Applied Physics Past Paper PDF 24PH101T 2024

Applied Physics Course Code: 24PH101T Applied Physics August 19, 2024 1 / 44 Outline 4 Divergence and Curl of Mangetic 1 Introduction – Magnetic Fields Field 2 Lorentz Force Law 3 Biot...

Applied Physics Course Code: 24PH101T Applied Physics August 19, 2024 1 / 44 Outline 4 Divergence and Curl of Mangetic 1 Introduction – Magnetic Fields Field 2 Lorentz Force Law 3 Biot Savart’s law 5 Problems (Magnetostatics) Applied Physics August 19, 2024 2 / 44 Introduction – (1) Two wires hang from the ceiling, a few centimeters apart When the current is turned on, so that it passes up one wire and back down the other, the wires jump apart. Suppose that the battery is actually charging up the wire, and that the force is simply due to the electrical repulsion of like charges. If a test charge is placed near these wires, there would be no force on it, for the wires are in fact electrically neutral. Applied Physics August 19, 2024 3 / 44 Introduction – (2) If the demonstration is set up so as to make the current flow up in both wires. The wires are found to attract Conclusion Whatever force accounts for the attraction of parallel currents and the repulsion of anti-parallel ones is not electrostatic in nature. Applied Physics August 19, 2024 4 / 44 Lorentz Force Law Magnetic Force The magnetic force on a charge Q, moving with velocity, ⃗v in a magnetic ⃗ is given by, field B   ⃗ = Q ⃗v × B Fmag ⃗ Lorentz Force In presence of both electric and magnetic field, the force on charge Q is, h  i F⃗ = Q E⃗ + ⃗v × B ⃗ There is no derivation. It is a fundamental axiom of the theory. It is a peculiar law, and it leads to some truly bizarre particle trajectories. Applied Physics August 19, 2024 5 / 44 Example 1 – Cyclotron Motion – (1) The archtypical motion of a charged particle in a magnetic field is circular, with the magnetic force providing the centripetal acceleration. A uniform magnetic field points into the page; if the charge Q moves counterclockwise, with speed ⃗v , around a circle of radius R. Lorentz Force   For circular motion, the mechanical force is ⃗ F⃗ = Q ⃗v × B given by, mv 2 F = QvB sinθ = QvB F = , here, m is mass of the particle, p R is its momentum. Applied Physics August 19, 2024 6 / 44 Example 1 – Cyclotron Motion – (2) Equating the two forces, mv 2 = QvB R p = QBR The charge moves in a plane It also suggests a simple ⃗ If it starts out perpendicular to B. experimental technique for finding with some additional speed ⃗v|| the momentum of a charged ⃗ this component of parallel to B, particle: send it through a region the motion is unaffected by the of known magnetic field, and magnetic field, and the particle measure the radius of its trajectory. moves in a helix Applied Physics August 19, 2024 7 / 44 Example 2 : “e/m” by Thomson’s Method – (1) In 1897, J. J. Thomson “discovered” the electron by measuring the charge-to-mass ratio of “cathode rays”. Part A: First he passed the beam through uniform crossed electric and magnetic fields E⃗ and B ⃗ (mutually perpendicular, and both of them perpendicular to the beam), and adjusted the electric field until he got zero deflection. What, then, was the speed of the particles (in terms of E⃗ and B)? ⃗ If the electron is travelling along Z-axis with velocity, ⃗v = v zb The electric field is along X-axis, E⃗ = E xb ⃗ = B yb The magnetic field is along Y-axis, B Applied Physics August 19, 2024 8 / 44 Example 2 : “e/m” by Thomson’s Method – (2) Using Lorentz force, h  i F⃗ = Q E⃗ + ⃗v × B⃗ = Q [E xb + (v zb × B yb)] = Q (E xb − vB yb) Since there is zero deflection, the total force is zero, E Q (E xb − vB yb) = 0 =⇒ v = B Part B Then he turned off the electric field, and measured the radius of curvature, R, of