# PHY PUC 2 MOVING CHARGE AND MAGNETISM

Bac 2 em no
29 de May de 2023
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### PHY PUC 2 MOVING CHARGE AND MAGNETISM

• 1. 4.Moving Charges and Magnetism Hans – Oersted Experiment: 1. Christian Oersted discovered that the conductor carrying current produces magnetic field around it. This effect is called magnetic effect of electric current. 2. Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a deflection in a nearby magnetic compass needle. Reversing the direction of the current reverses the orientation of the needle . The deflection increases on increasing the current or bringing the needle closer to the wire. Iron filings sprinkled around the wire arrange themselves in concentric circles Oersted concluded that “moving charges or currents produced a magnetic field in the surrounding space. Unlike a stationary charge which only produces electric field.” 3. The direction of magnetic field lines is given by amperes right hand classed rule. ELECTRICITY AND MAGNETISM WERE UNIFIED BY JAMES MAXWELL WHO THEN REALISED THAT LIGHT WAS ELECTROMAGNETIC WAVES. (a) The current emerges out of the plane of the paper, The orientation of the magnetic needles (b)The current moves into the plane of the paper. around current carrying conductor Sign convention:  A current or a field (electric or magnetic) emerging out of the plane of the paper is depicted by a dot ( )  A current or a field going into the plane of the paper is depicted by a cross (⊗) Right Hand Thumb Rule: If the conductor carrying current held in right hand such that thumb indicates the direction of current in the conductor, then direction of remaining fingers gives direction of magnetic field. a b with the wire as the . Page 69
• 2. 4.Moving Charges and Magnetism OBTAIN AN EXPRESSION FOR FORCE ON CHARGE MOVING IN UNIFORM MAGNETIC FIELD. The force acting on charge ‘q’ moving with velocity ‘v’ making an angle θ with uniform magnetic field ‘B’ is 1. Directly proportional to the magnitude of charge(q). 2. Directly proportional to the velocity of charge (v). 3. Directly proportional to magnetic field (B). 4. Directly proportional to sign of angle between ‘v’ & ‘B’ F ∝ qvBsinθ F = qvBsinθ In vector form 𝐅 ⃗ = 𝐪(𝐯 ⃗⃗ × 𝐁 ⃗⃗⃗). “The direction of force is given by right hand palm rule for positive charge and left hand palm rule for negative charge. Stretch all fingers of right/left hand in the direction of charge particle and curl all fingers in the direction of magnetic field except thumb, then thumb represents the direction of force.” Note:  The force acting on charge is maximum when it is moving ⊥er to field direction (θ = 900).  The force on charge is zero(minimum) when charge is moving parallel or antiparallel to magnetic field.( θ = 00 or 1800).  Charge is at rest (v=0) then F = 0.  Uncharged particle (q=0) like neutron then F = 0. What is S.I unit of magnetic field? Ans:Tesla (T), vector quantity, [MT2A-1] Define Tesla: If a charge is 1 coulomb is moving with velocity 1ms-1 perpendicular to magnetic field experiences a force of 1 Newton then the magnetic field is 1 Tesla. CGS unit of magnetic field is Gauss (1T = 10-4G). LORENTZ FORCE: Lorentz force is defined as the combination of the magnetic and electric force on a point charge due to electromagnetic fields. Force due to magnetic field, FB = qvB Force due to electric field, FE = qE 𝑭 ⃗ ⃗⃗ = 𝑭𝑩 + 𝑭𝑬 𝐹 ⃗ = 𝑞𝑣𝐵 + 𝑞𝐸 𝑭 ⃗ ⃗⃗ = 𝒒(𝒗 ⃗ ⃗⃗ × 𝑩 ⃗⃗⃗ + 𝑬 ⃗⃗⃗) vector form of Lorentz force. Page 70
