Physics Special Guide PDF

DDDD DD DEPARTMENT DEPARTMENT DDDDDDPOPOPPOP OF SCHOOL OF SCHOOL EDUCATION EDUCATION MADURAI DISTRICT MADURAI DISTRICT 12 Special Guide ...

DDDD DD DEPARTMENT DEPARTMENT DDDDDDPOPOPPOP OF SCHOOL OF SCHOOL EDUCATION EDUCATION MADURAI DISTRICT MADURAI DISTRICT 12 Special Guide  DEPARTMENT OF SCHOOL EDUCATION, MADURAI DISTRICT. 12 PHYSICS SPECIAL GUIDE CHAIRPERSON Thiru R. Swaminathan, Chief Educational Officer, Madurai. COORDINATOR : SUPERVISOR : Thiru V. Rajasekaran, Thiru P. Kumaresan, Headmaster ,Govt. HSS,Paravai. Convenor HSS & Headmaster , Govt. HSS, P.Ammapatti. Team Of Teachers Thiru S. Ponnaiah , Thiru R. Sivakumar , Govt. HSS - Uranganpatti. Elango Corp. HSS - Madurai. Thiru R. Saravanakumar, Thiru T. Senthilmurugan, Govt. HSS - E. Malampatti. Govt (B). HSS - T.Vadipatti. Tmt. P. Karpagam, Thiru S. Karthick, Govt ( G ). HSS- Alanganallur. Setupati HSS , Madurai. Tmt. J. Shalini Devi, Tmt. R. Jacquline Esther Rani, Govt. HSS - Muduvarpatti. Govt.HSS - Sedapatti. Tmt. E. Janet Pamila, Thiru P. Anbarasu, Govt ( G ).HSS - Y.Othakadai. Govt ( G ).HSS - Sholavanthan.  CONTENTS UNIT TITLE PAGE NO. 1 ELECTROSTATICS 1 2 CURRENT ELECTRICITY 11 3 MAGNETISM AND MAGNETIC EFFECT OF CURRENT 21 4 ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT 32 5 ELECTROMAGNETIC WAVES 41 6 RAY OPTICS 45 7 WAVE OPTICS 52 8 DUAL NATURE OF RADIATION AND MATTER 58 9 ATOMIC AND NUCLEAR PHYSICS 64 10 ELECTRONICS AND COMMUNICATION 72 11 RECENT DEVELOPMENTS IN PHYSICS 86  1. ELECTROSTATICS 2 MARK - QUESTIONS AND ANSWERS : 1. Write down Coulomb‟s law in vector form and mention what each term represents. 𝑞1 𝑞2  Coulomb’s law in vector form 𝐹 = 𝑘 𝑟 𝑟2  Here, 𝐹 - the force between point charges 𝑞1, 𝑞2 - magnitude of point charges r - distance between the two charges 𝑟 - the unit vector pointing along the line joining 𝑞1, 𝑎𝑛𝑑 𝑞2 State coulomb‟s law in electrostatics. The electrostatic force is directly proportional to the product of the magnitude of the two point charges and is 𝑞1𝑞2 inversely proportional to the square of the distance between them. i.e 𝐹 𝛼 𝑟2 2. State Gauss law in electrostatics. 𝑄  The total electric flux through a closed surface 𝜙𝐸 = ∈. 0  Here Q is the net charge enclosed by the surface. 3. What is an electric dipole? Give few examples.  Two equal and opposite charges separated by a small distance constitute an electric dipole.  Examples : Water (H2O), ammonia (NH3), HCl and CO. 4. What is the general definition of electric dipole moment?  The magnitude of the electric dipole moment is equal to the product of the magnitude of one of the charges and the distance between them. (i.e) 𝑝 = q 2a.  Its unit is coulomb meter (C m). 5. Define “Electrostatic potential”. Give its unit.  The electrostatic potential at a point is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to that point in the region of the external electric field.  Its unit is volt (V). 6. Define potential difference. Give its unit.  The electric potential difference is defined as the work done by an external force to bring unit positive charge from one point to another point against the electric field.  Its unit is volt (V) 7. Define Electrostatic potential energy. Give its unit.  Electrostatic potential energy for system of charges is equal to the work done to arrange the charges in the given configuration.  Its unit is joule ( J ). 1 8. What is an Equipotential Surface?  An equipotential surface is a surface on which all the points are at the same electric potential. 9. What is an electrostatic induction?  The phenomenon of charging without actual contact of charged body is called electrostatic induction. 10. What is dielectric (or) insulator?  A dielectric is a non-conducting material and has no free electrons.  The electrons in a dielectric are bound within the atoms. Examples: Ebonite, glass and mica. 11. What are non polar molecules ? Give examples.  A non polar molecule is one in which the centers of the positive and negative charges coincide.  It has no permanent dipole moment. Examples : O2, H2, CO2. 12. What are polar molecules ? Give examples.  A polar molecule is one in which the centers of the positive and the negative charges are separated even in the absence of an electric field.  They have a permanent dipole moment. Examples : N2O, H2O, HCl, NH3. 13. Define (electric) polarisation?  (Electric) Polarisation is defined as the total dipole moment per unit volume of the dielectric. 14. Define electric susceptibility. Give its unit.  Electric susceptibility is defined as polarization per unit external electric field. (i.e) 𝑃 = 𝜒𝑒 𝐸𝑒𝑥𝑡.  Its unit is 𝑪2 𝑵 -1𝒎 -2. 15. What is dielectric breakdown.?  When the external electric field applied to a dielectric is very large, it tears the atoms apart so that the bound charges become free charges.  Then the dielectric starts to conduct electricity. This is called dielectric breakdown. 16. What is dielectric strength?  The maximum electric field ,the dielectric can withstand before it breakdowns is called dielectric strength.  E.g. dielectric field strength of air is 3×106 V m-1. 17. What is corona discharge (or) action at points ?  Leakage of electric charges from the sharp edge of the charged conductor is called corona discharge or action at points. 18. The electric field lines never intersect. Justify.  If two lines cross at a point, then there will be two different electric field vectors at that point.  If a charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect. 19. Why is it safer to be inside a car than standing under a tree during lightning?  The metal body of the car provides electrostatic shielding, since the electric field inside is zero.  During lightning the electric discharge passes through the body of the car. 2 20. What is meant by quantisation of charges?  The charge in an object q = ne.  Here n = 0, 1, 2, 3, 4….. and e is electron charge. 21. Write short notes on superposition principle  The total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.  𝐹1𝑡𝑜𝑡 = 𝐹12 + 𝐹13 + 𝐹14 + ⋯ ….. +𝐹1𝑛 22. Define capacitance of a capacitor. Give its unit.  The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between them. (i.e) 𝐶=𝑄/𝑉  Its unit is farad ( F ) or C V -1. 23. Define electrostatic energy density.  The energy stored per unit volume of space is defined as energy density. 24. Define electric field. Give its unit.  The electric field at a point is defined as the force experienced by a unit charge placed at that point.  Its unit is N𝑪 −𝟏 (or) V m -1. 25. Define Electric flux. Give its unit.  The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux.  Its unit is 𝑵𝒎𝟐𝑪 − 𝟏 26. What are the properties of an equipotential surface?  The work done to move a charge between any two points on the equipotential surface is zero.  The electric field must always be normal to an equipotential surface. 27. Give the relation between electric field and electric potential. 𝑑𝑉  Electric field is the negative gradient of the electric potential. (i. e ) 𝐸 = − 𝑑𝑥. 28. Write a note on electrostatic shielding.  The electric field inside the charged spherical shell is zero.  