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Magnetic poles, north and south Chapter 21: Magnetic Forces and Fields • Magnetic fields Magnetic poles occur only as N-S pairs, unlike electrical charges that exist separately as + and – entities. • The force on a moving charge and on a current in a magnetic field • The torque on a coil in a magnetic field. Electric motors • Magnetic field produced by a current. Ampere’s law Existence of a “magnetic monopole”? • Omit 21.9, magnetic materials Only a theoretical possibility at present. Friday, February 2, 2007 1 Friday, February 2, 2007 The north pole of a compass needle points along the magnetic field lines toward the south pole of a magnet Compass needle points along magnetic field line toward the magnetic south pole 2 Magnetic field lines of the earth Magnetic field lines form a pattern similar to an electric dipole. ! the north pole of the earth is a magnetic south pole... Friday, February 2, 2007 3 Friday, February 2, 2007 4 Force on a charge moving in a magnetic field F = qvB sin ! ! q Force is at right angles to both !v and !B Right hand rule Magnetic field 21.C5: Determine whether each particle is positively or negatively charged, or neutral. !B #1: positive !F !v #2: neutral F in Newtons Palm (Push, i.e. force) Fingers (Field) #3: negative q in Coulombs v in m/s thuMb (Motion of charge) !F B in Tesla (T) Earth’s magnetic field ! 10-4 T Friday, February 2, 2007 5 21.5/2: B = 0.3 T, v = 45 m/s, q = 8.4 µC. Find the force on the charge in the three cases below. Friday, February 2, 2007 6 Motion of a charge in an electric field There is in general a component of the force in the direction of the displacement, so the force does work and the speed of the charge changes. Motion of a charge in a magnetic field The force is always perpendicular to the velocity (and displacement), so the force does no work and the charge moves at constant speed. The charge follows a circular path if the magnetic field is uniform. F = qvB sin !, qvB = 1.13 × 10−4 N a) ! = 30◦, F = 0.57 × 10−4 N, into screen b) ! = 90◦, F = 1.13 × 10−4 N, into screen c) ! = 150◦, F = 0.57 × 10−4 N, into screen Friday, February 2, 2007 7 Friday, February 2, 2007 8 Motion of a charge in a uniform magnetic field Prob. 21.11/12: a) Is q positive or negative? As B is at right angles to v: Right hand rule: q is negative F=qvB b) If the magnitude of q is 8.2"10-4 C, what is the mass of the particle? F always at right angles to velocity ! it does no work, v is constant The circular path has radius determined by the centripetal force provided by the magnetic field – mv2 Fc = = qvB r r= So, m = v = 140 m/s B = 0.48 T r = 960 m mv qB qBr 8.2 × 10−4 × 0.48 × 960 = v 140 = 2.7 × 10−3 kg mv So, r = qB Friday, February 2, 2007 9 Friday, February 2, 2007 10 Motion of a charge in a uniform magnetic field Force on a charge moving in a magnetic field ! Force is at right angles to both !v and B As B is at right angles to v: F! q F = qvB sin ! ! Right hand rule Palm (Push, i.e. force) Fingers (Field) F=qvB Magnetic field !v !B F always at right angles to velocity ! it does no work, v is constant F in Newtons The circular path has radius determined by the centripetal force provided by the magnetic field – mv2 Fc = = qvB r q in Coulombs v in m/s thuMb (Motion of charge) B in Tesla (T) So, r = Earth’s magnetic field ! 10-4 T Friday, February 2, 2007 11 Friday, February 2, 2007 mv qBr and, m = qB v 12 Mass spectrometer, to measure mass of ions Mass Spectrum In the magnetic field: Isotopes of neon qBr eBr = v v 1 2 and mv = eV 2 ! so, v = 2eV /m m= !F KE = eV m= m= eB2r2 2V eB2r2 2V –V m, q = e v!0 " atomic mass of isotopes of neon Removes an electron from neutral atoms, forming singly-charged positive ions 21.18 Friday, February 2, 2007 13 Friday, February 2, 2007 14 Velocity Selector Prob. 21.18: The ion source in a mass spectrometer produces both singly and doubly ionized species, X+ and X2+. The difference in mass is too small to detect. Magnetic Field The charge +q feels an upward force due to the magnetic field. Both species are accelerated through the same electric potential difference and experience the same magnetic field, which causes them to move on circular