2011/2012 E N G I N E E R I N G P H Y S I C S
SUBTOPICS S C I ● Magnets, magnetic poles and S Y magnetic field direction H P ● Magnetic field strength and magnetic G N force I R E ● Force and current-carrying conductor E N I ● Magnetic material: Ferromagnetic G N E 2011/2012
INTRODUCTION S C I When thinking about magnetism, most S Y people tends to think of an attraction H – certain things can be picked up with P a magnet. G N I R E E N I G N E 2011/2012
S C In physics, magnetism is one of the I S phenomena by which materials exert attractive Y or repulsive forces on other materials. H P Materials that exhibit easily detectable magnetic properties (called magnets ) are G nickel, iron and their alloys; however, all N materials are influenced to greater or lesser I R degree by the presence of a magnetic field . E E Magnetism also has other manifestations in N physics, particularly as one of the two I G components of electromagnetic waves such N as light . E 2011/2012
S Magnetostatics is the study of static C I magnetic fields. S Y H P In electrostatics, the charges are stationary, G whereas here, N I R the currents are stationary. As it turns out E magnetostatics is E N a good approximation even when the currents I G are not static N E as long as the currents do not alternate rapidly. 2011/2012
MAGNETS, MAGNETIC POLES S AND MAGNETIC FIELD C DIRECTION I S Y H Magnets have two P distinct types of G poles ; we refer to N them as north (N) I R and south (S) . E E N I G N E 2011/2012
By using two bar magnets, the nature of S forces acting between them C can be determined. I S - pole – force law or law of poles : - Y H Like magnetic poles repel and unlike poles P attract . G N I R E E N I G N E 2011/2012
S C T wo magnetic poles of opposite kind form a I S magnetic dipole . Y H All known magnets are dipoles (or higher P poles); magnetic monopoles could exist ( in G theory ) but have never been observed N ( experimentally ). I R E A magnet creates a magnetic field B : E The direction of a magnetic field (B) at any location is the N direction that the north pole of a compass would point if I G placed at that location. N E 2011/2012
S C North magnetic poles are attracted by south I S magnetic poles, so the magnetic field Y points from north poles to south poles . H P The magnetic field may be represented by G magnetic field lines. N The closer together (that is, the denser) the B field lines , I R the stronger the magnetic field . E At any location, the direction of the magnetic field is E tangent to the field line, or equivalently, the way the north N end of a compass points. I G N E 2011/2012
MAGNETIC FIELD STRENGTH S AND MAGNETIC FORCE C I A magnetic field B can exert a force on a S Y moving charged particle . H P G N I R E E N I When a charged particle G A horseshoe magnet enters a magnetic field, the N created by bending bar particle is acted on by a E magnet, produces a force whose direction is fairly uniform field obvious by the deflection of between its poles. the particle from its original 2011/2012 path.
S The magnitude of the force is proportional to C the particle’s charge and its speed. When I S the particle’s velocity v is perpendicular to Y the magnetic field B , the magnitude of the H field is P valid only when v is G perpendicular to B N I R E Where, F is force and q is charges E N SI unit of magnetic field: newton/ampere- I G meter (N/(A.m) or commonly used - tesla, N T. E 2011/2012
In general, a particle’s velocity may not be S perpendicular to the field. Then the C magnitude of the force depends on the sine of I S the angle θ between velocity vector and the Y magnetic field vector, thus it may be H represents as, P Magnetic force on G charged particle N I R The force is perpendicular to both the velocity E and to the field . E N NOTE: I The magnetic force F = 0 when v and B is G parallel (θ = 0° or 180°), N The maximum when v and B is perpendicular (θ = E 90°), where F = qvB sin (90°) = qvB 2011/2012
The right-hand Force Rule for moving S charges C I S Y H P G N I R E E N I G N E A right-hand rule gives the direction of the force. 2011/2012
EXAMPLE 1 S C I S y y y Y B B H P v x x x - + - G v N z v z B z I R (c) (a) (b) E E The figure above shows three situations in N which charged particle with velocity v travels I G through a uniform magnetic field B . In each N situation, what is the direction of the E magnetic force F B on the particle? 2011/2012
