetic substance (a) Decreases with temperature (b) Is not affected by temperature (c) Increases with temperature (d) First increases then decreases with temperature (iv) For a para magnetic material, the dependence of the magnetic susceptibility χ on the absolute temperature is given as (a) χ ∝ T (b) χ ∝ 1/T2 (c) χ ∝ 1/T (d) independent g. TWO MARK QUESTIONS: 1. A circular coil of ′N′ turns and radius ′R′ carries a current ′I′. It is unwound and rewound to make another coil of radius ′R/2′, current ′I′ remaining the same. Calculate the ratio of the magnetic moments of the new coil and original coil. 2. Explain the following: (i) Why do magnetic field lines from continuous closed loops? (ii) Why are the field lines repelled when a diamagnetic material is placed in an external uniform magnetic field? 3. Write three points of difference between para-, dia- and ferro- magnetic materials, giving one example for each. 4. Deduce the expression for the magnetic dipole moment of an electron orbiting around central nucleus. 5. The following figure shows the variation of intensity of magnetization versus the applied magnetic field intensity H for two magnetic materials A and B. (i) Identify the materials A and B. (ii)Why does the material B have a largest susceptibility then A for a given field at constant temperature? h. ASSERTION AND REASON QUESTIONS: XII/Physics-Support Material/Bengaluru Region/2024-25 Directions: These questions consist of two statements, each printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses. (a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. (b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion. (c) If the Assertion is correct but Reason is incorrect. (d) If both the Assertion and Reason are incorrect. 1. Assertion : We cannot think of a magnetic field configuration with three poles Reason : A bar magnet does exert a torque on itself due to its own field. 2. Assertion : Diamagnetic materials can exhibit magnetism. Reason : Diamagnetic materials have permanent magnetic dipole moment. 3. Assertion : Ferro-magnetic substances become paramagnetic above Curie temp. Reason : Domains are destroyed at high temperature. 4. Assertion : The poles of magnet can not be separated by breaking into two pieces. Reason : The magnetic moment will be reduced to half when a magnet is broken into two equal pieces. 5. Assertion : Basic difference between an electric line and magnetic line of force is that former is discontinuous and the latter is continuous or endless. Reason : No electric lines of forces exist inside a charged body but magnetic lines do exist inside a magnet. E. SELECT RESPONSE TYPE QUESTIONS b. MCQ (EASY): 1. The magnetic field lines of force inside a bar magnet: a) From S pole to N pole of the magnet b) Do not exist c) From N pole to S pole of the magnet d) Area of the cross-section of the magnet 2. The magnetic lines of force are: (a) Closed curves (b) Intersect far away from the poles (c) Always intersect (d)Do not pass through a vacuum 3. The magnetic dipole moment of a solenoid having N turns is given as - (a) NIA2 (b) NIA (c) NI2A (d) NI2A2 4. A freely suspended magnet aligns in which direction? (a) South-west (b)East-west (c) North-south (d) North-west 5. The SI unit of magnetic flux is XII/Physics-Support Material/Bengaluru Region/2024-25 (a) Dyne (b) Tesla (c) Weber (d) Ohm 6. The materials with magnetic susceptibility negative and small are called as (a) Paramagnetic (b) Diamagnetic (c) Ferromagnetic (d) None 7. Magnetic lines of force due to a bar magnet do not intersect because (a) a point always has a single net magnetic field (b) the lines have similar charges and so repel each other (c) the lines always diverge from a single force (d) the lines need magnetic lenses to be made to interest 8. If the magnetizing field on a ferromagnetic material is increased, its permeability (a) Decreased (b) Increased (c) Is unaffected (d) May be increased or decreased 9. A circular loop carrying current I is replaced by a bar magnet of equivalent magnetic dipole moment. The point on the loop is lying _____. (a) on equatorial plane of magnet (b) on axis of the magnet (c) A and B both (d) except equatorial plane or axis of bar magnet 10. In non-uniform magnetic field, a diamagnetic substance experiences a resultant force (a) which is zero (b) perpendicular to the magnetic field. (c) from the region of weak magnetic field to the region of strong magnetic field. (d) from the region of strong magnetic field to the region of weak magnetic field. F. CONSTRUCTED RESPONSE QUESTIONS: a. THREE MARK QUESTIONS: 1. A bar magnet is placed in uniform magnetic field with its magnetic moment at angle θ with the magnetic field. i. Find expression for torque acting on the magnet. ii. Define magnetic moment. 2. Derive the expression for induced emf produced by changing the area of the rectangular coil placed in perpendicular magnetic field. XII/Physics-Support Material/Bengaluru Region/2024-25 3. Differentiate between Diamagnetic, Paramagnetic and Ferromagnetic substances. 