the beam, as deflected by the magnetic field alone. In terms of E , B, and R, what is the charge-to-mass ratio (q/m) of the particles? Applied Physics August 19, 2024 9 / 44 Example 2 : “e/m” by Thomson’s Method – (3) If mass of the particle is M, the momentum is given by Mv The momentum as calculated by the radius of curvature, Mv = QBR Simplyfying, Q v = M BR Q E =⇒ = 2 M B R Applied Physics August 19, 2024 10 / 44 Currents – (1) Definition: The current in a wire is the charge per unit time passing a given point. Negative charges moving to the left count the same as positive ones to the right. In practice, it is ordinarily the negatively charged electrons that do the moving-in the direction opposite to the electric current. To avoid the petty complications this entails, it is often pretended that it’s the positive charges that move. Current is measured in coulombs-per-second, or amperes (A). Applied Physics August 19, 2024 11 / 44 Currents – (2) Current is actually a vector: ⃗I = λ⃗v , but the path of the flow is dictated by the shape of the wire, one doesn’t ordinarily bother to display the direction of I A line charge λ traveling explicitly. down a wire at speed v constitutes a current The magnetic force on a segment of current -carrying wire is, I = λv Z  Z    ⃗ F = ⃗ ⃗v × B dq = ⃗ λdl ⃗v × B because a segment of length v ∆t, carrying Z   charge λv ∆t, passes point F⃗ = ⃗I × B ⃗ dl P in a time interval ∆t. Applied Physics August 19, 2024 12 / 44 Currents – (3) In as much as ⃗I and d ⃗l both point in the same direction, it can be written this as, Z   ⃗ F = I d ⃗l × B ⃗ Typically, the current is constant (in magnitude) along the wire, and in that case I comes outside the integral: Z F = I d ⃗l × B ⃗ ⃗ Applied Physics August 19, 2024 13 / 44 Currents – (4) ⃗ is the current per unit width. K If the surface charge density is σ and its velocity is ⃗v , then ⃗ = σ⃗v K Surface current density Consider a “ribbon” of The magnetic force on the surface infinitesimal width dl⊥ , running current is, parallel to the flow with current Z   flowing through it, d ⃗I , F⃗ = ⃗ σda ⃗v × B ⃗ ⃗ = dI Z   K F⃗ = ⃗ ×B K ⃗ da dl⊥ Applied Physics August 19, 2024 14 / 44 Currents – (5) J⃗ is the current per unit area. If the volume charge density is ρ and the velocity is ⃗v , then J⃗ = ρ⃗v Volume current density Consider a “tube” of infinitesimal The magnetic force on the surface cross section da⊥ , running parallel current is, to the flow with the current d ⃗I Z   flowing in the tube, ⃗ F = ⃗ ρdτ ⃗v × B d ⃗I Z  J⃗ =  da⊥ F⃗ = J⃗ × B ⃗ dτ Applied Physics August 19, 2024 15 / 44 Example 3 Suppose the current density in the wire is proportional to the distance from the axis, J = ks zb (for some constant k). Find the total current in the wire. Because J varies with s, it Z must be integrated, ∴I = (ks) (s ds dϕ) Z Z a Z 2π 2 I = J⃗ · d⃗a⊥ =k s ds dϕ 0 3  0 a =k 2π 3 d⃗a⊥ = s ds dϕ zb 2πka3 = 3 Applied Physics August 19, 2024 16 / 44 Steady Currents Stationary Charges Steady Currents =⇒ Constant electric fields =⇒ Constant magnetic fields =⇒ ELECTROSTATICS =⇒ MAGNETOSTATICS Steady current means A continuous flow that has been going on forever, without change and without charge piling up anywhere. Formally, for electro/magnetostatics is the regime of ∂ρ ∂ J⃗ =0, =0 ∂t ∂t When a steady current flows in a wire, its magnitude I must be the same all along the line; otherwise, charge would be piling up somewhere, =⇒ ∇ ⃗ · J⃗ = 0 Applied Physics August 19, 2024 17 / 44 Biot Savart’s Law – (1) The magnetic field of a steady line current is given by r r Z ⃗ I ×c d ⃗l ′ × c Z ⃗ (⃗r ) = µ0 B ′ dl = µ0 I 4π r 2 4π r 2 The integration is along the current path, in the direction of the flow; d ⃗l ′ is an element of