Chapter 7 Magnetism 7.1 Introduction Magnetism has been known thousands of years dating back to the discovery recorded by the ancient Greek. 1900 Maxwell combine the theory of electric and magnetic to predict the electromagnetic wave theory. 7.2 Magnets, magnetic Poles and Magnetic Field Direction Poles of a magnet are the ends where objects are most strongly attractednorth and south poles. Like magnetic poles repel each other and unlike magnetic poles attract each other. Examples: bar magnet 7.21 Defining Magnetic Field Direction The direction of magnetic field (B) at any location is the direction that the north pole of a compass at that location would point.
7.22 Earth s Magnetic Field The Earth s magnetic field resembles that achieved by burying a huge bar magnet deep in the Earth s interior The Earth s geographic north pole corresponds to a magnetic south pole The Earth s geographic south pole corresponds to a magnetic north pole 7.3 Current and Magnetism There is a strong connection between electricity and magnetism 1820 Hans Christian Oerstead a professor of Physics at the University of Copenhagen discovered that electric current could indeed affect magnet field
7.31 Magnetic Forces on Electric Current. There will be a force perpendicular to a current carrying wire. The direction of the force is according to Fleming left hand rule (motor pinciple). F I I B B in F= B I L sin = Direction of current to the direction of magnetic field L = Length of wire in magnetic field Example 1. A wire 5.0 m length with 10.0 A current makes an angle of 30 0 with magnetic field of 0.3 T. Calculate force on it. Magnetic force F=LIB sin θ = 5.0x10.0x0.3 sin 30 0 = 7.5 N F B 30 0 I Example 2. A thin horizontal copper rod 1.0 m long has mass of 50 g. The interaction between electric current I with magnetic field of 2.0 T, let the rod floating on air. Determine the minimum current that result this phenomenon. F magnetic.x x x x x x x x x x x x x Bin Imin F gravity
F magnetic = F gravity L Imin B sin θ = mg Imin = mg/ L B sin θ = mg/ L B sin 90 0 = 0.05x9.81/1x2 =0.245 A 7.32 Torque on a Current Loop b/2 b/2 a B I From the formula of force, F= B I L sin = Force (F) x distance perpendicular = F x b/2 sin = B x I x a x b/2 sin Total torque = 1 + 2 = 2 x B x I x a x b/2 sin = B I A sin where a x b = A (area of coil) if the coil has N number of turns (loop) = N B I A sin we define magnetic dipole moment as = N I A = B sin
Example: The electric motor Example 3. A circular loop wire radius of 50 cm with current 2.0 A in a 0.4 T magnetic field region, calculate maximum torque on it. I clock. B rotation maximum torque τ max =μb sin 90 0 =NIAB = Iπr 2 B = 2 x 3.14 x 0.5 2 x 0.4 = 0.628 mn Example 4. Magnetohydrodynamic (MHD) propulsion is a type of vessel drive where thrust is generated through interaction of magnetic and electric field. MHD propulsion is preferable to the marine propellers because there are no mobile parts, no propeller noises and no vibration. Fthrust d E E X B X Ffoward The schematic design is as above. E is the electrical field between two plates 5.0x10 7 N/C, and B is the magnetic field supply from a superconducting magnet which can generate 10 tesla strength (into page). Given that sea water average resistance 0.04 ohm, and d = 15 cm. Fthrust(water) = F forward (ship) = LIB sin 90 0 = d (V/ R) B = d (E d) B/R = [0.15 2 x 5.0x10 7 x 10 / 0.04 ]N
7.33 Ammeter and Voltmeter Ammeter measures current through circuit elements and voltmeter measures voltage across circuit elements The basic component of an ammeter and voltmeter is a galvanometer. The resistance of a galvanometer is very small, since the coil consists of metal wires. It is a current-sensitive device which needle deflection (due to torque on current coil) is proportional to the current through the coil. (a) Ammeter To build an ammeter, a resistor has to be added in parallel to the galvanometer G I I g r Is R s V g =V s or I g r =I s R s I g r = (I-I g )R s I g = IR s r R s