A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside the cavity of a charged conductor. This is called electrostatic shielding. (e.g) Faraday cage. 29. What is meant by „electric field lines‟.  A set of continuous lines which are the visual representation of the electric field in some region of space is called electric field lines. 30. State conservation of electric charges.  The total electric charge in the universe is constant and charge can neither be created nor be destroyed  In any physical process, the net change in charge will be zero. 3 3 MARK - QUESTIONS AND ANSWERS : 1. Distinguish between Coulomb force and Gravitational force. S.N Coulomb force Gravitational force 1 It acts between two charges. It acts between two masses. 2 It can be attractive or repulsive. It is always attractive. 3 It is always greater in magnitude. It is always lesser in magnitude. 4 It depends on the nature of the medium. It is independent of the medium. 2. List the properties of electric field lines. i) They start from positive charge and end at negative charge. ii) The electric field vector at a point in space is tangential to the electric field line at that point. iii) The electric field lines are denser in a region where the electric field has larger magnitude and less dense in region where the electric field is of smaller magnitude. iv) No two electric field lines intersect each other. v) The number of electric field line is directly proportional to the magnitude of the charge. 3. Give the applications and disadvantage of capacitors Applications of capacitor: i) Flash capacitors are used in digital camera. ii) It is used in heart defibrillator to retrieve the normal heart function during cardiac arrest. iii) Capacitors are used in the ignition system of automobile engines to eliminate sparking. iv) Capacitors are used to reduce power fluctuations in power supplies and to increase the efficiency of power transmission. Disadvantage:  Even after the battery or power supply is removed, the capacitor stores charges and energy for some time. It causes unwanted electric shock. 4. Derive an expression for torque experienced by an electric dipole placed in the uniform electric field.  Consider an electric dipole AB placed in an uniform electric field at an angle .  The force on ‘+q’ = q 𝐸 ; The force on ‘-q’ = − q𝐸.  Then the total force acting on the dipole is zero.  Due to these two forces the dipole experience a torque which tends to rotate the dipole.  The magnitude of the torque  = magnitude of one of the force x perpendicular distance between forces  = qE x 2a sin ( p = q x 2a)  = pE sin.  In vector notation, 𝜏 = 𝑝 × 𝐸 4 5. Obtain an expression for electric potential at a point due to a point charge.  Consider a point charge +𝒒 at origin. ‘P’ be a point at a distance ‘r’ from origin. 𝑟  Electric potential at ‘P’ , 𝑉 = − ∞ 𝐸. 𝑑𝑟 1 𝑞  By definition, the electric field at ‘P’ , 𝐸 = 𝑟 4𝜋∈0 𝑟 2 𝑟 1 𝑞  𝑉=− ∞ 4𝜋 ∈0 𝑟 2 𝑟 𝑑𝑟 𝑞  𝑉= 4𝜋 ∈0 𝑟 6. Derive an expression for energy stored in capacitor.  The work done to transfer charge from one plate to other plate is stored as electrostatic energy in the capacitor.  The work done to transfer ′𝑑𝑄′ amount of charge 𝑄 𝑄 dW = V dQ = 𝑑𝑄 ∵ 𝑉 = 𝐶 𝐶  The total work done to charge a capacitor, 𝑄𝑄 𝑄2  𝑊= 0 𝐶 𝑑𝑄 = 2𝐶  This work done is stored as electrostatic energy of the capacitor, 𝑄2 1 𝑈= ( or ) U = 𝐶 𝑉2. 2𝐶 2 7. Derive an expression for capacitance of parallel plate capacitor.  Consider a capacitor consisting of two parallel plates each of area ‘A’ separated by a distance ‘d’. 𝑄  Let ‘𝝈′ be the surface charge density of the plates. (i.e) 𝜎 = 𝐴 𝜎 𝑄  The electric field between the plates 𝐸 = ∈ = 𝐴∈ 0 0 𝑄  The potential difference between the plates 𝑉 = 𝐸𝑑 = 𝑑 𝐴∈0 𝑄 𝑄  The capacitance of the capacitor 𝐶 = 𝑉 = 𝑄𝑑 𝐴 ∈0 ∈0 𝐴 𝐶=. 𝑑 8. Obtain Gauss‟s law from Coulomb‟s law.  Consider a point charge ‘+q’. C - is a point at a distance of ‘r’ from the charge. 𝐹  The electric field at this point, 𝐸 =. 𝑞0 is the test charge and q 0 = 1 C 𝑞0 1 𝑞 𝑞0  Force between the charges, 𝐹 = 4𝜋𝜖 0 𝑟2 𝑟. This is Coulomb’s law. 𝐹 1 𝑞  Therefore, Electric field at C , 𝐸 = = 𝑟 - - - - - - - (1) 𝑞0 4𝜋𝜖0 𝑟2  From the definition, the electric flux Φ𝐸 = 𝐸. 𝑑𝐴 = 𝐸 𝑑𝐴 - - - - - -(2) 1 𝑞 Substitute (1) in (2), Φ𝐸 = 4𝜋𝜖 0 𝑟 2 𝑟 𝑑𝐴 - - - - - - - ( 3 )  Put 𝑟 = 1 and 𝑑𝐴 = 4 𝜋𝑟 2 in (3) 𝑞  Φ𝐸 = 𝜖0. This is Gauss’s law. So we can able to derive Gauss’s law from Coulomb’s law. 5 5 MARK - QUESTIONS AND ANSWERS : 1. Calculate the electric field due to a dipole on its axial line.  Consider an electric dipole AB along X - axis. Let ‘C’ be the point at a distance ‘r’ from the mid point ‘O’ on its axial line. 1 𝑞  Electric field at C due to +q, 𝐸+ = 4𝜋𝜖 𝑝 0 (𝑟−𝑎 )2 1 𝑞  Electric field at C due to −𝒒, 𝐸− = − 4𝜋𝜖 2 𝑝 0 (𝑟+𝑎 )  The total electric field at ‘C’ due to dipole is 𝐸𝑡𝑜𝑡 = 𝐸+ + 𝐸− 1 𝑞 1 𝑞 = 4𝜋𝜖 2 𝑝 - 𝑝 0 (𝑟−𝑎 ) 4𝜋𝜖 0 (𝑟+𝑎 )2 𝑞 1 1 = − (𝑟+𝑎 )2 𝑝 4𝜋𝜖 0 (𝑟−𝑎 )2 𝑞 (𝑟+𝑎 )2 −(𝑟−𝑎 )2 = 𝑝 4𝜋𝜖 0 (𝑟 2 −𝑎 2 )2 𝑞 4𝑟𝑎 = 4𝜋𝜖 𝑝 0 𝑟 2 −𝑎 2 2  If r >> a, 𝑟 2 − 𝑎2 2 ≈ 𝑟4 2𝑝 𝐸𝑡𝑜𝑡 = 4𝜋𝜖 3 ( 𝑝 = 2𝑎𝑞𝑝) 0𝑟  The direction of 𝐸 is in the direction of 𝑝. 2. Calculate the electric field due to a dipole on its equatorial line.  Consider an electric dipole AB along X - axis. Let ‘C’ be the point at a distance ‘r’ from the mid point ‘O’ on its equatorial plane. 1 𝑞  The magnitude of electric field at C due to +𝒒, 𝐸+ = 4𝜋𝜖 - - - - - -(1) 0 (𝑟 2 +𝑎 2 ) 1 𝑞  The magnitude of electric field at C due to -𝒒 , 𝐸− = 4𝜋𝜖 - - - - - -(2) 0 (𝑟 2 +𝑎 2 )  Moreover, i) 𝐸+ = 𝐸− ii) The perpendicular components 𝐸+ sin 𝜃 , 𝐸− sin 𝜃 are equal and oppositely directed.So they cancel each other. iii) The parallel components 𝐸+ cos 𝜃 , 𝐸− cos 𝜃 are equal and same in direction.So they are added together.  𝐸𝑡𝑜𝑡 = − 2 𝐸+ cos 𝜃 𝑝 - - - - - -(3) 𝑎 Here, cos 𝜃 = - - - - - -(4) (𝑟 2 + 𝑎 2 )1/2 Substitute equation (1) ,(4) in (3) 1 𝑞 𝑎 𝐸𝑡𝑜𝑡 = − 2 𝑝 4𝜋𝜖 0 (𝑟 2 +𝑎 2 ) (𝑟 2 + 𝑎 2 )1/2 2𝑞 𝑎 =− 𝑝 4𝜋𝜖 0 (𝑟 2 + 𝑎 2 )3/2  If r >> a then 𝑝 𝐸𝑡𝑜𝑡 = − 4𝜋𝜖 3 𝑝 = 2𝑞𝑎𝑝. The direction of 𝐸𝑡𝑜𝑡 is opposite to the direction of 𝑝. 0𝑟 6 3. Derive an expression for electrostatic potential due to electric dipole.  Consider an electric dipole AB along X - axis. Let ‘P’ be the point at a distance ‘r’ from its midpoint ‘O’. 1 𝑞  Electric potential at P due to +q , 𝑉1 = - - - - - -(1) 4𝜋𝜖 0 𝑟1 −1 𝑞  Electric potential at P due to −q , 𝑉2 = - - - - - (2) 4𝜋𝜖 0 𝑟2 1 𝑞 1 𝑞  Electric potential at ‘P’ due to dipole is 𝑉 = − 4𝜋𝜖 0 𝑟1 4𝜋𝜖 0 𝑟2 q 1 1 𝑉= −r - - - - - -(3) 4πϵ0 r1 2 1 1 𝑎 cos 𝜃 1 1 𝑎 cos 𝜃  = 1+ and = 1− - - - - - -(4) 𝑟1 𝑟 𝑟 𝑟2 𝑟 𝑟 Substitute equation (4) in (3) q 1 𝑎 cos 𝜃 1 𝑎 cos 𝜃 𝑉= 1+ − 1− 4πϵ0 𝑟 𝑟 𝑟 𝑟 q 2𝑎𝑐𝑜𝑠𝜃 𝑉= 4πϵ0 𝑟2 P cos θ 𝟏 𝑷.𝒓 𝑉= ( P = 2qa ) (or) 𝑽 = 4πϵ0 r 2 𝟒𝝅𝝐𝟎 𝒓𝟐 P  =00 𝑉= 4πϵ0 r 2 Point P lies on the axial line of electric dipole, near +q − P  =1800 𝑉= 4πϵ0 r 2 Point P lies on the axial line of electric dipole, near - q  =900 𝑉=0 Point P lies on the equatorial line of electric dipole. 4. Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel. Capacitors in series Capacitors in parallel Consider three capacitors of capacitance C1, C2 and Consider three capacitors of capacitance C1, C2 C3 connected in series.CS - the equivalent capacitance and C3 connected in parallel.CP - the equivalent of capacitor in series connection. capacitance of capacitor in parallel connection Each capacitor has same amount of charge (Q). Each capacitor has same potential difference But potential difference across each capacitor will be (V). different. But charges on each capacitor will be different. V = V1 + V2 +V3 Q = Q1 + Q2 +Q3 𝑸 𝑄 𝑄 𝑄 Q = CPV 𝑉 = 𝑪 ; 𝑉1 = 𝐶 , 𝑉2 = 𝐶 ,𝑉3 = 𝐶 Q1=C1V ;Q2 = C2V;Q3 = C3V 𝑺 1 2 3 𝑄 𝑄 𝑄 𝑄 CPV = C1V +C2V + C3V = + + 𝐶𝑆 𝐶 1 𝐶2 𝐶3 1 1 1 1 CP = C1 +C2 + C3 = + + 𝐶𝑆 𝐶 1 𝐶2 𝐶3 The inverse of the equivalent capacitance of The equivalent capacitance of capacitors capacitors connected in series is equal to the sum connected in parallel is equal to the sum of of the inverses of each capacitance. the individual capacitances. 7 5.Obtain an expression for electric field due to an infinitely long charged wire. 𝑄𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 i) Consider an infinitely long straight wire of uniform linear charge density . 𝑖. 𝑒 𝜆 = 𝐿 So, total charge enclosed by the closed surface 𝑄𝑒𝑛𝑐𝑙 = 𝜆𝐿 - - - - ( 1 ) ii) Electric field : Let ‘E’ be the electric field at the point ‘P’ which is at a distance ‘r’ from the wire iii) Gaussian surface :A cylinder of length ‘L’ and radius ‘r’. iv) The electric flux for the curved surface: 𝜙𝐸 = 𝐸 𝑑𝐴 cos 𝜃 = E 2𝜋𝑟𝐿 [∵ 𝜃 = 0] Curved surface v) The electric flux for top and bottom surfaces: 𝜙𝐸 = 0 ( 𝐸 is perpendicular to 𝐴;. 𝑠𝑜 𝐸. 𝑑𝐴 = 0 ) vi) Then the total electric flux 𝜙𝐸 = E(2rL) - - - - - ( 2 ) 𝑄𝑒𝑛𝑐𝑙 vii) According to Gauss law, 𝜙𝐸 = - - - - -(3) ∈0 Substitute equation (1) and (2) in (3) 𝜆𝐿 𝐸 2𝜋𝑟𝐿 = ∈0 𝜆 1 𝜆 E = (OR) 𝐸 = 2𝜋𝜖 𝑟 2𝜋𝜖 0 𝑟 0 𝑟  If 𝜆> 0 , then the direction of 𝐸 is perpendicular to wire and pointing outward.  If 𝜆< 0 , then the direction of 𝐸 is perpendicular to wire and pointing inward. 6. Explain in detail the construction and working of Van de Graff generator. Principle : Electrostatic induction and Action at points Construction : i) ‘A’ is a hollow spherical conductor fixed on the insulating stand. ii) ‘B’and ‘C’ are pulleys and they are connected by a belt made up of silk. iii) ‘D’and ‘E’ are metallic comb shaped conductors fixed near the pulleys. iv) The comb ‘D’ is maintained at a positive potential of 10 4𝑉 by a power supply. v) The upper comb ‘E’ is connected to the inner side of the hollow metal sphere. Working of comb „D‟: i) Due to the high electric field near comb ‘D’, air gets ionized. ii) The positive charges are pushed towards the belt and negative charges are attracted towards the comb ‘D’. iii) The positive charges stick to the belt and reach comb ‘E’. Working of comb „E‟: i) Due to electrostatic induction ,the comb ‘E’ get negative charges and the sphere gets positive charges ii) Due to action at points at ‘E’ ,descending belt has no charge. 8 Charge leakage: i) Beyond the maximum potential difference of 107𝑉 of the sphere, the charges start leaking to the surroundings due to ionization of air. ii) It is prevented by enclosing the machine in a gas filled steel chamber at very high pressure. Application:  The high voltage (107𝑉) produced in the Van de Graff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations. 7. Obtain an expression for electric field due to an charged infinite plane sheet. 𝑄𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑  Consider an infinite plane sheet of uniform surface charge density ‘𝜎 ’ 𝑖. 𝑒  = 𝐴 So, total charge enclosed by the plane sheet 𝑄𝑒𝑛𝑐𝑙 = 𝜎𝐴 - - - - - - (1)  Electric field : Let ‘E’ be the electric field at ‘P’ which is at a distance ‘r’ from the sheet.  Gaussian surface : a cylindrical of length ‘2r’ and area of cross section ‘A’  The electric flux through the curved surface, 𝜙1 = 𝐸 𝑑𝐴 cos 𝜃 = 0 [∵ 𝜃 = 90] Curved surface  The electric flux through end surface P and P ’ 𝜙2 = 𝑝 𝐸 𝑑𝐴 + 𝑝′ 𝐸𝑑𝐴 [∵ 𝜃 = 0] 𝜙2 = 𝐸𝐴 + 𝐸𝐴 𝜙2 = 2𝐸𝐴 - - - - - - (2)  Total electric flux 𝜙𝐸 = 0 + 2𝐸𝐴 = 2𝐸𝐴 𝑄𝑒𝑛𝑐𝑙  According to Gauss law, 𝜙𝐸 = - - - - -(3) ∈0 Substitute equation (1) and (2) in (3) 𝜎𝐴 2EA = ∈0 𝜎 E = 2𝜖 0 8. Explain in detail the effect of introducing a dielectric medium between the plates of a parallel plate capacitor, when the capacitor is disconnected from the battery.  A parallel plate capacitor is charged by a battery of voltage ‘V o’. Qo - is the charge stored in the plates. Eo - is the electric field between the plates. 𝑄0  Capacitance of the capacitor without dielectric medium, 𝐶𝑜 =. 𝑉0  The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. Qo - is the charge stored in the plates. E - is the electric field between the plates. V - is the potential difference between the plates. ϵr − dielectric constant of the dielectric medium. 9   Effect of dielectric medium between the plates Quantity Value Effect of dielectric When ϵr > 1 E0 𝐸 < 𝐸0 Decreased Electric field 𝐸= ϵr V0 Potential difference 𝑉 = 𝑉 < 𝑉0 Decreased ϵr Capacitance 𝐶 = ϵr 𝐶0 𝐶 > 𝐶0 Increased U0 Energy 𝑈= 𝑈 < 𝑈0 Decreased ϵr 9. Obtain an expression for electric field due to an uniformly charged spherical shell. At a point on the surface of the At a point outside the shell At a point inside the shell shell Gaussian surface : Gaussian surface : Gaussian surface : Sphere with radius r Sphere with radius r Sphere with radius r R -radius of spherical shell R -radius of spherical shell R-radius of spherical shell Here r > R Here r = R Here r < R Substitute r =R According to Gauss’s law, According to Gauss’s 𝑄𝑒𝑛𝑐𝑙 law, 𝐸𝑑𝐴 cos 𝜃 = ∈0 𝑄𝑒𝑛𝑐𝑙 𝐸𝑑𝐴 cos 𝜃 = ∈0 Substitute Substitute 𝐸𝑑𝐴 cos 𝜃 = 𝐸 4𝜋𝑟 2 𝐸𝑑𝐴 cos 𝜃 = 𝐸 4𝜋𝑟 2 𝑄𝑒𝑛𝑐𝑙 = 𝑄 𝑄𝑒𝑛𝑐𝑙 = 0 𝑄 0 𝐸(4𝜋𝑟 2 ) = 𝑄 𝐸(4𝜋𝑟 2 ) = ∈0 𝐸= ∈0 𝑄 4𝜋 ∈0 𝑅2 𝐸= 𝐸=0 4𝜋 ∈0 𝑟 2 OTHER IMPORTANT QUESTION ( 3 - MARK ): 1. Define - 1 coulomb. S.Ponnaiah M.Sc., M.Phil., B.Ed., 2. Define - 1 volt. Govt. Hr. Sec. School , 3. Define - 1 farad. Uranganpatti , Madurai - 625 109. 4. Balloon sticks to the wall. Why? 5. What is the role of capacitor in a ceiling fan. 10  2. CURRENT ELECTRICITY 2 MARK -QUESTIONS AND ANSWERS: 1. Electric current is a scalar quantity why?  𝐼 = 𝐽. 𝐴  Even though current has particular direction and magnitude they will not obey vector laws. So current is a scalar quantity. 2. Distinguish between drift velocity and mobility. S.N Drift Velocity Mobility The average velocity acquired by the The magnitude of the drift velocity per unit 1 electrons inside the conductor when it is electric field. subjected to an electric field. 2 Its unit is 𝒎 𝒔−𝟏. Its unit is 𝒎𝟐 𝒗−𝟏 𝒔−𝟏. 3. Define current density and give its unit. 𝐼  The current density is defined as the current per unit area of cross section of the conductor. J = 𝐴.  Its unit is A 𝑚−2. 4.Give the microscopic form of ohm’s law.  Current density is directly proportional to the applied electric field.  J =𝝈 𝑬. Here J - current density. 𝜍- Conductivity 𝐸. - Electric field. 5. Give the macroscopic form of ohm’s law.  The macroscopic form of ohm’s law is V=IR.  Here ‘V’ - Potential difference, ‘I’ - Current and ‘R’ - Resistance. 