4. Show that the electron revolving around the nucleus in an orbit of radius ‘r’ with speed ‘v’ has magnetic moment evr/2. Hence using Bohr's postulate of angular momentum obtain the expression for magnetic moment of hydrogen atom in its ground state. b. FIVE MARK QUESTIONS: 1. (a) An iron ring of relative permeability ‘µ’ has windings of insulated copper wire of ‘n’ turns per metre. when the current in the winding is I, find the expression for magnetic field in the ring. (b) The susceptibility of magnetic material is 0.9853. Identify the type of magnetic material draw the modification of field pattern on keeping a piece of this material in a uniform magnetic field. 2. (a) Write the expression for equivalent magnetic moment of a planer current loop of area ‘A’, having ‘N’ turns and carrying current I use the expression to find the magnetic dipole moment of revolving electron. (b) A circular loop of radius ‘r’, having ‘N’ turns and carrying current ‘I’ is kept in X Y plane. It is then subjected to uniform magnetic field B = Bxi + Byj + Bzk. Obtain expression for the magnetic potential energy of the coil - magnetic field system. 3. (a) A long solenoid with air core has ‘n’ turns per unit length and carries a current ‘I’ using ampere circuital law derive an expression for magnetic field B at an interior point on its axis. Write an expression for magnetising field intensity ‘H’ in the interior of the solenoid. (b) Small bar magnet off material having magnetic susceptibility ‘χ’, is now put along the axis and near the centre, of the solenoid which is carrying a dc current through its coils. After sometime the bar is taken out and suspended freely with an unspun thread. Will the bar orient itself in the magnetic meridian if (a) ‘χ’< 0 (b) ‘χ’ > 1000? SELF ASSESSMENT TEST:- MCQ 1. A diamagnetic substance is brought near the north or south pole of a bar magnet. It will be : (a) repelled by both the poles. (b) attracted by both the poles. (c) repelled by the north pole and attracted by the south pole. (d) attracted by the north pole and repelled by the south pole. 2. In a permanent magnet at room temperature (a) magnetic moment of each molecule is zero. (b) the individual molecules have non-zero magnetic moment which are all perfectly aligned. (c) domains are partially aligned. (d) domains are all perfectly aligned. XII/Physics-Support Material/Bengaluru Region/2024-25 ASSERTION BASED QUESTIONS: (a) If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion. (b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion. (c) If the Assertion is correct but Reason is incorrect. (d) If both the Assertion and Reason are incorrect. 3. Assertion (A) : When a bar of copper is placed in an external magnetic field, the field lines get concentrated inside the bar. Reason (R) : Copper is a paramagnetic substance. 4.Assertion- diamagnetic materials can exhibit magnetism. Reason-diamagnetic materials have permanent magnetic dipole moment. i 5.Assertion :When radius of a circular loop carrying current is doubled, its magnetic moment becomes four times. Reason: Magnetic moment depends on the area of the loop. 6. Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify? 7. A circular coil of N turns and diameter ‘d’ carries a current ‘I’. It is unwound and rewound to make another coil of diameter ‘2d’, current T remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil. ANSWERS A. (COMPETENCY BASED QUESTIONS): e. MCQ (HOTS): 1.c 2.b 3.d 4.b 5. (d) Option (a) and (b) 6. c 7.d 8.b 9.c 10.c f. CASE BASED QUESTIONS: 1. i. b ii. a iii. b iv. c OR d 2. i. b ii. d iii. d Iv. b OR a 3 i. a ii. a iii. b iii. c g. TWO MARK QUESTIONS: XII/Physics-Support Material/Bengaluru Region/2024-25 1. As length of wire remains the Magnetic moment of a coil, m = NAI For the coil of radius R, magnetic moment Now, m2/m1 = 1:2 2. (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. 3. 4. Consider an electron revolving around a nucleus (N) in circular path of radius r with speed v. The revolving electron is equivalent to electric current XII/Physics-Support Material/Bengaluru Region/2024-25 I = e/T where T is period of revolution = 2πr/v Area of current loop (electron orbit), A = πr2 Magnetic moment due to orbital motion, 8. (i) Material A is Paramagnetic and Material B is Ferromagnetic. (ii)Since paramagnetic substances have a tendency to pull in magnetic field lines when placed in a magnetic field. h. ASSERTION AND REASONING: 9. d 10. c 11. a 12. b 13. a B. (SELECT RESPONSE TYPE QUESTIONS): a. MCQ (EASY): 1.a 2.a 3.b 4.c 5.c 6.b 7.a 8.a 9.a 10.d SELF ASSESMENT: 1. a 2. c 3. c 4. c 5.a Ans 6. Sol: (i) Magnetic susceptibility (χm) : It is the property of a material which determines how easily it can be magnetised when kept in a magnetising field. XII/Physics-Support