General Physics (PHY 2140) Lecture 12 Electricity and Magnetism Magnetism Magnetic fields and force Application of magnetic forces http://www.physics.wayne.edu/~apetrov/phy2140/ Chapter 19 1
Department of Physics and Astronomy announces the Fall 2003 opening of The Physics Resource Center on Monday, September 22 in Room 172 of Physics Research Building. Hours of operation: Mondays, Tuesdays, Wednesdays Thursdays and Fridays 11 AM to 6 PM 11 AM to 3 PM Undergraduate students taking PHY2130-2140 will be able to get assistance in this Center with their homework, labwork and other issues related to their physics course. The Center will be open: Monday, September 22 to Wednesday, December 10, 2003. 2
Lightning Review Last lecture: 1. DC circuits Kirchoff s rules RC circuit 2. Magnetism Magnets n I n = 0, V = 0 i i= 1 i= 1 t/ RC ( 1 ) q= Q e q = Qe Review Problem: The three light bulbs in the circuit all have the same resistance. Given that brightness is proportional to power dissipated, the brightness of bulbs B and C together, compared with the brightness of bulb A, is 1. twice as much. 2. the same. 3. half as much. t/ RC i 3
Last lecture: Magnetic Field Convenient to describe the interaction at a distance between magnets with the notion of magnetic field. Magnetic objects are surrounded a magnetic field. Moving electrical charges are also surrounded by a magnetic field (in addition to the electrical field). A vector quantity: magnitude and direction The letter B is used to represent magnetic fields. 4
Magnetic Field Direction The magnetic field direction (of a magnet bar) can studied with a small compass. N 1 S 5
Magnetic Field Lines N 1 S 6
Applications: A bit of history IBM introduced the first hard disk in 1957,, when data usually was stored on tapes. It consisted of 50 platters,, 24 inch diameter, and was twice the size of a refrigerator. It cost $35,000 annually in leasing fees (IBM would not sell it outright). It s total storage capacity was 5 MB,, a huge number for its time! 7
Magnetic Field of the Earth A small magnetic bar should be said to have north and south seeking poles. The north of the bar points towards the North of the Earth. The geographic north corresponds to a south magnetic pole and the geographic south corresponds to a magnetic north. The configuration of the Earth magnetic resemble that of a (big) magnetic bar one would put in its center. 8
Magnetic Field of the Earth 9
Magnetic Field of the Earth Near the ground, the field is NOT parallel to the surface of the Earth. The angle between the direction of the magnetic field and the horizontal is called dip angle. The north and south magnetic pole do not exactly correspond to the south and north geographic north. South magnetic pole found (in 1832) to be just north of Hudson bay in Canada 1300 miles from the north geographical pole. 10
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More on the Magnetic Field of the Earth The difference between the geographical north and the direction pointed at by a compass changes from point to point and is called the magnetic declination. Source of the field : charge-carrying carrying convection currents in the core of the earth. In part related to the rotation of the earth The orientation of the field flips and changes over time every few million years Basalt rocks Other planets (e.g. Jupiter) are found to have a magnetic field. 12
Mini-quiz You travel to Australia for a business trip and bring along your American-made made compass. Does the compass work correctly in Australia??? No problem using the compass in Australia. North pole of the compass will be attracted to the South geographic pole The vertical component of the field is different (opposite) but that cannot be detected with normal operation of the compass. 13
19.3 Magnetic Fields Stationary charged particles do NOT interact with a magnetic field. Charge moving through a magnetic field experience a magnetic force. Value of the force is maximum when the charge moves perpendicularly to the field lines. Value of the force is zero when the charge moves parallel to the field lines. 14
Magnetic Fields in analogy with Electric Fields Electric Field: Distribution of charge creates an electric field E(r) in the surrounding space. Field exerts a force F=q E(r) on a charge q at r Magnetic Field: Moving charge or current creates a magnetic field B(r) in the surrounding space. Field exerts a force F on a charge moving q at r 15
Strength of the Magnetic Field Define the magnetic field, B, at a given point in space in terms of the magnetic force imparted on a moving charge at that point. Observations show that the force is proportional to The field The charge The velocity of the particle The sine of the angle between the field and the direction of the particle s motion. 16
Strength and direction of the Magnetic Force on a charge in motion F F = qvbsinθ B +q v 17
Magnetic Field Magnitude B F = qv sin θ 18
Magnetic Field Units [F] = newton [v] = m/s [q] = C [B] = tesla (T). Also called weber (Wb)) per square meter. 1 T = 1 Wb/m 2. 1 T = 1 N s m - 1 C - 1. 1 T = 1 N A - 1 m - 1. CGS unit is the Gauss (G) 1 T = 10 4 G. 19
Right Hand Rule Provides a convenient trick to remember the spatial relationship between F, v, and B. Consider the motion of positive charge Direction of force reversed if negative charge. 20