6.What are ohmic and non-ohmic materials? S.N Ohmic materials Non-ohmic materials 1 V-I graph is a straight line V-I graph is non-linear 2 obey Ohm’s law Doesn’t obey Ohm’s law 3 They have constant resistance They do not have constant resistance 7.Define electrical resistivity and give its unit.  Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.  Its unit is Ω m (ohm meter). 8. Define temperature co-efficient of resistivity.  It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at To.  Its unit is per ℃. 11 9. What is known as superconductivity?  The resistance of certain material becomes zero below certain temperature.  The materials which exhibit this property are called superconductors.  The property of conducting current with zero resistance is called superconductivity. 10. What is electric energy and electric power ? Electric energy  Work has to be done by a cell to move the charge from one end to the other end of the conductor and this work done is called electric energy.  Its SI unit is joule (J). Its practical unit is kilowatt hour (kWh). Moreover 1 kWh=3.6× 106 J. Electric power  The rate at which the electrical potential energy is delivered is called electric power.  It SI unit is watt(W). Its practical unit is horse power(H.P). Moreover 1 H.P = 746W. 11.Derive the expression for power P=VI in electrical circuit.  The equation for electrical potential energy dU=V dQ 𝑑𝑈 𝑑𝑄  The rate at which the electrical potential energy is delivered is the electrical power. P = =V 𝑑𝑡 𝑑𝑡 𝑑𝑄  Since the electric current I = , electrical power P= VI. 𝑑𝑡 12. Write down the various forms of expression for power in electrical circuit.  Electrical power equation is P = VI.  According to ohm’s law V = IR. So electrical power P = (IR) I = 𝐼 2 R. 𝑉 𝑉  According to ohms law I = 𝑅 , so electrical power P= V 𝑅 = 𝑉 2 / R. 13. State Kirchhoff’s first rule (current rule or junction rule).  It states that the algebraic sum of the currents at any junction of a circuit is zero.(i.e) Σ𝐼 = 0. 14. State Kirchhoff’s second rule (voltage rule or loop rule).  It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. (i.e) Σ 𝑰 𝑹= Σ . 15. State the principle of potentiometer.  If I- current, r – resistance per unit length of the wire of potentiometer, 𝑙 – balancing length, then emf of the cell 𝜀 = 𝐼𝑟𝑙.  since I and r constant 𝜀 α 𝑙  The emf of the cell is directly proportional to the balancing length. 16. Define internal resistance of a battery.  The resistance offered by the electrolyte to the flow of charges within the battery is called internal resistance (r).  A freshly prepared cell has low internal resistance and it increases with ageing. 12 17. State Joule’s heating law.  Heat librated by Joule’s heating effect, H=𝑰𝟐 R t  The heat developed in an electrical circuit due to the flow of current varies directly to i) The square of the current (𝐻 𝛼 𝐼 2 ) ii) The resistance of the circuit (𝐻 𝛼 R) iii) The time of flow (𝐻 𝛼 t) 18.What is Seeback effect?  In a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures an emf (potential difference) is developed. This phenomenon is called Seeback effect. 19. What is Thomson effect?  If two points in a conductor are at different temperatures the density of electrons at these points will differ and as a result the potential difference is created between these two points.  Hence heat is evolved or absorbed throughout the conductor. This is called Thomson effect. 20. What is Peltier effect?  When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect. 21. State the applications of seeback effect.  Seeback effect is used in thermo electric generators. These generators are used in power plants to convert waste heat into electricity.  It is used in automobiles as automotive thermoelectric generators for increasing fuel efficiency.  It is used in thermocouples and thermopiles to measure the temperature difference between the two objects. 5 - MARK QUESTIONS AND ANSWERS : 1. Describe the microscopic model of current and obtain general form of ohm’s law.  Number of electrons per unit volume in a conductor = n  Cross sectional area of a conductor = A  Drift Velocity of an electron = 𝑣𝑑  Time taken to travel dx distance = dt.  The number of electrons available in the volume (Adx) = n A 𝑣𝑑 dt 13   charge of an electron = e.  Total charge in the volume element d𝑄= ( nA 𝑣𝑑 dt)e 𝑑𝑄  current I = 𝑑𝑡 I= ( nA 𝑣𝑑 dt)e / dt I = nA𝒗𝒅 e. 𝑰  current density 𝑱 = 𝑨 𝒏𝑨𝑽𝒅 𝒆 𝑱= = 𝒏𝒆𝒗𝒅 𝑨 𝑒𝜏 Substitute 𝑉𝑑 = − 𝑚 𝐸 𝒆𝝉 𝑱= -ne 𝑬 𝒎 𝒆𝟐𝝉 𝑱= -n 𝑬 or 𝒎 𝒏 𝒆𝟐 𝝉 𝑱= -𝝈𝑬 , Here 𝝈 = is conductivity. 𝒎  But conventionally we take the direction of current density as the direction of electric field. So, This is called macroscopic form of ohm’s law. 2.Obtain the macroscopic form of ohm’s law from its microscopic form.  Microscopic form of ohm’s law J=𝝈 E 𝑰 𝑽  Substitute current density J=𝑨 and electric field E = 𝒍 𝑰 𝑽  =𝝈 or 𝑨 𝒍 𝑙  𝑉=𝐼 , 𝜍𝐴 𝑙 Let 𝜍𝐴 is the resistance -R of a conductor then  V= IR. This is the macroscopic form of ohm’s law. 14 3.Explain the equivalent resistance of a series and parallel resistor network. S.no Series circuit Parallel circuit 1 Let R1, R2,R3 be the resistance of Let R1, R2, R3 be resistance of three 2 three resistors connected in series resistors connected in parallel The current flowing through all the The potential difference between all 3 resistors are equal the resistors are equal 4 Potential difference varies current varies 5 V = 𝑉1 +𝑉2 +𝑉3 𝐼 = 𝐼1 + 𝐼2 + 𝐼3 V = I𝑅𝑠 I = V/ RP 6 I1 = V / R1 ; I2 = V / R2 ; I3 = V / R3 𝑉1 =I 𝑅1 ; 𝑉=I 𝑅2 ;𝑉3 =I 𝑅3 V V V V 7 I 𝑅𝑆 = I 𝑅1 + I 𝑅2 + I 𝑅3 = + + RP R1 R 2 R 3 1 1 1 1 8 𝑅𝑆 =𝑅1 +𝑅2 +𝑅3 = + + RP R1 R 2 R 3 4. Explain the determination of the internal resistance of a cell using voltmeter.  When the electric circuit is open, the reading in voltmeter(v), is equal to the electro motive force (𝜀) V =𝜺 - - - - - - (1)  The external resistance R is included in the circuit. Current I is established in the circuit.  Potential drop across R is V =IR Or I R = V - - - - - - (2)  Due to internal resistance ‘r’ of the cell, the voltmeter reads a value ‘v’, which is less than the emf of cell 𝜺. Ir = ε - V -------- (3) 𝜺−𝐕 (3)  (2) Internal resistance 𝒓= 𝑹. 𝒗 15 5. Explain series and parallel connections in cell. S.no cells in series cells in parallel 1 2 ‘n ‘ Batteries having internal resistance ‘r’ ‘n’ batteries having internal resistance ‘r’ and emf ‘𝜺’ are connected in series and emf ‘𝜺’ are connected in parallel. 3 Total emf = n 𝜺 Total emf = 𝜺 𝑟 4 Total resistance = nr + R Total resistance = 𝑛 + R 𝑛𝜺 𝑛𝜺 5 Current in the circuit I = 𝑛𝑟 +𝑅 Current in the circuit I= 𝑟+𝑛𝑅 𝑛𝜺 𝜀 6 If r 𝑅, 𝐼 = 𝑟 If r>> 𝑅, 𝐼 = 𝑟 6. State and explain Kirchoff’s rules Kirchoff’s first rule  The algebraic sum of the current at any junction of a circuit is zero. (i.e) Σ𝐼 =0.  It is a statement of law of conservation of electric charges.  Current entering the junction is taken as positive and leaving the junction is taken as negative. 𝐼1 + 𝐼2 − 𝐼3 − 𝐼4 − 𝐼5 = 0 Kirchoff’s second rule:  In a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. (i.e) Σ 𝑰 𝑹= Σ   This rule follows from the law of conservation of energy for an isolated system. 16  1 The product of current and resistance is taken as positive when the direction of the current is followed. 