• 3. 4.Moving Charges and Magnetism 𝑭 = 𝑰𝑩𝒍𝒔𝒊𝒏𝜽 Where,  F is the force acting on the particle  q is the electric charge of the particle  v is the velocity  E is the external electric field  B is the magnetic field Applications of Lorentz force: Cyclotrons and other particle accelerators use Lorentz force. OBTAIN AN EXPRESSION FOR FORCE ON CURRENT CARRYING CONDUCTOR PLACED IN MAGNETIC FIELD Consider a conductor of length l, area of cross section ‘A’ carrying a current ‘I’ is placed ⊥er to uniform magnetic field ‘B’. Let ‘n’ be the free electrons in the conductor. Then total no. of free electrons in the conductor is ‘nAl’. The total charge of all electrons is nAlq. The force acting on all the charges is F = nAlq (Vd x B) Vd =drift velocity F = (AnqVd)l B F = IBl Where I = AnqVd (From chapter 3) For any arbitrary angle ‘θ’ Note: When the force on conductor carrying current placed in the uniform magnetic field I) maximum II) minimum I) Fmax: When conductor is perpendicular to magnetic field, θ = 900 II) Fmin: When conductor is parallel or antiparallel to magnetic field. ( θ = 00 or 1800). Solve Numerical number 4.1 from NCERT text book OBTAIN AN EXPRESSION FOR RADIUS OF CIRCULAR PATH OF CHARGE MOVING IN UNIFORM MAGNETIC FIELD. Consider a charge q moving with velocity ‘v’ ⊥er to uniform magnetic field ‘B’ Charge experience force FB= Bqvsinθ FB = Bqv (θ = 900) As a result charge perform circular motion. The centripetal force on charge is FC = 𝒎𝒗𝟐 𝒓 page 71
• 4. 4.Moving Charges and Magnetism For equilibrium ,force on charge due o magnetic field is balanced by centripetal force. Hence, FB= FC Bqv = 𝑚𝑣2 𝑟 Bq = 𝑚𝑣 𝑟 r = 𝒎𝒗 𝑩𝒒 This is expression for radius of circular path. Note: (NEET/CET) Other forms of radius , r = 𝒎𝑽 𝑩𝒒 = 𝑷 𝑩𝒒 = √𝟐𝒎𝑲 𝑩𝒒 = 𝟏 𝑩 √ 𝟐𝒎𝑽 𝒒 Time period(T): The time required to complete one rotation for a charge particle. Time period T = 𝟐𝝅𝒎 𝑩𝒒 Frequency(f): Number of rotation in one second. Frequency f = 𝑩𝒒 𝟐𝝅𝒎 Time period (or frequency) is independent of speed of particle and radius of the orbit and depends only on the field B and the nature HELICAL MOTION/SPRING LIKE MOTION Motion of charged particle in uniform magnetic field at arbitrary angle ‘θ’ Consider a charged particle makes an arbitrary angle ‘θ’ with the magnetic field (i.e. θ ≠ 900 or 00) as shown in figure. The velocity ‘v’ makes angle ‘θ’ with B which is along the x – axis. The velocity vector splits into two components. Page 72
• 5. 4.Moving Charges and Magnetism 1) vcosθ along horizontal ,responsible for forward motion 2) vsinθ along vertical, responsible for vertical circular motion Due to these two component, charge particle move vertical as well as horizontal at the same time. Hence the nature of this motion is spiral like known as Helix. The radius(r) and pitch(P) of the helical motion is given by r = 𝒎𝒗𝒔𝒊𝒏𝜽 𝒒𝑩 P = 𝟐𝝅𝒎𝑽𝒄𝒐𝒔𝜽 𝑩𝒒 Solve Numerical number 4.3 from NCERT text book CYCLOTRON The cyclotron is a machine to accelerate charged particles or ions (like, α- particles, deutrons etc.) to high energies. It was invented by E.O. Lawrence and M.S. Livingston. Principle: Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy i.e kinetic energy. Construction & Working: WORKING: During +ve (positive) half cycle of AC. D1 becomes +ve and D2 negative (- ve) . Charged