paths. F = qvB Electric Field The charge +q feels a downward force due to the electric field. Find the ratio of the radii of the paths. With both the B and E fields present, the net force on the charge is zero if: qvB = qE, that is, v = E/B " measurement of velocity, # independent of q, m Friday, February 2, 2007 15 Friday, February 2, 2007 F = qE 16 Force on a current-carrying conductor Force on a current-carrying conductor Current is a flow of charge. A charge $q flows at speed v along a length L of wire in time $t. Imagine + charges moving along the conductor in the direction of the current, I. I= !q !t and L = v$t The force on the charge #q is: ! " !q F = !q v B sin " = (v !t) B sin " !t There is a force F acting on the charges moving along the conductor that is due to the magnetic field. F = ILB sin ! I !v $q L The direction of the force is given by the right hand rule. Friday, February 2, 2007 17 Friday, February 2, 2007 Ex 13: Force per turn on lower end of loop: By how much does the weight of the coil of wire seem to increase when the current is turned on in the coil? 18 B = 0.2 T, I = 8.5 A 125 turns F = I L B, downwards There are N = 125 turns in the coil, so total force is: F = N I L B = 125 I L B I F B = 0.2 T, I = 8.5 A F = 125 " (8.5 A) " (0.015 m) " (0.2 T) = 3.19 N I F Friday, February 2, 2007 19 Friday, February 2, 2007 20 Prob. 21.30: The triangular loop of wire carries a current, I = 4.7 A and B = 1.8 T. Find the force acting on each side and the net force. Prob. 21.29: A single turn coil is placed in a uniform magnetic field of 0.25 T. Each side of the coil is of length L = 0.32 m. I = 12 A. Determine the magnetic force on each side of the coil. "F . F 0.32 m F=0 l F=0 L h L . F % = 550 I "F L = 0.32 m Friday, February 2, 2007 21 Friday, February 2, 2007 22 Force on a current-carrying conductor 21.34: B = 0.05 T, vertical m = 0.2 kg = mass of bar I L v constant A current I flows through the bar. The bar slides down the rails without friction at constant speed. What is I? In what direction does it flow? An alternating current in the voice coil causes the cone of the loudspeaker to move in and out and generate sound – an electrical signal is converted into a sound wave. • The component of the weight down the plane must be equal to the # component of the magnetic force up the plane... Friday, February 2, 2007 Ex 5 23 Friday, February 2, 2007 24 Force on a charge moving in a magnetic field Ex. 5: The voice coil of a speaker has a diameter d = 0.025 m, contains 55 turns of wire and is placed in a magnetic field of 0.1 T. The current in the coil is 2 A. !F F = qvB sin ! !B ! Magnetic field q Force is at right angles to both !v and !B a) Find the magnetic force that acts on the coil and cone. b) The voice coil and cone have a combined mass of 0.02 kg. Find their # acceleration. !v Force on a current in a magnetic field F = !q v B sin " = ! " !q (v !t) B sin " !t I !v $q L F = ILB sin ! Friday, February 2, 2007 25 Friday, February 2, 2007 26 Force on a charge moving in a magnetic field Charges inside the moving rod experience a force due to the magnetic field... !F F = qvB sin ! !B ! Magnetic field q Force is at right angles to both !v and !B ––– ++ + Force on a current in a magnetic field I !v I I !v I F = !q v B sin " = ! " !q (v !t) B sin " !t Conductor I The moving conductor acts as a generator. !v $q L F = ILB sin ! The basis of electromagnetic induction (next chapter) Friday, February 2, 2007 27 Friday, February 2, 2007 28 The Torque on a Current-Carrying Coil Magnetohydrodynamics (MHD) $ = angle between normal to coil and B Ions within the water allow a current to flow when a potential difference is applied. w L The moving ions feel a force due to the magnetic field that pushes the water out through the back of the boat. L w Area of coil, A = Lw The force on each vertical arm of the coil is: F = ILB Fw Fw The torque is: τ = sin φ + sin φ = I(Lw)B sin φ = IAB sin φ 2 2 (A = Lw) If the coil has N turns: ! = NIAB sin " The boat feels an equal and opposite reaction force to the front. Propulsion with no moving parts! Friday, February 2, 2007 29 Friday, February 2, 2007 30 Variation of torque with $ If the coil has N turns: ! = NIAB sin " Torque: ! = NIAB sin " The formula is valid for any shape of coil, not just rectangular. 