2 Suppose if the direction of current is opposite to the direction of the loop,then product of current and voltage across the resistor is negative. 3 The emf is considered positive when proceeding from the negative to the positive terminal. 4 The emf is considered negative when proceeding from the positive to the negative terminal. 7. Obtain the condition for bridge balance in wheatstone’s bridge.  The bridge consists of four resistances P,Q,R and S connected as shown in figure.  A galvanometer ‘G’ is connected between the points B and D.  The battery is connected between the points A and C.  The current through the galvanometer is 𝐼𝐺 and its resistance is G.  Applying Kirchhoff’s current rule to junction B and D respectively. 𝐼1 − 𝐼𝐺 − 𝐼3 = 0 - - - - - - (1) 𝐼2 + 𝐼𝐺 − 𝐼4 = 0 - - - - - - (2)  Applying Kirchhoff’s voltage rule to loop ABDA, 𝐼1 𝑃 + 𝐼𝐺 𝐺 − I2 𝑅 = 0 - - - - - - (3)  Applying Kirchhoff’s voltage rule to loop BCDB. 𝐼3 𝑄 − 𝐼𝐺 𝐺 − I4 𝑆 = 0 - - - - - - (4)  Substitute 𝐼𝐺 = 0 in equation (1) (2) (3) &(4) 𝐼1 = 𝐼3 - - - - - -(5) 𝐼2 = I4 - - - - - -(6) 𝐼1 𝑃 = I2 𝑅 - - - - - -(7) 𝐼3 𝑄 = I4 𝑆 - - - - - -(8) 𝐼1 𝑃 I2 𝑅  ( 7)  (8 ) 𝐼3 𝑄 = I4 𝑆 𝑷 𝑹  Using equations (5 ) and (6 ) = 𝑸 𝑺 17 8.Explain the determination of unknown resistance using metre bridge. Construction:  A uniform wire of manganin AB of one meter length is stretched along a metre scale on a wooden board between two copper strips.  In the gap 𝐺1 , unknown resistance ‘P’ and in the gap ‘𝐺2 ’ standard resistance 𝑄 are connected.  A jockey is connected to the terminal ‘E’ on the central strip through a galvanometer (G) and a high resistance (HR).  A Lechlanche cell and a key are connected between the ends of the bridge wire. Working:  The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection.  The resistances corresponding to AJ(𝑙1 ) and JB(𝑙2 ) of the bridge wire form the resistances ‘R’ and ‘S’ of the wheatstone’s bridge 𝑃 𝑅 𝑟. 𝐴𝐽 = = 𝑟 − 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑄 𝑆 𝑟.𝐽𝐵 𝑃 𝐴𝐽 𝑙1 = = 𝑄 𝐽𝐵 𝑙2 𝑙  Unknown resistance P= 𝑄 𝑙1 2  The end resistance due to the bridge wire soldered at the ends of the strips can be eliminated if another set of reading is taken with P and Q interchanged and average value of ‘P’ is found. 𝑃𝜋 𝑟 2  Specific resistance of the material of the wire 𝜌 = 𝑙 9.Explain the principle of a potentiometer  Primary circuit : The battery, key and the potentiometer wire are connected in series to form primary circuit.  Secondary circuit: The positive terminal of a primary cell of emf 𝜀 is connected to the point C and negative terminal is connected to the jockey through a galvanometer G and a high resistance HR. This forms the secondary circuit.  Contact is made at any point J on the wire.  If the potential difference across (J) is equal to the emf of the cell 𝜀, then no current will flow through the galvanometer and it will show zero deflection.  CJ is the balancing length. 18   potential difference across CJ is equal to 𝐼𝑟𝑙, whereas I – The current flowing through the conductor r – the resistance per unit length of wire.  emf of the cell = potential difference across CJ 𝜀 = 𝐼𝑟𝑙  If I and r constant, 𝜀 𝛼 𝑙  The emf of a cell is directly proportional to the balancing length. 10. How the emf of two cells are compared using potentiometer?  Primary circuit: Potentiometer wire (CD) is connected in series with battery (Bt), key(K) and rheostat (Rh).  Secondary circuit: The end C of potentiometer wire is connected to the terminal M of a DPDT switch and the other terminal N is connected to a jockey through a galvanometer G , a high resistance HR.  The cells whose emf 𝜺𝟏 and 𝜺𝟐 to be compared are connected to the terminals 𝑀1 𝑁1 and 𝑀2 𝑁2 of the DPDT switch.  I – Steady current passing through the potentiometer wire. r – Resistance per unit length of the potentiometer wire.  Procedure 1: i) Initially the cell of emf 𝜺𝟏 is included in the secondary circuit and the balancing length 𝑙1 is found by adjusting jockey for zero deflection. ii) According to the principle of potentiometer , 𝜺𝟏 = 𝐈𝐫𝒍𝟏 - - - - - - (1)  Procedure 2: i) Similarly the cell of emf 𝜺𝟐 is included in the secondary circuit and the balancing length 𝑙2 is found. ii) According to the principle of potentiometer , 𝜺𝟐 = 𝐈𝐫𝒍𝟐 - - - - - - (2) 𝜺𝟏 𝒍𝟏  (1) (2) = - - - - - - (3) 𝜺𝟐 𝒍𝟐 11.Explain the determination of the internal resistance of a cell using potentiometer.  Primary circuit: Potentiometer is connected in series with battery (Bt) and key 𝐾1.  Secondary circuit : The battery whose internal resistance is to be calculated is connected in parallel with resistance box(R) and Key(𝐾2 ). 19   Balancing length 𝑙1 is determined when key 𝐾2 is open. According to principle of potentiometer. 𝜺 𝜶 𝒍𝟏 - - - - - - (1)  When Key 𝐾2 is closed, the balancing length 𝑙2 is determined. 𝜺𝑹 𝜶 𝒍𝟐 - - - - - - (2) 𝑹+𝒓 𝒍𝟏 − 𝒍𝟐  (1) ÷(2) 𝑟= R - - - - - (3) 𝒍𝟐 Substituting R, 𝒍𝟏 , 𝒍𝟐 in equation (3) the internal resistance of a cell can be calculated. Annexure 2 - MARK QUESTIONS AND ANSWERS: 1.Why nichrome is used as heating element in electric heaters? i) It has a high specific resistance. ii) It has high melting point. iii) It can be heated to very high temperature without Oxidation. 2. Differentiate Joule heating effect and peltier effect. S.no Joule heating effect Peltier effect 1 It is irreversible It is reversible 2 Heat energy developed is directly Rate of heat energy developed is proportional to the square of directly proportional to the current. current ( H 𝛼 𝐼 2 ) ( 𝑑𝐻 𝛼𝐼) 𝑑𝑡 3 It does not depend on the direction It depends on the direction of of current. current. 3. Define electric current and give its unit.  The electric current in a conductor is defined as the rate of flow of charges through a given cross sectional area.  The SI unit of current is ampere (A). 4. What is called mean free time?  The average time between two successive collisions is called the mean free time. E.JANET PAMILA M.Sc.,M.Phil.,B.Ed., Govt.(G) Hr.Sec.School, Y.Othakadai, Madurai - 625 107. 20  3.MAGNETISM AND MAGNETIC EFFECTS OF ELECTRIC CURRENT 2 MARK - QUESTIONS AND ANSWERS : 1. What are the elements of the Earth’s Magnetic field? i) Magnetic declination (D) ii) Magnetic dip or inclination (I) iii) The horizontal component of the Earth’s magnetic field (BH) 2. Define Magnetic Dipole Moment. Give its unit.  The product of its pole strength (𝑞𝑚) and magnetic length.(2𝑙).  The magnitude of magnetic dipole moment 𝒑𝒎 = 𝟐𝒍𝒒𝒎.  SI unit is A m 2. 3. Define Magnetic flux. Give its unit.  Magnetic flux is defined as the number of magnetic field lines crossing any area normally.  𝝓𝑩 = 𝑩. 𝑨  SI unit is weber. Dimension is M L 2 T-2 A-1 4. State Coulomb’s inverse law of magnetism. The force of attraction or repulsion between two magnetic poles is  directly proportional to the product of their pole strengths  inversely proportional to the square of the distance between them. 𝒒𝒎𝑨 𝒒𝒎𝑩  𝑭= 𝒌 𝒓 𝒓𝟐 5. State Tangent law.  When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.  𝑩 = 𝑩𝑯 𝐭𝐚𝐧 𝜽 6. Define Intensity of Magnetisation. Give its unit. 𝒑𝒎  The net magnetic moment per unit volume of the material. 