particles experiences an electrical force and accelerate towards D2 . Once the charge particle reach into hallow dees ,the magnetic field, however, acts on the particle and makes it go round in a circular path inside a dee. Every time the particle moves from one dee to another it is acted upon by the electric field. This ensures that It consist of two D shaped hollow metal disc placed opposite with gap, they are called dees. Dees D1 & D2 are connected to AC oscillator. A uniform magnetic field is applied ⊥er to the plane of dees. A charged particle to be accelerated is produced at the of dees in the gap. Page 73
• 6. 4.Moving Charges and Magnetism Solve Numerical number 4.4 from NCERT text book the particle is always accelerated by the electric field. Each time the acceleration increases the energy of the particle. Frequency of cyclotron is 𝒇𝒄 = 𝟐𝝅𝒎 𝑩𝒒 The kinetic energy of the ions is, 𝑲𝒎𝒂𝒙 = 𝒒𝟐𝑩𝟐𝑹𝟐 𝟐𝒎 What are the applications of cyclotron? 1) The accelerated charged particles are used to study structure of elements. 2) To implant ions into solids to change their properties. 3) To produce radioactive substances in hospitals for treatment. 4) The cyclotron is used to bombard nuclei with energetic particles, so accelerated by it, and study the resulting nuclear reactions. Assignment: "Write note on accelerators in India." Describe the motion of charged particle in combined electric & magnetic field. Consider a charge q moving with velocity 'v' along x – axis and an uniform electric field ‘E’ is applied along y – axis and uniform magnetic field ‘B’ is along z – axis. There for charge is acted by both forces. Then, total force on the charge is zero and the charge will move in the fields undeflected. This happens when, qE = qvB E = vB v = 𝑬 𝑩 [Velocity selector Formula] The particle moving with velocity E/B do not experiences any force and travel straight. Thus the particles of particular velocity can be separated from stream of particles using crossed fields and is called velocity selector. This method was used by J.J. Thomson to calculate e/m (charge ratio to the mass of e - ). It is also used in mass spectrometer. Page 74
• 7. 4.Moving Charges and Magnetism MAGNETIC FIELD DUE TO A CURRENT ELEMENT,BIOT-SAVART LAW I. Directly proportional to current(I) through element. II. Directly proportional to length of element(dl). III. Directly proportional to sinθ. IV. Inversely proportional to square of distance( 𝑟2 ) dB ∝ 𝐼𝑑𝑙𝑠𝑖𝑛𝜃 𝑟2 dB = ( 𝜇0 4𝜋 ) 𝐼𝑑𝑙𝑠𝑖𝑛𝜃 𝑟2 where ( 𝜇0 4𝜋 ) is constant of proportionality µo ---is called permeability of free space(vacuum) and 𝜇0 4𝜋 = 10-7 TmA-1 Biot-Savart law in vector: 𝒅𝑩 ⃗⃗⃗⃗⃗⃗⃗ = ( 𝝁𝟎 𝟒𝝅 ) 𝑰(𝒅𝒍 ⃗⃗⃗⃗⃗×𝒓 ⃗⃗) 𝒓𝟑 ⃗⃗⃗⃗⃗ Similarities and differences between Biot-Savart law and Coulomb’s Law  The current element produces a magnetic field, whereas a point charge produces an electric field.  The magnitude of magnetic field varies as the inverse square of the distance from the current element, as does the electric field due to a point charge  The electric field created by a point charge is radial, but the magnetic field created by a current element is perpendicular to both the length element Consider an element of length dl carrying current ‘I’. Let ‘P’ be a point at distance ‘r’ from According to Biot-Savart law magnetic field 'dB' at ‘P’ due to current element is of current element. Page 75