length along the wire. The constant µ0 is called the permeability of free space, µ0 = 4π × 10−7 NA−2. Applied Physics August 19, 2024 18 / 44 Biot Savart’s Law – (2) ⃗ Unit of B For Surface Currents, Newtons per ampere-meter – N/ (A · m) or Teslas, T. r ′   ⃗ r⃗′ × c Z K ⃗ (⃗r ) = µ0 B da As the starting point for 4π r2 magnetostatics, the Biot-Savart law plays a role analogous to For Volume Currents, Coulomb’s law in electrostatics. r ′   Z J⃗ r⃗′ × c ⃗ (⃗r ) = µ0 The 1/ r 2 dependence is common B 4π r2 dτ to both laws. The superposition principle applies to magnetic fields just as it does to electric fields. Applied Physics August 19, 2024 19 / 44 Example 4 – (1) Find the magnetic field a distance s from a long straight wire carrying a steady current I. Biot Savart’s Law r d ⃗l ′ × c d ⃗l ′ × r⃗ Z Z ⃗ = µ0 I B = µ0 I ′ ⃗r = s sb , r⃗′ = l zb 4π r 2 4π r 3 d ⃗l ′ = dl ′ zb r⃗ = ⃗r − r⃗′ = s sb − l ′ zb Approaching the problem in Cylindrical Coordinates, assuming the wire is placed r = s 2 + l ′2 1/2
along the Z-axis and the point P is on XY-plane. r 3 = s 2 + l ′2 3/2
Applied Physics August 19, 2024 20 / 44 Example 4 – (2) d ⃗l ′ × r⃗ = dl ′ zb × s sb − l ′ zb
= dl ′ s ϕb Substituting... dl ′ s ϕb Z ⃗ = µ0 I B 4π (s 2 + l ′2 )3/2 If a zb and b zb substitute an angle of θ1 and θ2 , as shown in Figure. If a zb and b zb are two points on the line with a < b l ′ = s tanθ µ0 Is b Z dl ′ s ⃗ B= ϕb dl ′ = s sec2 θ dθ = dθ 4π a (s 2 + l ′2 )3/2 cos2 θ Applied Physics August 19, 2024 21 / 44 Example 4 – (3) Z θ2 ⃗ = µ0 Is ∴B 1 s dθ ϕb 4π θ1 (s 2 + s 2 tan2 θ) 3/2 cos2 θ θ2 µ0 Is 2 Z µ0 I = cosθ dθ ϕb = (sinθ2 − sinθ1 ) ϕb 4πs 3 θ1 4πs This is a solution for a line segment located on a wire. For infinite wire, θ1 = −π/2 and θ2 = π/2. ⃗= µ0 I b ∴B ϕ 4πs Applied Physics August 19, 2024 22 / 44 Example 5 – (1) Find the magnetic field a distance z above the center of a circular loop of radius R, which carries a steady current I. Idea is... ⃗ attributable to the segment d ⃗l ′ The field d B points as shown. As it is integrated around the ⃗ sweeps out a cone. The horizontal loop, d B components cancel, and the vertical components combine along Z-axis. Approaching the problem in Cylindrical coordinates... Applied Physics August 19, 2024 23 / 44 Example 5 – (2) Biot Savart’s Law d ⃗l ′ × r⃗ d ⃗l ′ × r⃗ = Rdϕ ϕb × (−R sb + z zb) Z ⃗ = µ0 I B 4π r 3 = R 2 dϕ zb From figure Substituting... r⃗′ = R sb Z 2π ⃗ µ0 I R2 ⃗r = z zb B= dϕ zb 4π (R 2 + z 2 )3/2 0 r⃗ = −R sb + z zb µ0 I R2 d ⃗l ′ = Rdϕ ϕb = 2π zb 4π (R 2 + z 2 )3/2 r = R 2 + z 2 1/2
µ0 I R2 = zb r 3 = R 2 + z 2 3/2 2 (R 2 + z 2 )3/2
Applied Physics August 19, 2024 24 / 44 Straight Line Currents – (1) The magnetic field of an infinite straight wire is shown. At a glance, it is clear that this field has a nonzero curl. ⃗ around a circular The integral of B path of radius s, I I I ⃗ · d ⃗l = µ0 I µ0 I B dl = dl = µ0 I 2πs 2πs Applied Physics August 19, 2024 25 / 44 Straight Line Currents – (2) Notice that the answer is independent of s; that’s because B⃗ Computing integral, decreases at the same rate as the I I circumference increases. ⃗ ⃗ µ0 I 1 B · dl = sdϕ 2π s Using Cylindrical coordinates, µ0 I 2π Z = dϕ = µ0 I 2π 0 d ⃗l = dsb s + sdϕϕb + dz zb This assumes the loop encircles Magnetic field for current flowing the wire exactly once; if it went along z-axis, around twice, then ϕ would run from 0 to 4π. ⃗ = µ0 I ϕb B 2πs Applied Physics August 19, 2024 26 / 44 Straight Line Currents – (3) R If it didn’t enclose the wire at all, then dϕ = 0. There is a bundle of straight wires. Each wire that passes through the loop contributes µ0 I , and those outside contribute nothing. The line integral will then be, I ⃗ · d ⃗l = µ0 Ienc B If the flow of charge