𝑴 =. 𝑽  SI unit is Am-1. It is a vector quantity. 7. Define Magnetic susceptibility.  The ratio of the intensity of magnetisation (M) induced in the material to the magnetising field (H) 𝑴  m = 𝑯 8. What is Meissner effect?  The expulsion of magnetic flux from a dia magnetic material during its transition to the super conducting state. 21 9. Define Curie’s law.  Magnetic susceptibility of paramagnetic materials decreases with increase in temperature. 𝟏  m 𝛼. This relation is called Curie’s law. 𝑻 10. What is curie temperature?  At a particular temperature, ferromagnetic material becomes paramagnetic material.  This temperature is known as Curie temperature. 11. What is Hysteresis?  The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. 12. What are the Soft and Hard Ferromagnetic materials?  Soft Ferromagnetic materials: Area of the Hysteresis loop is small. Example: Soft iron , Mumetal.  Hard Ferromagnetic materials: Area of the Hysteresis loop is Large. Example: Steel, Alnico. 13. State right hand thumb rule. If we hold the current carrying conductor in our right hand,  The thumb points the direction of current flow  The fingers encircling the conductor point the direction of the magnetic field lines produced. 14. State Maxwell’s right hand cork screw rule. If we rotate a right-handed screw using a screw driver,  The direction of current is same as the direction in which screw advances  The direction of rotation of the screw gives the direction of the magnetic field. 15. Define Magnetic dipole moment of current loop.  Magnetic dipole moment (𝑝𝑚 ) of any current loop is equal to the product of the current (I) and area of the loop (A).  𝑝𝑚 = 𝐼𝐴 16. What are the limitations of cyclotron? i) The speed of the ion is limited. ii) Electron cannot be accelerated. iii) Uncharged particles cannot be accelerated. 17. State Fleming’s left hand rule.  Stretch out forefinger, the middle finger and the thumb of the left hand such that they are in three mutually perpendicular directions.  Forefinger - direction of magnetic field Middle finger - direction of the electric current Thumb - direction of the force experienced by the conductor. 22 18. Define current sensitivity of a galvanometer. 𝜽 𝑵𝑩𝑨  The deflection produced by per unit current flowing through galvanometer. (i.e) = 𝑰 𝑲 19. How the current sensitivity of a galvanometer can be increased?  By increasing the number of turns (N )  By increasing the magnetic induction (B )  By increasing the area of the coil (A)  By decreasing the couple per unit twist of the suspension wire. 20. Why Phosphor-Bronze is used as suspension wire in galvanometer?  Phosphor-Bronze wire has very small couple per unit twist. 21.Define Voltage sensitivity of the galvanometer. 𝜽 𝑵𝑩𝑨  Deflection produced per unit voltage applied across the galvanometer. (i.e) = 𝑽 𝑲𝑹𝒈 22. Define magnetic declination and inclination.  The angle between magnetic meridian at a point and geographical meridian is called the declination.  The angle subtended by the Earth’s total magnetic field 𝐵 with the horizontal direction in the magnetic meridian is called dip or magnetic inclination at that point. 23. Explain the concept of velocity selector.  By the proper choice of electric and magnetic fields, the particle with particular speed can be selected. Such an arrangement of field is called a velocity selector.  For a given magnitude of Electric field 𝐸 and magnetic field 𝐵 the forces act only on the E particle moving with particular speed v0 = B. 24. Define - 1 ampere.  One ampere is defined as that constant current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one metre apart in vacuum causes each conductor to experience a force of 2 × 10−7 newton per metre length of the conductor. 3 - MARK QUESTIONS AND ANSWERS : 1. Give properties of magnetic field lines. i) Magnetic field lines are continuous closed curves. ii) The direction of magnetic field lines is from North pole to South pole outside the magnet and South pole to North pole inside the magnet. iii) The direction of magnetic field at any point on the curve is known by drawing tangent to the magnetic field lines at that point. iv) Magnetic field lines never intersect each other. v) The magnetic field is strong where magnetic field lines are crowded and weak where magnetic field lines are separated. 23 2. Calculate the torque acting on a bar magnet in uniform magnetic field. i) Consider a magnet of length 2l of pole strength qm kept in a uniform magnetic field. ii) The force experienced by north pole , 𝐹𝑁 = 𝑞𝑚 𝐵 The force experienced by south pole, 𝐹𝑆 = −𝑞𝑚 𝐵 iii) The net force acting on the dipole becomes zero. iv) The moment of force or torque , 𝜏 = 𝑞𝑚 𝐵 × 2𝑙𝑠𝑖𝑛 𝜃 𝜏 = 𝑝𝑚 𝐵 sin 𝜃 ( 𝑝𝑚 = 2𝑙𝑞𝑚 ) v) In vector form 𝜏 = 𝑝𝑚 × 𝐵. 3. List the properties of Diamagnetic materials. i) Magnetic susceptibility is negative. ii) Relative permeability is slightly less than unity. iii) The magnetic field lines are repelled or expelled by it when placed in a magnetic field. iv) Susceptibility is temperature independent v) Examples: Bismuth, Copper and Water. 4. List the properties of paramagnetic materials. i) Magnetic susceptibility is positive and small. ii) Relative permeability is greater than unity. iii) The magnetic field lines are attracted into the paramagnetic materials when placed in a magnetic field. iv) Susceptibility is inversely proportional to temperature. v) Example: Aluminium, Platinum, Chromium and Oxygen. 5. List the properties of Ferro magnetic materials. i) Magnetic susceptibility is positive and large. ii) Relative permeability is very large. iii) The magnetic field lines are strongly attracted into the ferromagnetic materials when placed in a magnetic field. iv) Susceptibility is inversely proportional to temperature. v) Example: Iron, Cobalt and Nikel. 24 6. State and explain Biot savart law. According to Biot savart law, the magnitude of magnetic field 𝑑𝑩 is i) directly as the strength of the current ( I ) ii) directly as the magnitude of the length of the element ( 𝑑𝑙 ) iii) directly as the sine of the angle (θ) between 𝑑𝑙 and 𝑟 iv) inversely as the square of the distance r between the point P and length element 𝑑𝑙. 𝜇 𝐼𝑑𝑙 sin 𝜃 𝑖. 𝑒 𝑑𝐵 = 4𝜋0 𝑟2 𝝁 𝑰 𝒅𝒍×𝒓 v) In a vector form , 𝑑𝑩 = 𝟒𝝅𝟎 𝒓𝟐 7. How is a galvanometer converted into an ammeter? i) A galvanometer is converted into an ammeter by connecting a low resistance in parallel with it. ii) ( I - IG ) - current flowing through the shunt resistance. S - shunt resistance. IG - current flows through the galvanometer. RG - galvanometer resistance. iii) Vgalvanometer = Vshunt (𝑰 − 𝑰𝑮 ) 𝑺 = 𝑰𝑮 𝑹𝑮 𝑰𝑮 𝑹𝑮 𝑺= (𝑰−𝑰𝑮 ) 𝟏 𝟏 𝟏 𝑅𝑎 − Resistance of Ammeter =𝑹 +𝑺 𝑹𝒂 𝑮 iv) An ammeter is a low resistance instrument and it is always connected in series to the circuit. v) An ideal ammeter has zero resistance. 8. How is galvanometer converted into a voltmeter ? i) A galvanometer is converted into a voltmeter by connecting high resistance (Rh) in series with it. ii) V - Voltage to be measured. IG - current flowing through the galvanometer. ( RG + Rh ) - Total resistance. iii) 𝑽 = 𝑰𝑮 𝑹𝒉 + 𝑹𝑮 𝑽 𝑹𝒉 = − 𝑹𝑮 𝑰𝑮  Total resistance of the voltmeter Rv = ( RG + Rh ) iv) Voltmeter is a high resistance instrument and it is always connected in parallel with the circuit. v) An ideal voltmeter has infinite resistance. 