• 8. 4.Moving Charges and Magnetism The direction of magnetic field is determined with the help of the following simple laws : (1) MAXWELL’S CORK SCREW RULE According to this rule, if we imagine a right handed screw placed along the current carrying linear conductor, be rotated such that the screw moves in the direction of flow of current, then the direction of rotation of the thumb gives the direction of magnetic lines of force. (2) RIGHT HAND THUMB RULE According to this rule if a current carrying conductor is held in the right hand such that the thumb of the hand represents the direction of current flow, then the direction of folding fingers will represent the direction of magnetic lines of force. (3) RIGHT HAND THUMB RULE OF CIRCULAR CURRENTS According to this rule if the direction of current in circular conducting coil is in the direction of folding fingers of right hand, then the direction of magnetic field will be in the direction of stretched thumb. Page 76
• 9. 4.Moving Charges and Magnetism OBTAIN AN EXPRESSION FOR MAGNETIC FIELD A POINT ON AXIS OF CURRENT CARRYING CIRCULAR LOOP USING BIOT-SAVART LAW.(5marks) Consider a circular loop of radius ‘R’ carrying current ‘I’. Let 'P' be a point on axis at of length dl. According to Biot-Savart law Magnetic field at P due to current element is dB = ( 𝜇0 4𝜋 ) 𝐼𝑑𝑙𝑠𝑖𝑛𝜃 𝑟2 from figure, 𝑥2 + 𝑅2 = 𝑟2 & 𝑠𝑖𝑛𝜃 = 1 𝑑𝐵 = ( 𝜇0 4𝜋 ) 𝐼𝑑𝑙 (𝑥2+𝑅2) Resolve dB into component. Vertical component due to diametrically opposite elements get cancelled where as the component along x-axis 'dB cosθ' get added hence total magnetic field due to loop is 𝑑𝐵𝑥=𝛴𝑑𝐵𝑐𝑜𝑠𝜃 But 𝑐𝑜𝑠𝜃 = 𝑅 (𝑥 2 +𝑅2 ) 1 2 ⁄ B = ( 𝜇0 4𝜋 ) 𝐼 (𝑥2+𝑅2) 𝑅 (𝑥2+𝑅2) 1 2 ⁄ 𝛴𝑑𝑙 The summation of elements dl over the loop yields 2ПR, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is B = ( 𝜇0 4𝜋 ) 𝐼𝑅 (𝑥2+𝑅2) 3 2 ⁄ × 2𝜋𝑅 B = ( 𝜇0 4𝜋 ) 2π𝐼𝑅2 (𝑥2+𝑅2) 3 2 ⁄ distance ‘x’ from ‘O’. Consider diametrically opposite element Page 77
• 10. 4.Moving Charges and Magnetism Solve Numerical number 4.7 from NCERT text book 𝐵 = μ0𝐼𝑅2 2(𝑥2+𝑅2) 3 2 ⁄ Note: B = 𝝁𝒐𝑰 𝟐𝑹 2) If There are ‘N’ number of turns in coil then B = 𝝁𝒐𝑰𝑵 𝟐𝑹 Note: Ratio of Bcentre and Baxis , 𝑩𝒄𝒆𝒏𝒕𝒓𝒆 𝑩𝒂𝒙𝒊𝒔 = (𝟏 + 𝒙𝟐 𝑹𝟐 ) 𝟑 𝟐 ⁄ AMPERES CIRCUTAL LAW State and explains Amperes circutal law. Statement: Line integral of 'Bdl' for a closed surface having current through it is ′μo′ times the total current(I) through the surface. ∮ 𝐵𝑑𝑙 = 𝜇𝑜𝐼 1) Using Amperes circutal law obtain an expression for magnetic field due to straight infinite length wire carrying current.(3marks) Consider an infinitely long straight wire carrying current ‘I’. Let ‘P’ be a point at distance r from wire. Imagine a circle of amperian loop. Amperian loop has current ‘I’ through it by amperes circutal law the magnetic field at distance ‘r’ from conductor is ∮ 𝐵𝑑𝑙 = 𝜇𝑜𝐼 B. 2πr = 𝜇𝑜𝐼 B = 𝜇𝑜𝐼 2𝜋𝑟 radius ‘r’ having on wire as an 1) Magnetic field at of circular current loop i.e x=0, Page 78