Material/Bengaluru Region/2024-25 Also, it is the ratio of intensity of magnetisation (I) produced in the material to the intensity of magnetising field (H) (ii) Positive susceptibility : para-magnetic material Example: Al, Ca. Negative susceptibility : diamagnetic material Example: Bi, Cu. Ans7. Magnetic moment of the coil is given by M = NIA But as given, Ist coil is rewound to make new coil. ************************************************************************* XII/Physics-Support Material/Bengaluru Region/2024-25 6. ELECTROMAGNETIC INDUCTION SYLLABUS: Electromagnetic induction; Faraday's laws, induced EMF and current; Lenz's Law, Self and mutual induction. GIST MAGNETIC FLUX: The number of magnetic field lines crossing a surface normally is called magnetic flux (Φ𝐵) linked with the surface. Φ𝐵 = ⃗⃗⃗ 𝐁. ⃗⃗⃗ 𝐀 = 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. The SI, unit of magnetic flux is weber (Wb). 1 weber = 108 maxwell 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. 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. XII/Physics-Support Material/Bengaluru Region/2024-25 3. The magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux. Induced e.m.f., e = − = ⅆ𝛟 𝛟𝟐 −𝛟𝟏 ⅆ𝐭 𝐭 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. MOTIONAL E.M.F When a conductor of length l 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. e=Blv FLEMING'S RIGHT-HAND RULE It is used to find the direction of flow of the induced current. It states that if the thumb, forefinger and the central finger of the right hand are kept perpendicular to each other, so that the forefinger points in the direction of the field and the thumb in the direction of motion of the conductor, then the induced current flows in the direction of the central finger. EDDY CURRENTS XII/Physics-Support Material/Bengaluru Region/2024-25 The currents induced in the body of a conductor, when the magnetic flux linked with the conductor changes, are called eddy currents (or Foucault’s currents). The direction of the eddy currents set up in the conductor can be found by applying Lenz’s law or Fleming’s right-hand rule. SELF-INDUCTION The phenomenon according to which an opposing induced e.m.f. is produced in a coil as a result of change in current or magnetic flux linked with it, is called self- induction. COEFFICIENT OF SELF-INDUCTION The coefficient of self-induction or simply self-inductance (L) of a coil is numerically equal to the magnetic flux (Φ) linked with the coil, when a unit current flows through it. Φ =LI SELF-INDUCTANCE The Self-Inductance of a coil is numerically equal to the induced e.m.f. produced in the coil, when the rate of change of current in the coil is unity. e= − L ⅆ𝐈 ⅆ𝐭 The induced emf is also called back emf. Self-induction is also called inertia of electricity. The SI unit of self-inductance is Henry (H). XII/Physics-Support Material/Bengaluru Region/2024-25 The self-inductance of a coil is said to be one henry, if a rate of change of current of 1 ampere per second induces an e.m.f. of 1 volt in it. ENERGY STORED IN AN INDUCTOR When a current! flows through an inductor of self-inductance L, the energy stored in the inductor is given by 𝟏 𝟐 𝐔= 𝐋𝐈 𝟐 The energy resides in the inductor in the form of magnetic field. SELF-INDUCTANCE OF A LONG SOLENOID The self-inductance of a long solenoid of length /, area of cross-section A and number of turns per unit length 1 is given by L = 𝝁𝑶𝒏𝟐𝒍 𝑨 ENERGY STORED IN A SOLENOID When a current is passed through a solenoid, the energy is stored inside it in the form of magnetic field. If the current builds up a magnetic field of induction B, then the energy stored in the solenoid is given by U = 𝟐𝝁 B2A𝒍 𝟏 𝟎 where l is length and A, the area of cross-section of the solenoid. XII/Physics-Support Material/Bengaluru Region/2024-25 MUTUAL INDUCTION The phenomenon according to which an opposing induced e.m.f. is produced in a coil as a result of change in current or magnetic flux linked with a neighbouring coil is called mutual induction. COEFFICIENT OF MUTUAL INDUCTION The coefficient of mutual induction or simply mutual inductance (M) of the two coils is numerically equal to the magnetic flux (p) linked with one coil, when a unit current flows through the neighbouring coil. Φ = MI MUTUAL INDUCTANCE The mutual inductance of two coils is also numerically equal to the induced e.m.f. produced in one coil, when the rate of change of current is unity in the other coil. e = -M ⅆ𝐭 ⅆ𝐈 The SI unit of mutual inductance is henry (H). The mutual inductance of two coils is said to be one henry, if a rate of change of current of 1 ampere per second in one coil induces an e.m.f. of 1 volt in the neighbouring coil. MUTUAL INDUCTANCE OF TWO LONG SOLENOIDS When over a solenoid S1, of length l, area of cross-section A and number of turns per unit length n1, another solenoid S2, of same length and number of turns XII/Physics-Support Material/Bengaluru