is represented ⃗ the by a volume current density J, ⃗ R enclosed current is, Ienc = J · d⃗a. Applied Physics August 19, 2024 27 / 44 Curl of Magnetic Field Using the formula of enclosed current shown as integral of volume current density, in the equation of closed loop integral, I Z B · d l = µ0 J⃗ · d⃗a ⃗ ⃗ Applying Stoke’s law, I Z   Z ⃗ ⃗ B · dl = ∇ × B · d⃗a = µ0 J⃗ · d⃗a ⃗ ⃗ S Hence, ⃗ = µ0 J⃗ ⃗ ×B ∇ Applied Physics August 19, 2024 28 / 44 Ampere’s Law The equation for the curl of B, ⃗ = µ0 J, ⃗ ×B ∇ ⃗ is called Ampere’s law (in differential form). Taking a surface integral, The line integral is along Rthe Z   Z boundary of the surface. S J⃗ · d⃗a ∇ × B · d⃗a = µ0 J⃗ · d⃗a ⃗ ⃗ is the total current passing S S through the cross section area. Applying Stoke’s law to LHS, I I Z ⃗ · d ⃗l = µ0 Ienc B B · d l = µ0 J⃗ · d⃗a ⃗ ⃗ S Ampere’s Law (integral form) Applied Physics August 19, 2024 29 / 44 Ampere’s Law (Discussion) The integral form of Ampere’s Law inherits the sign ambiguity of the Stoke’s theorem. Which way around the loop is the integration supposed to be calculated? And which direction through the surface corresponds to a “ positive” current? The answer, as always, is the right-hand rule. If the fingers of the right hand indicate the direction of integration around the boundary, then your thumb defines the direction of a positive current. For currents with appropriate symmetry, Ampere’s law in integral form offers a lovely and extraordinarily efficient way of calculating the magnetic field. Applied Physics August 19, 2024 30 / 44 Example 6 Find the magnetic field a distance s from a long straight wire, carrying a steady current I. By Right Hand Rule... ⃗ is found to be “circumferential” The direction of B I Amperian Loop is a circle of radius ⃗ · d ⃗l = µ0 Ienc B s around the wire. Z 2π B s dϕ = µ0 I Using Cylindrical System, 0 B2πs = µ0 I ⃗ = B ϕb , d ⃗l = s dϕ ϕb B µ0 I B= 2πs Applied Physics August 19, 2024 31 / 44 Example 7 – (1) Find the magnetic field of an infinite ⃗ = K xb, uniform surface current K flowing over the XY-plane. ⃗ What is the direction of B? Biot Savart’s Law ⃗ can only be having Y- and B Z-components. r ′   ⃗ r⃗′ × c Z K ⃗ = By yb + Bz zb ⃗ (⃗r ) = µ0 B B da 4π r2 Amperian Loop to be put up in ⃗ ⊥K =⇒ B ⃗ YZ-plane. Applied Physics August 19, 2024 32 / 44 Example 7 – (2) Amperian loop is a rectangle with four components. d l⃗1 = −dy yb – for line segment above the XY-plane. d l⃗2 = −dz zb – for line segment left of the XZ-plane. d l⃗3 = dy yb – for line segment below the XY-plane. d l⃗4 = dz zb – for line segment right of the XZ-plane. I Z Z Z Z ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ∴ B · d l = B · d l1 + B · d l2 + B · d l3 + B ⃗ ⃗ · d l⃗4 1 2 3 4 For path 2 For path 4 ⃗ · d l⃗2 = 0 B ⃗ · d l⃗4 = 0 B This shows that for the vertical line, the contribution from Current cancels each other out. Applied Physics August 19, 2024 33 / 44 Example 7 – (3) For path 1 Z Z ⃗ · d l⃗1 = By dy =⇒ B ⃗ · d l⃗1 = B By dy = By l 1 1 For path 3 ⃗ = −By yb Here, B Z Z ⃗ · d l⃗3 = By dy =⇒ B ⃗ · d l⃗3 = B By dy = By l 3 1 Applied Physics August 19, 2024 34 / 44 Example 7 – (4) Using Ampere’s Law, I ∴ ⃗ · d ⃗l = 2 By l = µ0 K l B µ0 K ∴ By = 2 Hence the Magnetic field is given by,  + µ0 K yb,  for z > 0 ⃗ B= 2 µ K − 0  yb, for z < 0 2 Applied Physics August 19, 2024 35 / 44 Example 8 – (1) Find the magnetic field of a very long solenoid, consisting of n closely wound turns per unit length on a cylinder of radius R, each carrying a steady current I. Magnetic field is of the form, ⃗ = Bs sb + Bϕ ϕb + Bz zb B d ⃗l = ds sb + s dϕ ϕb + dz zb ⃗ · d ⃗l = Bs ds + Bϕ s dϕ + Bz dz ∴B Applied Physics August 19, 2024 36 / 44 Example 8 – (2) ⃗ ⊥K Using B ⃗ Since, the