Example: Proton traveling in Earth s magnetic field. A proton moves with a speed of 1.0 x 10 5 m/s through the Earth s magnetic field which has a value of 55 µt a particular location. When the proton moves eastward, the magnetic force acting on it is a maximum, and when it moves northward, no magnetic force acts on it. What is the strength of the magnetic force? And what is the direction of the magnetic field? V = 1.0 x 10 5 m/s B = 55 µt F= qvbsinθ ( 1.6 10 19 )( 8.0 10 6 / )( 2.5 )( o sin60 ) F= C m s T = 2.8 10 12 N Northward or southward. 21
19.4 Magnetic Force on Current-carrying conductor. A magnetic force is exerted on a single charge in motion through a magnetic field. That implies a force should also be exerted on a collection of charges in motion through a conductor I.e. a current. And it does!!! The force on a current is the sum of all elementary forces exerted on all charge carriers in motion. 22
19.4 Magnetic Force on Current If B is directed into the page we use blue crosses representing the tail of arrows indicating the direction of the field, If B is directed out of the page, we use dots. If B is in the page, we use lines with arrow heads. x x x x x x x x x x x x x x x x x x x x x x x x........................ 23
Force on a wire carrying current in a magnetic field. B in x x x x x x x x x x x x x x x x x x x x x x x x B in x x x x x x x x x x x x x x x x x x x x x x x x B in x x x x x x x x x x x x x x x x x x x x x x x x I = 0 I I 24
Force on a wire carrying current in a magnetic field. A x x x x x x x x x x x x vx x x x d x x q x x x x x x x x x x x x x x x x x x x x x x x x ( )( ) F = qv B nal max d I = Fmax nqv A = d BIl Magnetic Field and Current at right angle from each other. 25
Force on a wire carrying current in a magnetic field. General Case: field at angle θ relative to current. F = BIlsinθ max B θ B sin θ I 26
Voice Coil 27
Mini-Quiz In a lightning strike, there is a rapid flow of negative charges from a cloud to the ground. In what direction is a lightning strike deflected by the Earth s magnetic field? Reasoning: Negative charge flow down. Positive Current upward. B field direction Geo South to Geo North Answer: Force towards the west. I 28
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Example: Wire in Earth s B Field A wire carries a current of 22 A from east to west. Assume that at this location the magnetic field of the earth is horizontal and directed from south to north, and has a magnitude of 0.50 x 10-4 T. Find the magnetic force on a 36-m m length of wire. What happens if the direction of the current is reversed? B=0.50 x 10-4 T. I = 22 A l = 36 m F max = BIl F max = BIl ( 4 0.50 10 T)( 22A)( 36m) = = 4.0 10 2 N 30
19.5 Torque on a Current Loop Imagine a current loop in a magnetic field as follows: B I B F F b a/2 a F F 31
B I B F F b a/2 a F F F1 = F2 = BIb ( ) ( ) τ = F + F = BIb + BIb a a a a max 1 2 2 2 2 2 τ max = BIba = BIA τ = BIAsinθ 32
In a motor, one has N loops of current τ = NBIAsinθ 33
Example: Torque on a circular loop in a magnetic field A circular loop of radius 50.0 cm is oriented at an angle of 30.0 o to a magnetic field of 0.50 T. The current in the loop is 2.0 A. Find the magnitude of the torque. B 30.0 o r = 0.500 m θ = 30 o B = 0.50 T I = 2.0 A N = 1 τ = = τ = NBIAsinθ ( 0.50T)( 2.0A) π ( 0.5 m) 2 0.39Nm 0 sin30.0 o 34
19.6 Galvanometer/Applications Device used in the construction of ammeters and voltmeters. Scale Current loop or coil Magnet Spring 35
Galvanometer used as Ammeter Typical galvanometer have an internal resistance of the order of 60 W - that could significantly disturb (reduce) a current measurement. Built to have full scale for small current ~ 1 ma or less. Must therefore be mounted in parallel with a small resistor or shunt resistor. Galvanometer 60 Ω R p 36
Galvanometer 60 Ω R p Let s convert a 60 W, 1 ma full scale galvanometer to an ammeter that can measure up to 2 A current. R p must be selected such that when 2 A passes through the ammeter, only 0.001 A goes through the galvanometer. ( 0.001A)( 60Ω ) = ( 1.999 ) R p = 0.03002Ω AR R p is rather small! The equivalent resistance of the circuit is also small! p 37
Galvanometer used as Voltmeter Finite internal resistance of a galvanometer must also addressed if one wishes to use it as voltmeter. Must mounted a large resistor in series to limit the current going though the voltmeter to 1 ma. Must also have a large resistance to avoid disturbing circuit when measured in parallel. R s Galvanometer 60 Ω 38
R s Galvanometer 60 Ω Maximum voltage across galvanometer: ( )( ) Vmax = 0.001A 60Ω = 0.06V Suppose one wish to have a voltmeter that can measure voltage difference up to 100 V: ( )( ) p 100V = 0.001A R + 60Ω R p = 99940Ω Large resistance 39
19.7 Motion of Charged Particle in magnetic field Consider positively charge particle moving in a uniform magnetic field. Suppose the initial velocity of the particle is perpendicular to the direction of the field. Then a magnetic force will be exerted on the particle and make follow a circular path. B in q v r F 40
The magnetic force produces a centripetal acceleration. F = qvb= 2 mv r The particle travels on a circular trajectory with a radius: r = mv qb 41
Example: Proton moving in uniform magnetic field A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of magnitude 0.35 T, directed perpendicular to the velocity of the proton. Find the orbital speed of the proton. r = 0.14 m B = 0.35 T m = 1.67x10-27 kg q = 1.6 x 10-19 C r = mv qb v qbr = m = ( 19 )( )( 2 1.6 10 C 0.35T 14 10 m) ( 27 1.67 10 kg) 6 4.7 10 m s = 42