SCS CS 139 39 App Appli lied ed Ph Physics ysics II II Dr. - PowerPoint PPT Presentation

Magnetic Force on a Current-Carrying Wire  Note that it does not matter whether we consider negative charges drifting downward in the wire (the actual case) or positive charges drifting upward. The direction of the deflecting force on the wire is the same.  If the wire is not straight or the field is not uniform,  we can imagine the wire broken up into small straight segments.  In the differential limit, we can write   dF idL B  The force on the wire as a whole is then the vector sum of all the forces on the segments that make it up. 39

Torque on a Current Loop  Electric motor.  A rectangular loop of wire, carrying a current and free to Side 4 rotate about a fixed axis, is placed in a magnetic field. Side 2  Magnetic forces on the wire produce a torque that rotates the loop. 40

Normal vector  To define the orientation of the loop in the magnetic field, we use a normal vector 𝒐 that is perpendicular to the plane of the loop.  Useful Fact: The loop will rotate so that 𝑜 has the same direction Right-hand rule: To find the direction as 𝐶 . of 𝑜 , point or curl the fingers of your right hand in the direction of the current at any point on the loop. Your extended thumb then points in the direction of the normal vector . 41

Forces The normal vector 𝑜 is shown at an arbitrary angle  to the direction of 𝐶 . 1 = 𝐺 3 = 𝑗𝑏𝐶 𝐺 2 = 𝐺 𝐺 4 = 𝑗𝑐𝐶 sin 90 − 𝜄 = 𝑗𝑐𝐶 cos 𝜄 2 and 𝐺  𝐺 4 have the same magnitude but opposite directions. Thus, they cancel out exactly. Their net force is zero. 3 have the same magnitude but opposite directions.  𝐺 1 and 𝐺 Thus, they do not tend to move the loop up or down. 42

Torques 2 and 𝐺  𝐺 4 : Their common line of action is through the center of the loop. Therefore, their net torque is also zero. 1 and 𝐺 3 : Do not share the  𝐺 same line of action; so they produce a net torque.     b           2 sin sin sin   r F  iaB  iabB iAB     2 Area enclosed by the coil 43

General formula  Use a coil of N loops or turns  Flat coil assumption: Assume that the turns are wound tightly enough that they can be approximated as all having the same dimensions and lying in a plane.  Total torque    sin NiAB Area enclosed by the coil  The formula holds for all flat coils no matter what their shape. 2 sin      For the circular coil, we have Ni r B 44

Strong Magnets Can be Dangerous magnet Think of the pencil as your finger 45

Strong Magnets Can be Dangerous 46

SCS CS 139 39 II.2 .2 Ma Magne netic tic For orce ces s an and Fields elds (I (II) Dr. Prapun Suksompong   ˆ i ds r  prapun@siit.tu.ac.th 0 dB  2 4 r     B ds i 0 enc Office Hours: Library (Rangsit) Mon 16:20-16:50 BKD 3601-7 Wed 9:20-11:20 1

Reference  Principles of Physics  Ninth Edition, International Student Version  David Halliday, Robert Resnick, and JearlWalker  Chapter 29 Magnetic Fields due to Currents  29-2 Calculating the Magnetic Field Due to a Current  29-3 Force Between Two Parallel Currents  29- 4 Ampere’s Law  29-6 A Current-Carrying Coil as a Magnetic Dipole 2

Magnetic Field Due to a Current in a Long Straight Wire  Fact: When a current flows through a wire, it can produce a magnetic field.  For long (infinite) straight wire carrying a current i , i  B R Perpendicular distance 3

Another Right-Hand Rule Wrap your right hand around the wire with your thumb in the direction of the current. The fingers reveal the field vector’s direction, which is tangent to a circle. 4

Application: Magnetic field of two wires  Computer cables, or cables for audio-video equipment, create little or no magnetic field.  This is because within each cable, closely spaced wires carry current in both directions along the length of the cable. The magnetic fields from these opposing currents cancel each other. 2

Contribution of a small piece Biot-Savart Law : The contribution 𝑒 𝐶 to the magnetic field produced by a current-length element 𝑗 𝑒 𝑡 at a point P located a distance r from the current element can be found by: (experimentally deduced) permeability constant = 4  ×10 -7 T∙m /A  1.26×10 -6 T∙ m/A   ˆ i ds r  0 dB  2 4 r   sin ids  0 dB  2 4 r 5