current is going around a circle, it can be written as, K⃗ = Ks sb + Kϕ ϕb This information concludes that the magnetic field is ⃗ = Bz zb. along Z-direction. ∴ B ⃗ · d ⃗l Only the 3rd term survives in the B To find the magnetic field inside and outside the solenoid two Amperian loops are used. One entirely outside and another partially inside and partially outside. Applied Physics August 19, 2024 37 / 44 Example 8 – (3) For Loop 1 I ⃗ · d ⃗l = [Bz (a) − Bz (b)] L = µ0 Ienc = 0 B =⇒ Bz (a) = Bz (b) Evidently the field outside does not depend on the distance from the axis. For Loop 2 I ⃗ · d ⃗l = Bz L = µ0 Ienc = µ0 n I L =⇒ Bz = µ0 nI , ∴ B B ⃗ = µ0 nI zb Applied Physics August 19, 2024 38 / 44 Comparison of Electrostatics and Magnetostatics Electrostatics Magnetostatics =⇒ Stationary Charges =⇒ Steady Currents =⇒ Constant electric fields =⇒ Constant magnetic fields =⇒ Coulomb’s Law =⇒ Biot Savart’s Law =⇒ Gauss’s Law =⇒ Ampere’s Law ⃗ · E⃗ = ρ ∇ ⃗ = µ0 J⃗ ⃗ ×B ϵ0 ∇ I I Qenc ⃗ · d ⃗l = µ0 Ienc E⃗ · d⃗a = B S ϵ0 ⃗ × E⃗ = 0 =⇒ ∇ ⃗ =0 ⃗ ·B =⇒ ∇ Applied Physics August 19, 2024 39 / 44 Problems for Homework – (1) Problem 1 A particle of charge q enters a region of uniform magnetic field B⃗ (pointing into the page). The field deflects the particle a distance d above the original line of flight. Is the charge positive or negative? In terms ⃗ and q, find the momentum of of a, d, B the particle. Problem 2 Suppose that the magnetic field in some region has the form B ⃗ = kz zb (where k is a constant). Find the force on a square loop (side a), lying in the YZ–plane and centered at the origin, if it carries a current I , flowing counterclockwise, when you look down the X–axis. Applied Physics August 19, 2024 40 / 44 Problems for Homework – (2) Problem 3 A current I flows down a wire of radius a. 1 If it is uniformly distributed over the surface, what is the surface current density K ⃗? 2 If it is distributed in such a way that the volume current density is inversely proportional to the distance from the axis, what is J⃗ (s)? Problem 4 Find the magnetic field at the center of a square loop, which carries a steady current I. Let R be the distance from center to side. Applied Physics August 19, 2024 41 / 44 Problems for Homework – (3) Problem 5 Find the magnetic field at point P on the axis of a tightly wound solenoid (helical coil) consisting of n turns per unit length wrapped around a cylindrical tube of radius a and carrying current I. Express your answer in terms of θ1 and θ2 (it’s easiest that way). Consider the turns to be essentially circular, and use the result of Example 6. What is the field on the axis of an infinite solenoid (infinite in both directions)? Applied Physics August 19, 2024 42 / 44 Problems for Homework – (4) Problem 6 A steady current I flows down a long cylindrical wire of radius a. Find the magnetic field, both inside and outside the wire, if 1 The current is uniformly distributed over the outside surface of the wire. 2 The current is distributed in such a way that J⃗ is proportional to s, the distance from the axis. Problem 7 A thick slab extending from z = −a to z = +a (and infinite in the X and Y directions) carries a uniform volume current J⃗ = J xb. Find the magnetic field, as a function of z, both inside and outside the slab. Applied Physics August 19, 2024 43 / 44 Problems for Homework – (5) Problem 8 Analyze the motion of a particle (charge q, mass m) in the magnetic field of a long straight wire carrying a steady current I. 1 Is its kinetic energy conserved? 2 Find the force on the particle, in cylindrical coordinates, with I along the Z axis. 3 Obtain the equations of motion. dz 4 Suppose dt is constant. Describe the motion. Applied Physics August 19, 2024 44 / 44