Region/2024-25 per unit length n2, is wound, then mutual inductance between the two solenoids is given by M = 𝝁𝑶𝒏𝟏𝒏𝟐 𝒍 𝑨 ALTERNATING CURRENT GENERATOR It is a device used to obtain a supply of alternating e.m.f. by converting mechanical energy into electrical energy. It is based on the phenomenon of electromagnetic induction. The instantaneous value of e.m.f. produced is given by e = nBAω sin ωt where n is number of turns of the coil, A is the area of coil and ω is angular frequency of rotation of the coil inside a magnetic field strength B. XII/Physics-Support Material/Bengaluru Region/2024-25 MIND MAP XII/Physics-Support Material/Bengaluru Region/2024-25 COMPETENCY BASED QUESTIONS 1 A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then, (a) a direct current flows in the ammeter A (b) no current flows through the ammeter A (c) an alternating sinusoidal current flow through the ammeter A with a time period T = 2π/𝜔 (d) a time varying non-sinusoidal current flows through the ammeter A. ANS b 2 There are two coils A and B as shown in figure. A current start flowing in B as shown, when A is moved towards B and stops when A stops moving. The current in A is counter clockwise. B is kept stationary when A moves. We can infer that (a) there is a constant current in the clockwise direction in A (b) there is a varying current in A (c) there is no current in A (d) there is a constant current in the counter clockwise direction in A ansDd d 3 The self-inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as XII/Physics-Support Material/Bengaluru Region/2024-25 (a) l and A increase (b) l decreases and A increases (c) l increases and A decreases (d) both l and A decrease ANS b 4 A metal plate is getting heated. It cannot be because (a) a direct current is passing through the plate (b) it is placed in a time varying magnetic field (c) it is placed in a space varying magnetic field, but does not vary with time (d) a current (either direct or alternating) is passing through the plate ANS c 5 An emf is produced in a coil, which is not connected to an external voltage source. This cannot be due to (a) the coil being in a time varying magnetic field (b) the coil moving in a time varying magnetic field (c) the coil moving in a constant magnetic field (d) the coil is stationary in external spatially varying magnetic field, which does not change with time ANS d 6 A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of XII/Physics-Support Material/Bengaluru Region/2024-25 turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be (a) halved (b) the same (c) doubled (d) quadrupled ANS b 7 Two identical circular loops of metal wire are lying on a table without touching each other. Loop A carries a current which increases with time. In response the loop B (a) remains stationary (b) is attracted by loop A (c) is repelled by loop A (d) rotates about is CM with CM fixed ANS c 8 A small bar magnet is being slowly inserted with constant velocity inside a solenoid as shown in figure. Which graph best represents the relationship between emf induced with time (a) (b) XII/Physics-Support Material/Bengaluru Region/2024-25 (c) (d) ANS c 9 The coils in resistance boxes are made from doubled insulated wire to nullify the effect of (a) heating (b) magnetism (c) pressure (d) self- induced e.m.f. ANS d 10 The north pole of a long bar magnet was pushed slowly into a short solenoid connected to a short galvanometer. The magnet was held stationary for a few seconds with the north pole in the middle of the solenoid and then withdrawn rapidly. The maximum deflection of the galvanometer was observed when the magnet was (a) moving towards the solenoid (b) moving into the solenoid (c) at rest inside the solenoid (d) moving out of the solenoid ANS d ONE MARK QUESTIONS 1 A plot of magnetic flux (ϕ) versus current (I) is shown in the figure for two inductors A and B. Which of the two has larger value of self-inductance? ANS Since ϕ = LI ∴ L = ϕ /I= slope Slope of A is greater than slope of B ∴ Inductor A has larger value of self-inductance than inductor B. XII/Physics-Support Material/Bengaluru Region/2024-25 2 Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily. ANS In metal ring 1, the induced current flows in the clockwise direction. In metal ring 2, the induced current flows in the anticlockwise direction 3 A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the directions of induced current in each coil. ANS By Lenz’s law, the ends of both the coils closer to the magnet behave as south pole. Hence the current induced in both the coils will flow clockwise when seen from the magnet side. 4 The motion of copper plates is damped when it is allowed to oscillate between the two poles of a magnet. If slots are cut in the plate, how will the damping be affected? ANS Eddy current will decrease due to which damping reduces 5 Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely from the same h