1 NIA is known as the “magnetic moment” of the coil. ! = 90◦ From above sin(pi*phi/180) 0.8 Torque, & $ = 0o 0.6 0.4 0.2 00 0 1800 3600 5400 ! -0.2 -0.4 -0.6 -0.8 -1 0 100 200 300 400 500 600 700 Torque changes sign! – rotation reverses, no good for an electric motor Friday, February 2, 2007 31 Friday, February 2, 2007 32 Reverse the current at the right times... Electric motor ! = 0◦ ...the torque is always in the same direction and the motor continues to rotate 1 sin(pi*phi/180) 1 0.8 2 1 2 Torque, &0.6 0.4 0.2 00 0 1800 3600 5400 ! -0.2 -0.4 -0.6 -0.8 -1 0 100 200 300 400 500 600 700 Friday, February 2, 2007 Commutator reverses direction of current when $ = 0o, 180o... Torque continues in the same direction 33 Prob. 21.41: The loop is free to rotate about the z-axis. Find the torque on the loop and whether the 35º angle will increase or decrease. Friday, February 2, 2007 34 Prob. 21.44: A square coil and a rectangular coil are made from the same length of wire. Each contains a single turn. The long sides in the rectangle are twice as long as the short sides. Find the ratio of torques the coils experience in the same magnetic field. I = 4.4 A B = 1.8 T • Find the relation between the lengths of the sides of the square and the # rectangle, given that the perimeters are of the same length. • Work out the areas of the two shapes and relate to the torques. a = 2b l b l $ = 90º – 35º Friday, February 2, 2007 35 Friday, February 2, 2007 36 Magnetic field produced by currents Magnetic field produced by currents At a distance r from an infinitely long, straight wire: I B= µ0I 2!r Can be proved using Ampere’s law (later) µ0 = “permeability of free space” = 4! × 10−7 T.m/A B Right hand rule: the magnetic field wraps around the wire in the direction indicated by the fingers of the right hand. Friday, February 2, 2007 37 21.C18: Currents of the same magnitude flow in each of the wires that run perpendicular to the plane of the page. Choose the direction of the current for each wire, so that when any single current is turned off, the total magnetic field at point P at the centre is directed toward a corner of the square. Friday, February 2, 2007 38 Prob. 21.49/47: In a lightning bolt, 15 C of charge flows in 1.5 ms. Assuming the lightning bolt can be represented as a long, straight line of current, what is the magnitude of the magnetic field at a distance of 25 m from the bolt? X • What is the current? !B4 !B1 + !B3 = 0 !B3 !2 + B !4 = 0 B !B1 !2 B X Friday, February 2, 2007 Currents at opposite corners should be in the same direction 39 Friday, February 2, 2007 40 Prob. 21.59: The picture shows the end view of three wires that are perpendicular to the screen. The currents in the wires are I1 = I2 = I. Prob. 21.72: Two long, straight wires are separated by 0.12 m. They carry currents of 8 A in opposite directions. Find the magnetic field at A and at B. What current I3 is needed in wire 3 to make the magnetic field at the empty corner equal to zero? !B3 a I1 45º 45º !B2 45º 45º l a !B1 !B1 + !B2 a |!B3| = |!B1 + !B2| I3 Friday, February 2, 2007 41 Magnetic field produced by currents a I2 I Friday, February 2, 2007 42 1 Force on a charge moving parallel to a conductor At a distance r from an infinitely long, straight wire: B= µ0I 2!r µ0 = “permeability of free space” = 4! × 10−7 T.m/A I B Fingers show direction in which the magnetic field wraps around the wire F = q0vB Direction of force on the charge Friday, February 2, 2007 µ0 I 2πr Direction in which the B magnetic field wraps around the wire I B B= 43 Friday, February 2, 2007 Replace the moving charge by a current ! attractive force between currents in same direction 44 Force on a moving charge 21.C14: The two outer wires are held in place. ! attractive force between currents in same direction I=3A q0 = 6.5"10-6 C, Which way will the middle wire move? !F23 !F12 The charge is moving parallel to the wire at a distance r = 0.05 m at speed v = 280 m/s. r 1 F = q0vB sin% = q0vB µI At position of qo: B = o 2!r µ0I So, F = q0v × 2!r F = (6.5 × 10−6 C) × (280 m/s) × F = 2.2 × 10−8 N 2 3 4! × 10−7 × (3 A) 2! × (0.05 m) Friday, February 2, 2007 45 Force between long current-carrying wires !B1 Friday, February 2, 2007 46 Prob. 21.54: Find the net force acting on the wire loop that is beside the long, straight wire. !B1 L d d L I1 = 12 A !F1 µ0I1 µ0I1I2 At wire 2: F = I2LB1 = I2L = L 2!d 2!d (1) I2 = 25 A " !B1 (2) B1 is due to I1 (Due to I1 in wire 1, at position of wire 2) !F2 Force per unit length of wire 2 is: µ0I1I2 2!d I2 = 25 A " !B1 Equal and opposite force on wire 1 due to current in wire 2 Friday, February 2, 2007 47 Friday, February 2, 2007 48 Magnetic field from a current loop of radius R 21.