• 11. 4.Moving Charges and Magnetism THE SOLENOID AND THE TOROID The solenoid and the toroid are two pieces of equipment which generate magnetic fields. The television uses the solenoid to generate magnetic fields needed. * What is solenoid: Solenoid is a helical cell of N turns of insulated wire wound closely. * What is an ideal Solenoid: If length of solenoid is very much longer than its radius, then it is called ideal solenoid (l >>>5r). * How is magnetic field inside given solenoid made strong: It is done by inserting laminated Core inside the solenoid. Obtain expression magnetic field due to solenoid carrying current.(3marks) Let n be number of turns per unit length of solenoid carrying ‘I’ for ideal solenoid (large length). The field outside the solenoid approaches zero. We shall assume that the field outside is zero. The field inside becomes everywhere parallel to the axis. To find magnetic field inside solenoid imagine a rectangular amperian loop ‘abcd’ of length ‘h’. The number of turns inside amperian loop is ‘nh’. Total current through amperian loop is ‘nhI’ by ampere circuital loop ∮ 𝐵𝑑𝑙 = 𝜇𝑜𝐼 Bh = µonhI Page 79
• 12. 4.Moving Charges and Magnetism B = µonI The direction of the field is given by the right-hand rule. The solenoid is commonly used to obtain a uniform magnetic field How the magnetic field along axial distance vary in case of a long solenoid. Why solenoid tend to contract when a current passes through it: The parallel current in adjacent turns create opposite magnetic poles which start attracting another and solenoid tend to contract. 𝐵 = 𝜇0𝑛𝐼 2 Page 80
• 13. 4.Moving Charges and Magnetism TOROID What is toroid Toroid is an endless solenoid bend inform of circle. Using ampere circutal law expression for magnetic field inside toroid. Toroid is an endless solenoid bend inform of circle. It is used to produce very high magnetic field. Let 'n' be the number of turns per unit length of toroid. ‘I’ is current through toroid. I) Magnetic field at a point in a free space inside toroid. Let 'P' be a point at a distance 𝑟1 r1 element through it. By ampere circutal law ∮ 𝐵𝑑𝑙 = µoI B2π𝑟1 = µo(O) B = O Hence Magnetic field at a point in a free space inside toroid is ZERO.(1marks) II) Magnetic field outside toroid: Let ‘Q’ be a point at distance r3 Imagine a circle of radius r3 inside amperian loop is equal to the current leaving the loop. i.e. Net current I = 0 by ampere circuital law ∮ 𝐵𝑑𝑙 = µoI B = O Hence Magnetic field outside toroid is ZERO. from ‘o’ of toroid. Imagine a circle of radius and ‘o’ as an amperian loop. This amperian loop do not have current from of toroid. and ‘o’ has an amperian loop. The current entering Page 81
• 14. 4.Moving Charges and Magnetism Solve Numerical number 4.9 from NCERT text book III) Magnetic field inside toroid: Let ‘S’ be a point at distance r2 toroid. Imagine a circle of radius r2 total no.of turns in toroid is N = nl N = 2πrn Total current entering the toroid is I(total)=2πrIn by ampere circuital law ∮ 𝐵𝑑𝑙 = µo(total) = 𝜇02𝜋𝑟𝐼𝑛 B. 2πr = µo2πrIn B = µonI FORCE BETWEEN TWO PARALLEL CURRENTS, THE AMPERE We know that a current carrying conductor creates magnetic field around it and magnetic field can exert a force. Hence if two conductor carrying current placed side by side they experience a force on each other. DERIVATION: Let two long parallel conductors 'a' and 'b' separated by a distance 'd' and carrying (parallel) currents Ia and Ib, respectively. The conductor ‘a’ produces, the same magnetic field Ba at all points along the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards. Its magnitude is given by Ampere’s circuital law, 𝐵𝑎 = 𝜇0𝐼𝑎 2𝜋𝑑 The conductor ‘b’ carrying a current Ib will experience force due to the field Ba. The direction of this force is towards the conductor ‘a’. We label this force as Fba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by 𝐹𝑏𝑎 = 𝐼𝑏𝐵𝑎𝐿 (Sinθ=1) 𝐹𝑏𝑎 = 𝜇0𝐼𝑎𝐼𝑏 2𝜋𝑑 𝐿 Similarly the force on ‘a’ due to ‘b’ is equal but opposite in direction. Fba Ba and ‘o’ as amperian loop. This loop has current entering into it. The Page 82