25 5 - MARK QUESTIONS AND ANSWERS : 1.Calculate the magnetic field at a point on the axial line of a bar magnet.  Let N be the north pole and S be the south pole of the bar magnet, each of pole strength 𝑞𝑚 and are separated by a distance of 2l. C be the point on the axial line at a distance r from the centre of magnet. 𝝁𝟎 𝒒𝒎  The magnetic field at C due to the north pole, 𝑩𝑵 = 𝒊 𝟒𝝅 𝒓−𝒍 𝟐 𝝁𝟎 𝒒𝒎  The magnetic field at C due to the south pole, 𝑩𝑺 = − 𝒊 𝟒𝝅 𝒓+𝒍 𝟐  The net magnetic field due to the magnetic dipole at point C , 𝑩 = 𝑩𝑵 + 𝑩𝑺 𝝁𝟎 𝒒𝒎 𝝁𝟎 𝒒𝒎 𝑩= 𝒊 − 𝒊 𝟒𝝅 𝒓−𝒍 𝟐 𝟒𝝅 𝒓+𝒍 𝟐 𝝁𝟎 𝒒𝒎 𝟒𝒓𝒍 𝑩= 𝒊 𝟒𝝅 (𝒓𝟐 −𝒍𝟐 )𝟐 𝝁𝟎 𝟐 𝑷𝒎  If 𝑟≫𝑙 , 𝑩 = (∵ 𝑷𝒎 = 𝒒𝒎 × 𝟐𝒍 𝒂𝒏𝒅 𝑷𝒎 𝒊 = 𝑷𝒎 ) 𝟒𝝅 𝒓𝟑 2. Obtain the magnetic field at a point on the equatorial line of a bar magnet.  Let N be the north pole and S be the south pole of the bar magnet, each of pole strength 𝑞𝑚 and are separated by a distance of 2l. C be the point on the axial line at a distance r from the centre of magnet.  Each pole of the bar magnet is in equal distance from a point C. Therefore magnitude of magnetic field due to each pole of the bar magnet are equal. 𝝁𝟎 𝒒𝒎  𝑩𝑵 = 𝑩𝑺 = - - - - - - (1) 𝟒𝝅 (𝒓 +𝒍𝟐 ) 𝟐  Vertical component cancel each other. The horizontal components are add up.  Total magnetic field at C , 𝑩 = 𝑩𝑵 + 𝑩𝑺 = − 𝟐 𝑩𝑵 𝐜𝐨𝐬 𝜽 𝒊 - - - - - - (2) ( ∵ 𝑩𝑵 = 𝑩𝑺 ) 𝝁𝟎 𝒒𝒎 Sub (1) in (2) 𝑩 = −𝟐 𝐜𝐨𝐬 𝜽 𝒊 - - - - - - (3) 𝟒𝝅 (𝒓𝟐 +𝒍𝟐 ) 𝒍  From the right angle triangle NOC , 𝐜𝐨𝐬 𝜽 = 𝟏 - - - - - - (4) (𝒓𝟐+𝒍𝟐 ) 𝟐 𝝁 𝒒𝒎 × (𝟐𝒍) Sub (4) in (3) 𝑩 = − 𝟒𝝅𝟎 𝟑 𝒊 (𝒓𝟐 +𝒍𝟐 ) 𝟐 𝝁 𝑷𝒎  If 𝑟≫𝑙 , 𝑩 = − 𝟒𝝅𝟎 ∵ 𝑷𝒎 = 𝒒𝒎 × 𝟐𝒍 𝒂𝒏𝒅 𝑷𝒎 𝒊 = 𝑷𝒎 𝒓𝟑 26 3. Describe the principle, Construction and working of cyclotron. Principle :  When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force. Construction:  𝐷1 and 𝐷2 are two semi-circular metal containers called Dees. The two Dees are separated with a gap.  The source S is placed at the center in the gap between the Dees.  The direction of magnetic field is normal to the plane of the Dees.  Dees are connected to high frequency alternating potential difference. Working  The ion ejected from source S is positively charged is accelerated towards a Dee (𝐷1)which has negative potential at that time.  Since the magnetic field is normal to the plane of the Dees, the ion moves in a circular path.  After one semi-circular path inside Dee-1, the ion reaches the gap between Dees.  At this time, the polarities of the Dees are reversed. So that the ion is now accelerated towards Dee- 2 with a greater velocity. Calculation:  For this circular motion, the centripetal force on the charged particle is provided by Lorentz force. Magnetic force = Centripetal force 𝒎𝒗𝟐 𝒒𝒗𝑩 = 𝒓 𝒎 𝒓= 𝒗 𝒒𝑩 𝒓𝜶𝒗 ( The increase in velocity increases the radius of circular path.)  When the ion reaches near the edge ,it is taken out with help of deflector plate and allowed to hit the target T. Resonance Condition of cyclotron operation: 𝒒𝑩  The frequency of the positive ion circulates in the magnetic field , 𝒇 =. 𝟐𝝅𝒎 𝑞𝐵  It must be equal to the constant frequency of the electrical oscillator , 𝑓𝑜𝑠𝑐 =. 2𝜋𝑚 𝟐𝝅𝒎  Time period of oscillation, 𝑻 = 𝒒𝑩 27 4. Deduce the relation for magnetic field at a point due to an infinitely long straight conductor carrying current.  I be the steady current flowing through an infinitely long straight conductor of YY1.  Point P is at a distance of ‘a’ from the centre of a wire O.  Consider a small line element dl of the wire at a distance l from point O.  The magnetic field at a point P due to current element I dl 𝝁𝟎 𝑰 𝒅𝒍 𝐬𝐢𝐧 𝜽 𝑑𝐵 = 𝟒𝝅 𝒓𝟐 𝒏 - - - - - - - - (1) 𝑎  Substitute dl sin𝜃= 𝑟 𝑑𝜙 and 𝑟 = 𝑐𝑜𝑠𝜙 in eqn (1) 𝜇0 I 𝑑𝐵 = 4𝜋𝑎 cosϕ dϕ 𝑛  The total magnetic field at P due to the conductor YY1 𝜙2 𝜙2 𝜇 0I 𝐵= −𝜙 1 𝑑𝐵 = −𝜙 1 4πa cosϕ dϕ 𝑛 0𝜇 𝐼 𝜙2 0𝜇 I 𝐵 = 4𝜋𝑎 sin 𝜙 −𝜙 1 𝑛 = 4𝜋𝑎 sin 𝜙1 + sin 𝜙2 𝑛  If 𝜙1 = 𝜙2 = 90° then sin 𝜙1 + sin 𝜙2 = 2 𝜇0 I 𝐵= ×2 𝑛 4𝜋𝑎 𝜇 I  𝐵 = 2𝜋𝑎 0 𝑛 5. Obtain a relation for the magnetic field at a point along the axis of a circular coil carrying current.  Consider a current carrying circular loop of radius R , I be the current flowing through the wire.  The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O.  Magnetic field B is computed by taking two diametrically opposite line elements 𝑑𝑙 of the coil each of length at C and D.  Let 𝑟 be the vector joining the current element ( 𝐼𝑑𝑙 ) at C and the point P.  According to Biot-Savart’s law, the magnetic field at P due to the current elements are equal in magnitude. The magnitude of 𝑑𝐵 is 𝝁𝟎 𝑰 𝒅𝒍 𝒅𝑩 = [ ∵ 𝜃 = 900 ] - - - - - - - - (1) 𝟒𝝅 𝒓𝟐  The magnetic field 𝑑𝐵 due to each current element is resolved into two components.  The horizontal components cancel out while the vertical components (dB sinϕ 𝑘) alone contribute to the net magnetic field at the point P. 𝐵= 𝑑𝐵 = 𝑑𝐵 sin ∅ 𝑘 - - - - - - - - ( 2 ) 28 Substitute (1) in (2) 𝜇 0𝐼 𝑑𝑙 𝐵= sin ∅ 𝑘 - - - - - - - - (3) 4𝜋 𝑟2 𝑅 sin ∅ = 1 𝑅 2 +𝑍 2 2 Substitute in eqn (3) 𝑟2 = 𝑅 + 𝑧 2 2 dl = 2πR 𝜇 0𝐼 2𝜋𝑅 𝑅  𝐵= 1 𝑘 4𝜋 𝑅 2 +𝑧 2 𝑅 2 +𝑍 2 2 𝜇0 𝐼 𝑅2 𝐵= 𝑘 2 (𝑅2 + 𝑍 2 )32  If the circular coil contains N turns 𝜇0 𝑁 𝐼 𝑅2 𝐵= 3 𝑘 2 (𝑅2 + 𝑍 2 ) 2 𝜇 0 𝐼𝑁  The magnetic field at the centre of the coil 𝐵 = 𝑘 [ ∵ z = 0 ] 2𝑅 6. Obtain an expression for magnetic field due to the current carrying wire of infinite length using Ampere’s law.  Let I be the current flowing through in a infinite length of current carrying wire.  We construct an Amperian loop in the form of a circular shape at a distance r from the centre of the conductor.  From the Ampere’s law 𝑪 𝑩. 𝒅𝒍 = 𝝁𝟎 𝑰 𝐶 𝐵𝑑𝑙 = 𝜇0 𝐼 (∵ 𝜃 = 0)  If 𝐶 𝑑𝑙 = 2𝜋𝑟 𝐵 2𝜋𝑟 = 𝜇0 𝐼 𝜇0 𝐼 𝐵= 2𝜋𝑟 𝜇 0𝐼  In vector form 𝐵 = 𝑛 2𝜋𝑟 (𝑛 is the unit vector along the tangent to the amperian loop.) 29 7. Obtain an expression for magnetic field due to long current carrying solenoid.  Consider a solenoid of length L having N turns.  Consider a rectangular loop ‘abcd’ inside the solenoid in the magnetic field at any point.  From Ampère’s circuital law 𝑪 𝑩. 𝒅𝒍 = 𝝁𝟎 𝑰𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 - - - - - - - (1)  L.H.S of the equation 𝒃 𝒄 𝒅 𝒂 𝑪 𝑩. 𝒅𝒍 = 𝒂 𝑩. 𝒅𝒍 + 𝒃 𝑩. 𝒅𝒍 + 𝒄 𝑩. 𝒅𝒍 + 𝒅 𝑩. 𝒅𝒍 𝒄 𝒂  𝒃 𝑩. 𝒅𝒍 = 𝒅 𝑩. 𝒅𝒍 =𝟎 (∵ 𝐵 and 𝑑𝑙 are perpendicular ) 𝒅 𝒄 𝑩. 𝒅𝒍 =0 (∵ element present outside the solenoid ) 𝑏  So , 𝐶 𝐵. 𝑑𝑙 = 𝑎 𝐵. 𝑑𝑙 = 𝐵𝐿 − − − − − (2)  Let I be the current passing the solenoid of N turns, 𝑰𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 = 𝑵 𝑰 − − − − − (3)  Substitute (2) and (3) in (1)  𝑩𝑳 = 𝝁𝟎 𝑵𝑰 − − − − − (4) 𝑵  𝑩 = 𝝁𝟎 𝑰 −− −−−(5) 𝑳 𝑁  Let 𝑛= then 𝐿  Magnetic field due to long current carrying solenoid, 𝐵 = 𝜇0 𝑛 𝐼. 8. Obtain an expression for the force on a current carrying conductor placed in a magnetic field.  Consider a wire of length L, with cross- sectional area A placed in a magnetic field. I be the current flowing through the wire.  Consider a small segment of wire of length 𝑑𝑙.  