Recipe To find the magnetic field produced at a point  Step 1: Find the contribution from each (single) current- length element   ˆ i ds r  0 dB  2 4 r  Step 2: Integrate (superimposing/adding/summing the contributions from all current-length elements) to find the net field produced by all the current-length elements.   B dB 6

Ex 1: Long (infinite) straight wire   sin( ) i ds  0 dB  2 4 r      sin( ) i i R r     0 0 2 B ds ds   2 2 4 4 r r  0     1 1 iR iR     0 0 ds ds   3 3   2 2 r  2 2 0 0 2 s R   1 iR i   0 0   2 2 2 R R 7 Difficult. Easier to derive via the Ampere’s law. (TBD)

Ex 2: Easy case  Consider a wire consists of two straight sections (1 and 2) and a circular arc (3), and carries current i .  What magnetic field (magnitude and direction) does the current produce at C?  For now, focus on the two straight sections (1 and 2).  Their extensions intersect the center C of the arc. Conclusion: Current directly toward or away from C does not create any magnetic field there. 8

Ex 3: At center of a circular arc of wire  Consider only point P at the center of the arc (point C)     sin(90 ) i ds i ds i R d    0 0 0 dB    2 2 2 4 4 4 r R R  i   0 B  4 R 9

Ex 2: (a revisit)  Consider a wire consists of two straight sections (1 and 2) and a circular arc (3), and carries current i .  What magnetic field (magnitude and direction) does the current produce at C?   1 i i     0 0 0 0 B 2 4 8 R R 10

Ex 4: Magnetic Field of a Coil Consider point P on the central perpendicular axis of the loop  2 o iR  B 3    2 2 2 2 R z A Current-Carrying Coil as a Magnetic Dipole : Observation: One side of the loop acts as a north pole and the other side as a south pole, as suggested by the lightly drawn magnet. 11

Ex 4: Derivation From the symmetry, the vector sum of all the perpendicular components due to all the loop elements is zero.     sin 90 i ds i   0 0 dB ds   2 2 4 4 r r  i         0 cos cos dB d B ds  // 2 4 r   iR iR   0 0 ds ds  3 3   4 r R       2 2 4 cos R z 2 r    2 iR i R iR          0 0 0 2 B dB ds R // 3 3 3            2 2 2 2 2 2 4 4 2 2 2 2 R z R z R z 12

Force Between Two Parallel Currents  Two long parallel wires carrying currents exert forces on each other.       i Li i    0 0 a a b sin 90 F i LB i L     ba b a b   2 2 d d 13

Force Between Two Parallel Currents  Two long parallel wires carrying currents exert forces on each other.  Two wires with parallel currents attract each other.  Two wires with antiparallel currents repel each other. 14

Applications  Basis for the definition of ampere  The ampere is that constant current which, if maintained in two straight, parallel conductors of infinite length, of negligible circular cross section, and placed 1 m apart in vacuum, would produce on each of these conductors a force of magnitude 2×10 -7 N/m of wire length.  Rail gun 15

Exercise  Three long straight parallel equally-spaced wires with identical currents either into or out of the page.  Rank the wires according to the magnitude of the force on each due to the currents in the other two wires, greatest first. 16

Net 𝐶  Net magnetic field due to the three currents: i 1 i 3 𝑪 i 2 17

Ampere’s Law Net current encircled by (passing through) the loop     B ds i 0 enc    cos d B s The field component tangent to the loop 18

i enc  Use the curled – straight right-hand rule to determine the signs for currents   i i i 1 2 enc  May have to integrate the current density.    i J dA enc  For conducting cylinder,       2 i J r r dr enc Remark: The integration is over the area encircled by the loop. 19

Exercise 20

Choosing the Amperian Loop  Use one that create symmetry.  Draw the loop so that 𝐶cos𝜄 is constant.    In which case,    cos B ds B ds In addition, if the magnetic field is always tangent to the loop     B ds B ds 21

Example: Long Straight Wire with Current  B outside:        2 B r B ds i 0  i  0 B  2 r  B inside : With uniformly distributed current,    2 r        2   B r B ds i  0 2   R  i r  0 B  2 2 R 22