• 15. 4.Moving Charges and Magnetism Solve Numerical number 4.10 from NCERT text book 𝐹𝑏𝑎 = 𝐹𝑎𝑏 = 𝐹 = 𝜇0𝐼𝑎𝐼𝑏 2𝜋𝑑 L Hence force per unit length of conductor is 𝐹 𝐿 = 𝜇0𝐼𝑎𝐼𝑏 2𝜋𝑑 The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 × 10–7 newtons per metre of length. NOTE:If conductors carries current in same direction, then force between them will be attractive. If conductor carries current in opposite direction, then force between them will be repulsive. Page 83
• 16. 4.Moving Charges and Magnetism B m ϴ 𝒂 𝟐 ⁄ 𝒂 𝟐 ⁄ 𝒂 𝟐 𝐬𝐢𝐧 𝜽 𝑭𝟐 𝑭𝟏 TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE Consider a rectangular loop ABCD such that it carries a current of magnitude I. This loop is placed in a magnetic field, it experiences a torque but no net force. It quite similar to what an electric dipole experiences in a uniform electric field. When the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. The field exerts no force on the two arms AD and BC of the loop. The arm AB of the loop experience a force F1 ,Its magnitude is, F1 = IbB Similarly, a force F2 on the arm CD, Its magnitude is F2 = IbB Hence from above two equation we say, F1= F2 Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces F1 and F2 hence loop rotates anticlockwise direction. Net torque on rectangular loop is, 𝝉 = 𝒇𝒐𝒓𝒄𝒆 × 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝜏 = 𝐹1 × 𝑎 2 + 𝐹2 × 𝑎 2 𝜏 = 𝐼𝑏𝐵 × 𝑎 2 + 𝐼𝑏𝐵 × 𝑎 2 𝜏 = 𝐼(𝑎𝑏)𝐵 but Area of loop (A) = ab 𝝉 = 𝑰𝑨𝑩 When the plane of the loop, is not along the magnetic field, but makes an angle(θ) with it, then torque is given by 𝝉 = 𝑰𝑨𝑩 𝐬𝐢𝐧 𝜽 Page 84
• 17. 4.Moving Charges and Magnetism For N number of turns 𝝉 = 𝑵𝑰𝑨𝑩 𝐬𝐢𝐧 𝜽 (i) 𝝉 is zero when θ = 0, i.e., when the plane of the coil is perpendicular to the field. (ii) 𝝉 is maximum when θ= 90 , i.e., the plane of the coil is parallel to the field. 𝜏𝑚𝑎𝑥 =NBIA Where 'm=IA' in above equation is called as "magnetic moment defined as the product of area of the loop and current through the loop". Unit of magnetic moment is --Am2 and Dimention formula is [AL2] Vectorially torque is, 𝝉 ⃗⃗ = 𝒎 ⃗⃗⃗⃗ × 𝑩 ⃗⃗⃗ Page 85
• 18. 4.Moving Charges and Magnetism **THE MAGNETIC DIPOLE MOMENT OF A REVOLVING ELECTRON** Obtain the expression for magnetic dipole moment of electron revolving around nucleus. Consider a electron of charge 'e' mass ‘m’ is revolving around nucleus in an orbit of radius ‘ r’ current in orbit is I = 𝑒 𝑇 where T = Time period of electron. T = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 T = 2𝜋𝑟 𝑣 I = 𝑒 ( 2𝜋𝑟 𝑣 ) = 𝑒𝑣 2𝜋𝑟 magnetic dipole moment of electron is µe = IA = 𝑒𝑣 2𝜋𝑟 𝜋𝑟2 µe = 𝑒𝑣𝑟 2 -----------------(1) Multiplied & divide eqn (1) by 'm' µe = 𝑒𝑚𝑣𝑟 2𝑚 µe = 𝑒.𝑙 2𝑚 where l = mvr (angular momentum) "The ratio of magnetic dipole moment of electron to that of angular momentum is called as Gyromagnetic ratio" i.e. 𝝁𝒆 𝒍 = 𝒆 𝟐𝒎 = 8.8x1010 C /kg by Bohr’s angular momentum condition l = 𝑛ℎ 2𝜋 µe = 𝑒 2𝑚 × 𝑛ℎ 2𝜋 µe = 𝒏𝒆𝒉 𝟒𝝅𝒎 This is magnetic dipole moment of electron in first (Ist) orbit of hydrogen atom for n = 1 and is called Bohr magneton.(value of Bohr magneton µe = 9.27 x 10-27Am2) Page 86