The relation between current I and drift velocity 𝑣𝑑 , 𝐼 = 𝑛𝐴𝑒𝑣𝑑 − − − (1)  Current element in the conductor , 𝐼𝑑𝑙 = − 𝑛𝐴𝑒 𝑣𝑑 𝑑𝑙 − − − − − (2)  Average force experienced by a electron in the wire , 𝑓 = −𝑒 (𝑣𝑑 × 𝐵)  Total number of free electrons in the small element 𝑁 = 𝑛 𝐴 𝑑𝑙.  Lorentz force on the wire of length dl , 𝑑𝐹 = − 𝑛𝐴𝑑𝑙 𝑒( 𝑣𝑑 × 𝐵) − − − − − (3) Sub (2) in (3) 𝑑𝐹 = (𝐼𝑑𝑙 × 𝐵) − − − − − (4)  The force on the wire of length l , 𝐹 = (𝐼𝑙 × 𝐵)  In magnitude, F= 𝑩𝐈𝒍𝐬𝐢𝐧𝜽 − − − − − (5) special case : i) If the conductor is placed along the direction of the magnetic field, θ = 00 then F = 0. ii) If the conductor is placed perpendicular to the magnetic field, θ =90 0 then F= BIl. 30 9. Obtain a force between two long parallel current carrying conductors. Hence define ampere.  Two long straight parallel current carrying conductors A and B separated by a distance r are kept in air. Conductor A Conductor B Current in the conductor I1 Current in the conductor I2 𝜇 𝐼 0 1 𝜇 0 𝐼2 Magnetic filed at r, 𝐵1 = − 2𝜋𝑟 𝑖 Magnetic filed at r, 𝐵2 = 𝑖 2𝜋𝑟 Force on the element dl of conductor B Force on the element dl of conductor A 𝜇0 𝐼1 𝐼2 𝑑𝑙 𝜇0 𝐼1 𝐼2 𝑑𝑙 𝑑𝐹 = 𝐼2 𝑙 × 𝐵1 = − 𝑗 𝑑𝐹 = 𝐼1 𝑙 × 𝐵2 = 𝑗 2𝜋𝑟 2𝜋𝑟 Force per unit length of conductor B due to A Force per unit length of conductor B due to A 𝐹 𝜇0 𝐼1 𝐼2 𝐹 𝜇0 𝐼1 𝐼2 = − 𝑗 = 𝑗 𝑙 2𝜋𝑟 𝑙 2𝜋𝑟  current in the conductors same direction - attractive force.  current in the conductors opposite direction - repulsive force.  One ampere is defined as that constant current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one metre apart in vacuum causes each conductor to experience a force of 2 × 10−7 newton per metre length of conductor. R. JACQULINE ESTHER RANI M.Sc., M.Ed., M.Phil., Ph.D., Govt. Hr. Sec. School, Sedapatti, Madurai - 625 527. 31  4. ELECTROMAGNETICINDUCTION AND ALTERNATING CURRENT 2 -MARK QUESTIONS AND ANSWERS: 1. What is electromagnetic induction or Faraday’s I law?  Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced in the circuit. 2. Write Faraday’s laws of electromagnetic induction?  First law : Whenever magnetic flux linked with a closed circuit changes, an emf is induced in the circuit.  Second law :The magnitude of induced emf in a closed circuit is equal to the time rate of change of 𝑑𝜑 𝐵 magnetic flux linked with the circuit. 𝑖. 𝑒 𝜀 = 𝑑𝑡 3. State Lenz’s law.  The direction of the induced current always opposes the cause responsible for its 𝑑𝜑 𝐵 production. 𝑖. 𝑒 𝜀 = − 𝑑𝑡 4. State Fleming’s right hand rule.  The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions.  The Index finger - The direction of the magnetic field The thumb -The direction of motion of the conductor The middle finger -The direction of the induced current. 5. What are eddy currents? (or) What are Foucault currents? How do they flow in a conductor?  when magnetic flux linked with a sheet or plate changes, electric currents are induced. As these electric currents resemble eddies of water, these are known as Eddy currents (or) Foucault currents.  The induced currents flow in concentric circular paths. 6. Mention the ways of producing induced emf. i) By changing the magnetic field (B) ii) By changing the area (A) of the coil and iii) By changing the relative orientation (θ) of the coil with magnetic field 7. Define Q factor  It is defined as the ratio of voltage across L or C at resonance to the applied voltage. 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝐿 𝑜𝑟 𝐶 𝑎𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒  𝑄 − 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝐴𝑝𝑝𝑙𝑖𝑒𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 8. What are advantages of three phase AC generator? Three phase AC generator has i) Higher power output. ii) Smaller in size. iii) Transmission system is cheaper. 32 9. Define power factor.  Power factor is the ratio of true power of AC circuits to apparent power of it. 𝑇𝑟𝑢𝑒 𝑝𝑜𝑤𝑒𝑟  𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝐴𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑝𝑜𝑤𝑒𝑟 10. Define the efficiency of transformer.  Efficiency of transformer is the ratio of the useful output power to the input power. 𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟  𝜂= × 100% 𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 11. Why capacitor blocks DC?  For a steady current, frequency f = 0. So XC = ∞.  Thus a capacitive circuit offers infinite resistance to the steady current. 12. What are the uses of RLC circuits? i) RLC circuits are used in filter circuits, oscillators, voltage multipliers etc. ii) It is used in tuning circuits of radio and TV systems. 13. Distinguish step-up and step-down transformer S.N Step-up transformer Step-down transformer 1 Increases potential difference. Decreases potential difference. 2 Decreases current. Increases current. 3 Number of turns in secondary is high. Number of turns in secondary is low. voltage transformation ratio ( K) is voltage transformation ratio (K) is 4 greater than 1. lesser than 1. 14. What is wattless current?  The current in an AC circuit is said to be wattless current if the power consumed by it is zero. 15. What do you mean by self-induction?  When an electric current passing through a coil changes , magnetic flux linked with that same coil also changes then an emf is induced in the same coil. 16. What is meant by mutual induction?  When an electric current passing through a coil changes , an emf is induced in the neighbouring coil. 17. Define average value of an alternating current.  The average of all values of current over a positive half-cycle or negative half-cycle. 18. Define RMS value of an alternating current.  The square root of the mean of the squares of all currents over one cycle. 19.What are phasors?  A sinusoidal alternating voltage (or) current can be represented by a vector which rotates about the origin in anti-clockwise direction at a constant angular velocity.  Such a rotating vector is called a phasor 33 20. Define electric resonance.  When the frequency of the applied alternating source is equal to the natural frequency of the RLC circuit, then the circuit is said to be in electrical resonance. 21. What do you mean by resonant frequency?  The frequency at which current in the RLC circuit is maximum is called resonant frequency. 22. Write the principle used in AC generator (Alternator)?  The principle used in AC generator is Electromagnetic induction.  The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf. 23. What are LC oscillations?  Whenever energy is given to a LC circuit, the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations. 24. Define the unit of self - inductance (or) Define one henry.  The inductance of the coil is said to be one henry if a current changing at the rate of 1 A s -1 induces an opposing emf of 1 V in it. 25. An inductor blocks AC but it allows DC. Why? and How?  Inductive reactance of an inductor is directly proportional to the frequency of AC.When AC flows through an inductor produces time varying magnetic field which in turn induces back emf. This back emf opposes any change in the Ac current and hence inductor blocks AC.  The frequency of the DC , f = 0. So XL = 0. Hence there is no self induction and self induced emf (opposing emf). So