(b) Voltmeter To build a voltmeter, a resistor have to be added in series with the galvanometer. I g R m G r V = V g +V m =I g r + I g R m = I g (r+r m ) I g = V r R m 7.34 Magnetic Forces on Moving Charged Particle A current in a wire consists of moving charges. I = t q Since a current carrying wire may experience a force when placed in a magnetic field, it is not surprising that a moving charge that is not confine within a wire may also experience a force due to magnetic field. F= B I L sin F = B t q L sin Since v (velocity) = Length (L) / time F = q v B sin
Motion of free charge in a magnetic field v B in F q When the charge particle enter the magnetic field in perpendicular direction, it will move in a circular motion. The particle will experience a centripetal force F c. F E = q v B sin since the angle is 90 o, sin 90 o = 1 F c = F E mv2 = B q v r mv r = qb Period of Rotation T Since, Velocity = Distance / Time T = 2 r v T = 2 r Br q m = 2 m qb (period for 1 rotation)
Example 5. A positive ion mass of 2.5 x 10-26 kg been accelerated through p.d of 250 V. It enters perpendicular into region of 0.5 T field. Determine the path radius. +p v x x x x x Bin 250 V PE = qv KE = 0.5mv 2 r v a-clock circular motion.r = mv/qb = m (qv/0.5m) 0.5 /qb = 1.77 cm Cyclotron An accelerator which used a magnetic field to bend the path of particles into nearly circular orbits. Oscillating Voltage Source X X X X X X s B IN X X X X X X S-ion source within the `dees experiences changes electric potential difference between the dees passes with velocity v. This will result the ion with circular path of radius r within of the dees. 7.4 Ampere s Law Ampere s Law states that the summation of the contributions around the entire loop is proportional to the net current I through the loop. B L cos = o I
7.41 Magnetic field of a long straight wire For the magnetic field surrounding a wire; I l r B B = o I 2 r o = 4 x 10-7 Tm/A o = 4 x 10-7 N/A 2 Example 6. Two wires with 3.0 A and 5.0 A separated 20 cm each other (both currents out of page). Calculate the field at P which just above the 5.0 A wire distance of 20 cm. B1 45 0 B1 =μ 0 I 1 /2π R1 = 2.12 x 10-6 T B2 P B2 =μ 0 I 2 /2π R2 = 5.00 x 10-6 T B P = B1 + B2 = (-6.5 x + 1.5 y )10-6 T R1 R2 Bp =6.67 x 10-6 T (at 77 0 left of the vertical axis) wire1 wire2 3 A 5A Bp 77 0 from Pythagoras R1 =0.283 m
7.42 Magnetic field at the center of the coil I(a-clockwise) B out = oi 2r 7.43 Magnetic Field at the center of a solenoid B axis N turns L I B = o n I where n = N/L (no of turns per unit length) 7.44 Magnetic field of a toroid Toroid- think as solenoid bent into a circle of radius r (doughnut-like shape) r NI o B= 2 r
7.5 Forces between two current carrying wires I 1 I 2 F 1 F 2 B 1 = I o 1 2 d F 2 = B 1 I 2 L o F 2 = I 1 2 d I 2 L Force per unit length on wire 2: F 2 = L oi1i 2 d 2 Example 7. Two parallel wires 10.0 cm apart each carries 10 A current in same direction. Force per unit length F/L = μ 0 I 2 / 2πd = 2.0 x 10-4 N/m attracted each other 7.6 Magnetic materials The direction of a magnetic material is determined by its resultant magnetic domain. The magnetic domain arises from the electron spin in an atom. In atoms with two or more electron, the electron usually is arranged in pairs with their spin oppositely aligned. The magnetic field then cancels each
other, and the material is not magnetic. Aluminum (Diamagnetism) is an example. In unmagnetized ferromagnetic materials, the domains are randomly oriented and there is no magnetization. But when it is placed in an external magnetic field a. Domain boundaries change and the domain with magnetic orientations in the direction of the external field grow at the expense of the other domain. b. The magnetic orientation of some domain may change slightly so as to be more aligned with the field.