-/53: !B2 !B1 " B1 = B2 I2 = 6.6 I1 At the centre of the loop, B = 0. What is H? Due to the loop: B1 = µ0I B= N 2R for a coil of N turns For B1 = B2 : µ0I1 2R I1 I2 = R !H Due to the wire: B2 = So, H = R µ0I2 2!H I2 6.6 = R = 2.1R !I1 ! 21.55 Friday, February 2, 2007 49 Friday, February 2, 2007 50 Magnetic field from a solenoid 21.55: Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns or wire, has a radius r1 = 0.015 m and carries a current I1 = 7.2 A. The outer coil contains 180 turns and had a radius r2 = 0.023 m. What must be the current in the outer coil so that the net magnetic field at the common centre is zero? I1 = 7.2 A P I2 = ? r2 = 0.023 m N2 = 180 r1 = 0.015 m Inside the solenoid B = µ0nI n = number of turns per unit length of the solenoid N1 = 140 Friday, February 2, 2007 51 Friday, February 2, 2007 52 21.C15: Attractive or repulsive forces? S Construct a closed path around the two wires. S 21.C16: Attractive or repulsive forces? Ampère’s Law N S N NS Split the path into short segments of length #l, measure the magnetic field, B|| in the direction of each segment. Add up all of the B|| #l N Then: ! B||"l = µ0I N S S Friday, February 2, 2007 I is the net current passing through the surface, (I1 + I2). N 53 Field from a long, straight wire, using Ampère’s Law Choose a circular path of radius r around the wire. Friday, February 2, 2007 54 Prob. 21.61/60: A uniform magnetic field is everywhere perpendicular to the page. A circular path is drawn on the page. Use Ampere’s law to show there can be no net current passing through the circular surface. ! B||"l = µ0I around the circle B is the same all the way around. !B so, ! B||"l = B × 2#r = µ0I B= #l µ0I 2!r !B B|| = 0 I = 0, zero net current passes through the surface As before! 21.60 Friday, February 2, 2007 55 Friday, February 2, 2007 56 Prob. 21.34: A thin, uniform rod 0.45 m long and of mass 94 g is attached to the floor by a hinge at point P. A uniform magnetic field B = 0.36 T is directed into the screen. The current in the rod is I = 4.1 A. The rod is in equilibrium. What is the angle %? (Hint: the magnetic force can be taken to act at the centre of gravity of the rod). Prob. 21.62: A very long, hollow cylinder is formed by rolling up a thin sheet of copper. Electric charges flow along the copper sheet parallel to the axis of the cylinder, forming a hollow tube carrying current I. a) Use Ampere’s law to show that the magnetic field outside the cylinder at # a distance r from the axis is: B= F = ILB µ0I 2!r b) Show that the field is zero inside the cylinder. F There are forces at the hinge L/2 90º I Centre of gravity L/2 Floor mg Friday, February 2, 2007 57 Friday, February 2, 2007 58 Summary of chapter 21 Prob. 21.43: Coil: N = 410 turns; area per turn, A = 3.1 " 10-3 m2; current I = 0.26 A. Magnetic field, B = 0.23 T. Force on a moving charge: # # A brake shoe is pressed against the shaft to keep the coil from turning. The coefficient of static friction between the shaft and the brake shoe is µs = 0.76. The radius of the shaft is 0.012 m. What is the magnitude of the minimum normal force that the brake shoe exerts on the shaft? # F = qvB sin% Force on a current-carrying conductor:# # F = ILB sin% Torque on a coil:# # & = NIAB sin$ # # # # # # # # Magnetic field from a long, straight wire:# Brake shoe v, I B = µo I/2!r Magnetic field at centre of a circular loop:# B = µo NI/2R Magnetic field at centre of a solenoid:# • Find out the maximum torque exerted by the coil Ampere’s law: # B = µo nI ! B||"l = µ0I • Equate it to the torque exerted on the shaft by the brake shoe Friday, February 2, 2007 59 Friday, February 2, 2007 60