EXAMPLE 2 S C I S Y A uniform magnetic field B , with magnitude H 1.2 mT, points vertically upward throughout P the volume of a laboratory chamber. A proton G with kinetic energy 5.3 MeV enters the N chamber, moving horizontally from south to I R north. What magnetic deflecting force acts on E proton as it enters the chamber? E [the proton mass, m = 1.67 x 10^-27kg] N I G N E 2011/2012
Solution: S C The magnetic deflecting force depends on the I speed of the proton, S which we can find from K = ½ mv 2, solving for v, Y H we find kinetic energy ,k = 1 2 2 mv P find velocity ,v G N m = − 13 J / MeV v = I 2 5.3MeV 1.60x 10 2k R − 27 kg E 1.67x 10 E 7 m / s v = 3.2x10 N using F B = qvB sin I G − 19 3.2x10 − 3 T sin 90 7 m / s 1.2x10 o B = 1.60x10 F N − 15 N E B = 6.1x10 F It may seem like a small force, but it acts on a particle of small 2011/2012 mass, producing a large acceleration
EXAMPLE 3 S C An electron propagates with velocity of 4 x 10^7 I m/s perpendicular with B = 1.5 T esla. Calculate S the electron’s orbit radius? Y H Where mass of electron, m = 9.11x10^-31 kg P Solution: G N I R E E N I G N E 2011/2012
Applications: Charged Particles in S Magnetic Fields C I A cathode-ray tube, such as a television or S computer monitor, uses a magnet to direct a Y beam of electrons to different spots on a H fluorescent screen , creating an image. P G N I R E E N I G N E 2011/2012
S C I A velocity selector consists of an electric and magnetic field at S right angles to each other. Ions entering the selector will Y experience an electric force: H P G N and a magnetic force: I R These two forces will be E perpendicular to each other. E N E I G N E B 2011/2012
MAGNETIC FORCES ON S CURRENT-CARRYING WIRES C The magnetic force on a current-carrying wire is a I S consequence of the forces on the charges . Y The force on an infinitely long wire would be H infinite; the force on a length L of wire is: P G N I θ is the angle between I and B . R E E N If current, I , and magnetic I G field, B , is perpendicular to each other (θ = 90°), thus N F = ILB E 2011/2012
The direction of the force is given by a right- S hand rule: C When the fingers of the right hand are pointed in the direction of I S the conventional current I and then curled toward the vector B, the Y extended thumb points in the direction of the magnetic force on the H wire. P G N I R E E N I G N E Can you used “Left-Hand Rule”? 2011/2012
Additional: Fleming’s Left-Hand Rule S C Fleming's left hand rule ( for electric motors ) I S shows the direction of the Y thrust ( F ) on a conductor carrying a current ( I ) H in a magnetic field ( B ). P G N I R E E N John Ambrose Fleming (1849 – 1949) I G N E Named after British engineer John Ambrose Fleming who invented them. 2011/2012
EXAMPLE 4 S C A straight, horizontal stretch of copper wire I S has a current I = 28A through it. What are the Y magnitude and direction of the minimum H magnetic field B needed to suspend the wire, P that is, balance its weight? Its linear density is G 46.6 g/m. N I R F B E E N B Length, L I G N mg E [side view] 2011/2012
Solution: S C The wire with length L , and the current is out of I S page. If the field is to be minimal, the force F B Y that is exerts on the section must be upward. B H to be horizontal. P In order to balance the weight of the section, F B G must have the N magnitude F B = mg , then I R The direction of F B is related to the direction of E E B and wire's length, L, thus N iLB sin = mg I G find magnetic field B , N B = mg E iL sin 2011/2012
Solution: S C − 3 kg / m 9.8 m / s 2 B = 46.6x 10 I S o 28A sin 90 Y H − 2 T B = 1.6x10 P G N I R E E N I G N E 2011/2012
Applications: Current-Carrying Wires S in Magnetic Fields C I S The Galvanometer : The Foundation of the Y Ammeter & Voltmeter H P A galvanometer has a coil in a magnetic field. G When current flows in the coil, the deflection is N proportional to the current. I R E E N I G N E Sensitive Large Volt/Ampere Galvanometer Galvanometer 2011/2012
The dc Motor S C An electric motor converts electric energy I S into mechanical energy, using the torque on a Y current loop. H P G N I R E E N I G N E 2011/2012
Recommend
More recommend