Application  Magnetic fields are associated with a signal-carrying coaxial cable.  If the current is the same magnitude in each direction, the magnetic field outside the coaxial cable is zero.  The absence of 𝐶 fields around a coaxial cable results in no interference in nearby electrical equipment and wires 23

Ex: Magnetic Field of a Solenoid  The field inside the coil is fairly strong and uniform over the cross section of the coil.  At points inside and reasonably far from the wire, 𝐶 is approximately A solenoid carrying current i. parallel to the (central) solenoid axis.  The external field, however, is relatively weak. A vertical cross section through the central axis of a “stretched -out ” solenoid. 24

Ex: Long Ideal Solenoid  Infinitely long  Consist of tightly packed (close-packed) turns of square wire  The magnetic field outside the solenoid is zero.  This is practically holds for real solenoid if  its length is much greater than its diameter  we consider points that are well away from the solenoid ends     B ds i 0 enc #turns per unit   length   Bh i h n 0   B i n 0 25

SCS CS 139 39 II.3 .3 Induction duction an and Inductance ductance Dr. Prapun Suksompong  d   prapun@siit.tu.ac.th B dt di   L dt L Office Hours: Library (Rangsit) Mon 16:20-16:50 BKD 3601-7 Wed 9:20-11:20 1

Reference  Principles of Physics  Ninth Edition, International Student Version  David Halliday, Robert Resnick, and Jearl Walker  Chapter 30  30-2 Two Experiments  30- 3 Faraday’s Law of Induction  30- 4 Lenz’s Law  30-7 Inductors and Inductance  30-8 Self-Induction  30-9 RL Circuits 2

Review + New Fact  Review  Force occurs when a charged particle moves through a magnetic field.  Force occurs when a current-carrying wire is placed in a magnetic field.  Magnetic field is found around a current-carrying wire.  New Fact: Change in magnetic field can produce ( induce ) a current in a loop of wire 3

Application: TMS  TMS = Transcranial Magnetic Stimulation  A technique for studying (or stimulating or deactivating (suppressing)) the function of various parts of the brain.  A coil held to the subject’s head carries a varying electric current, and so produces a varying magnetic field. This field causes an induced emf, and that triggers electric activity in the region of the brain underneath the coil. 14

Experiment 1  Moving a magnet bar toward or away from a (conducting) loop of wire can produces (induces) a current in the loop.  The current produced in the loop is called an induced current .  Observation:  Current only occurs when there is a relative motion between the loop and the magnet.  Faster motion produces a greater current  Direction (CW or CCW) of the (induced) current depends on the direction of motion and polarity of the magnet. 4

Experiment 2  When the switch is suddenly closed (i.e. current flows through the right-hand loop) the ammeter will show a brief current appearing in the left-hand loop.  When the switch is suddenly opened (i.e. no current flows through the right-hand loop) the ammeter will again show a brief current appearing in the left-hand loop, but in the opposite direction. 5

Induction  The current produced in the loop is called an induced current .  The work done per unit charge to produce that current (to move the conduction electrons that constitute the current) is called an induced emf .  The process of producing the current and emf is called induction .  Faraday’s law of induction : An emf is induced in a loop when the (number of magnetic field line) amount of magnetic field that passes through the loop is changing. 6

Magnetic flux  Need to quantify the amount of magnetic field that passes through the loop  Magnetic flux through a loop enclosing an area A Vector of magnitude dA that is Dot product perpendicular to a differential area dA     B d A B  Unit: weber (Wb)  1 weber = 1 Wb = 1 T·m 2 7

Changing magnetic flux through a coil Here are the general means by which we can change the magnetic flux through a coil:  Change the magnitude B of the magnetic field within the coil.  Change either the total area of the coil or the portion of that area that lies within the magnetic field (for example, by expanding the coil or sliding it into or out of the field).  Change the angle between the direction of the magnetic field and the plane of the coil (for example, by rotating the coil 𝐶 is first perpendicular to the plane of the coil so that field 𝐶 and then is along that plane). 8

Faraday’s law of induction  d   B dt  The magnitude of the emf induced in a conducting loop is equal to the rate at which the magnetic flux 𝛸 𝐶 through the loop changes with time.  The negative sign is there because the induced emf tends to oppose the flux change. (TBD)  If we change the magnetic flux through a coil of N turns , an induced emf appears in every turn and the total emf induced in the coil is the sum of these individual induced emfs Assume that the coil is  tightly wound (closely d   B N packed), so that the same dt magnetic flux passes 9 through all the turns.