• 19. 4.Moving Charges and Magnetism MOVING COIL GALVANOMETER * What is moving coil galvanometer (MCG) ? Moving coil galvanometer (MCG) is a device to detect current. Principle :- When the current carrying coil is suspended in a uniform magnetic field, then torque acts on coil which deflect it. The deflection is directly proportional to the magnitude of the current. Construction :- The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis in a uniform radial magnetic field. There is a cylindrical soft iron core which increases the strength of the magnetic field. It consists a rectangular coil of number of turns of insulated wire pivoted between poles of magnet. It rotates over semi-circular scale with central ‘0’(zero). Working :- When current is passed through it is deflected and experiences torque is given by τ = nIAB Where n = number of turns, I = current in coil A = Area of coil, B = Magnetic field Due to deflection the spring is twisted and tends to bring the coil to original position. Rotational effect of twist is called restoring couple whose moment is directly proportional to deflection of coil. 𝜏 ∝ 𝜃 𝜏 = K 𝜃 Where K is couple per unit twist in equilibrium nBIA = K 𝜃 I = ( 𝐾 𝑛𝐵𝐴 ) 𝜃 I ∝ 𝜃 The galvanometer cannot as such be used as an ammeter to measure the value of the current in a given circuit. This is for two reasons: Page 87
• 20. 4.Moving Charges and Magnetism h (i) Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of μA. (ii) For measuring currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit. Explain the conversion of galvanometer to voltmeter.(3 marks) Voltmeter is potential difference(PD) measuring device. It must be connected in parallel in the circuit. It’s resistance must be high. Galvanometer cannot be used to measure PD directly because its resistance is low. Galvanometer can be converted into voltmeter by connecting high resistance in series with it. Let ‘R’ be the high resistance connected in series with galvanometer and Ig is current required to produce full scale deflection in galvanometer then, PD across combination = Ig (R+G) V = Ig(R+G) 𝑉 𝐼𝑔 = R+G R=S = 𝑽 𝑰𝒈 – G Voltage sensitivity: Deflection per unit volt is called voltage sensitivity of galvanometer. Unit: Division /volt Page 88
• 21. 4.Moving Charges and Magnetism Solve Numerical number 4.13 from NCERT text book Explain the conversion of Galvanometer into ammeter.( 3marks) Ammeter is current measuring device. It must be connected in series. It’s resistance must be low. Galvanometer cannot be used to measure current directly because its resistance is not very low. Galvanometer can be converted into ammeter by connecting low resistance in ‖el with it. Let ‘R’ be the resistance(shunt resistance) connected in ‖el with galvanometer and Ig is current require to produce full scale deflection in galvanometer then, PD across galvanometer = PD cross resistance IgG = (I – Ig) S S = 𝑰𝒈𝑮 𝑰−𝑰𝒈 Current sensitivity: Deflection per unit current is called current sensitivity. Unit: division/A Note: 1) Ideal voltmeter resistance is ′∞′ (infinite). 2) Ideal ammeter has resistance '0'(zero). Distinguish b/w ammeter and voltmeter Ammeter Voltmeter 1. It is current measuring device. 1. It is PD measuring device. 2. Ammeter is a galvanometer with low resistance in parallel 2. Voltmeter is a galvanometer with high resistance in series. 3. Ideal ammeter has zero resistance 3. Ideal voltmeter has resistance infinity Page 89