Exercise  The graph gives the magnitude B(t) of a uniform magnetic field that exists throughout a conducting loop,with the direction of the field perpendicular to the plane of the loop.  Rank the five regions of the graph according to the magnitude of the emf induced in the loop, greatest first. 10

Ex: Induced emf due to a solenoid  A long solenoid has 220 turns/cm and carries a current i =1.5 A; its diameter D is 3.2 cm.  At the center we place a 130 turn closely packed coil C of diameter d = 2.1 cm. The current in the long solenoid is reduced to zero at a constant rate in 25 ms.  What is the magnitude of the induced emf in coil C? 11

Lenz’s law  “An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current.” 12

Ex: Lenz’s law . Note carefully that the flux of 𝐶 𝑗𝑜𝑒 always opposes the change in the flux of 𝐶 . Does not mean that 𝐶 𝑗𝑜𝑒 always points opposite 𝐶 13

Ex. Induction due to a changing B  Consider a conducting loop consisting of a half-circle of radius r = 0.20 m and three straight sections. The half-circle lies in a uniform magnetic field that is directed out of the page; the field magnitude is given by B = 4.0 t 2 +2.0 t + 3.0, with B in teslas and t in seconds.  An ideal battery with emf ℰ 𝑐𝑏𝑢 = 2.0 V is connected to the loop.  The resistance of the loop is 2.0 Ω 14

Ex. Induction due to a changing B  r = 0.20 m, B = 4.0 t 2 +2.0 t + 3.0, ℰ 𝑐𝑏𝑢 = 2.0 V , R = 2.0 Ω What are the magnitude and direction of the emf ℰ 𝑗𝑜𝑒 a) at t = 10 s? induced around the loop by field 𝐶 What is the current in the loop at t = 10 s? b) 15

Self-Induction  An induced emf appears in any coil in which the current is changing.  This process is called self-induction .  The emf that appears is called a self-induced emf .  Still obeys Faraday’s law and Lenz’s law. 16

Inductors and Inductance  An inductor is an electrical component typically made by coiling a conductor around a core.  Solenoid is our basic type of inductor.  The inductance of the inductor is the number of turns  N  B L i  Unit: henry B N       1 henry = 1 H = 1 T∙m 2 /A. n in A  0 L solenoid i   2 n A 0 17

Inductor: self-induced emf  Let’s combine self induction and inductance.  In any inductor (such as a solenoid) a self-induced emf appears whenever the current changes with time.       d N d Li d di         B B N L L dt d t dt dt  N  Faraday’s law Inductance B L i  The direction can be obtained by Lenz’s law .  For an ideal inductor with negligible resistance, the magnitude of the potential difference V L across the inductor is equal to the magnitude of the self-induced emf ℰ 𝑀 . 18

RL Circuit  Kirchhoff’s voltage law 19

Example. RL L ci circu cuit it  Consider a circuit that contains three identical resistors with resistance R = 9.0 , two identical inductors with inductance L = 2.0 mH, and an ideal battery with emf ℰ = 18 V .  What is the current i through the battery just after the switch is closed? 20

Example. RL L ci circu cuit it  What is the current i through the battery long after the switch has been closed? 21

SCS CS 139 39 II.4 .4 Alt lterna ernati ting ng Cur urrent rent  v i R Dr. Prapun Suksompong R R dv prapun@siit.tu.ac.th  C i C dt C di  L v L dt L Office Hours: Library (Rangsit) Mon 16:20-16:50 BKD 3601-7 Wed 9:20-11:20 1

Reference  Principles of Physics  Ninth Edition, International Student Version  David Halliday, Robert Resnick, and JearlWalker  Chapter 31  31-6 Alternating Current  31-7 Forced Oscillations  31-8 Three Simple Circuits 2

Alternating-Current Generator  A conducting loop rotates (with constant angular speed  ) in an external (uniform and constant) magnetic field.  Connections from each end of the loop to the external circuit are made by means of that end’s slip ring. 3