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- Title: Week7_2
- Type: Notes
- School: Drexel
- Course: PHYS 102
- Term: Spring
the Consider current in the length of wire shown in Figure 22.24. Rank the points A, B, and C, in terms of magnitude of the magnetic field due to the current in the length element shown, from greatest to least. 1. 2. 3. A>B>C B>C>A C>A>B 33% 33% 33% 10 r r I ds r ^ o dB = 4 r 2 C A > > B > C > A B C > A > B 1 Total Magnetic Field To find the total field, you need to sum up the contributions from all the current elements You need to evaluate the field by integrating over the entire current distribution The magnitude of the field in an infinitely long straight wire will be oI B= 2 r 2 B Compared to E Distance The magnitude of the magnetic field varies as the inverse square of the distance from the source The electric field due to a point charge also varies as the inverse square of the distance from the charge r r F q e ^ E= = ke 2 r qo r r r I ds r ^ o dB = 4 r 2 3 B Compared to E, 2 Direction The electric field created by a point charge is radial in direction The magnetic field created by a current element isrperpendicular to both the length ^ element ds and the unit vector r r r F q e ^ E= = ke 2 r qo r r r I ds r ^ o dB = 4 r 2 4 B Compared to E, 3 Source An electric field is established by an isolated electric charge The current element that produces a magnetic field must be part of an extended current distribution Therefore you must integrate over the entire current distribution r r F q e ^ E= = ke 2 r qo r r r I ds r ^ o dB = 4 r 2 5 B for a Long, Straight Conductor, Direction The magnetic field lines are circles concentric with the wire The field lines lie in planes perpendicular to to wire The magnitude of the field is constant on any circle of radius a The right hand rule for determining the direction of the field is shown o I B= 2 r 6 B for a Circular Current Loop The loop has a radius of R and carries a steady current of I r Find B at point P oIR 2 Bx = 3 2 2 2( x + R ) 2 r r I ds r ^ o dB = 4 r 2 7 Field at the Center of a Loop Consider the field at the center of the current loop At this special point, x = 0 Then, Bx = oIR 2 2 x2 + R2 ( ) 3 = 2 o I 2R 8 Magnetic Field Lines for a Loop Figure a shows the magnetic field lines surrounding a current loop Figure b shows the field lines in the iron filings Figure c compares the field lines to that of a bar magnet 9 Problem 22.27. A conductor consists of a circular loop of radius R and two straight, long sections as shown in Figure P22.27. The wire lies in the plane of the paper and carries a current I. Find an expression for the vector magnetic field at the center of the loop. 10 Magnetic Force Between Two Parallel Conductors Two parallel wires each carry a steady current r The field B2 due to the current in wire 2 exerts a force on wire 1 of F1 = I1l B2 11 Magnetic Force Between Two Parallel Conductors, cont Substituting the equation for B2 gives oI1I2 l F1 = 2 a Parallel conductors carrying currents in the same direction attract each other Parallel conductors carrying current in opposite directions repel each other 12 Magnetic Force Between Two Parallel Conductors, final The result is often expressed as the magnetic force between the two wires, FB This can also be given as the force per unit length, FB/l FB oI1I2 = 2a l 13 Problem 22.35. In Figure P22.35, the current in the long, straight wire is I1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries the current I2 = 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. 14 Definition of the Ampere The force between two parallel wires can be used to define the ampere When the magnitude of the force per unit length between two long parallel wires that carry identical currents and are separated by 1 m is 2 x 10-7 N/m, the current in each wire is defined to be 1 A 15 Definition of the Coulomb The SI unit of charge, the coulomb, is defined in terms of the ampere When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross section of the conductor in 1 s is 1 C 16 A loose spiral spring is hung from the ceiling and a large current is sent through it. Do the coils 1. 2. 3. move closer together move farther apart, or not move at all? er to g 33% 33% 33% or et he al l? no tm ov e at r ov e m m ov e fa rth e cl os ra 10 17 pa rt , Magnetic Field of a Wire A compass can be used to detect the magnetic field When there is no current in the wire, there is no field due to the current The compass needles all point toward the earth's north pole Due to the earth's magnetic field 18 Magnetic Field of a Wire, 2 The wire carries a strong current The compass needles deflect in a direction tangent to the circle This shows the direction of the magnetic field produced by the wire 19 Magnetic Field of a Wire, 3 The circular magnetic field around the wire is shown by the iron filings 20 -Marie Andr Amp re 1775 1836 Credited with the discovery of electromagnetism The relationship between electric currents and magnetic fields Died of pneumonia 21 Ampere's Law r r The product of B ds can be evaluated for r small length elements ds on the circular path defined by the compass needles for the long straight wire Ampere's Law states that the line integral of r r B ds around any closed path equals oI where I is the total steady current passing through any surface bounded by the closed r r path B ds = o I 22 r r Rank the values of B d s for the closed paths in Figure 22.29, from smallest to largest. 1. 2. 3. b, d, a, c A, b, d ,c C, d, a, b 33% 33% 33% 10 ,c d, ,b b, A C ,d ,a ,b a, c ,d 23 r r Rank the values of B d s for the closed paths in Figure 22.30, from smallest to largest. 1. 2. 3. B, a=c=d A=c=d, b a < b< c <d 33% 33% 33% 10 =c =d d, b =c = B ,a A a < b< c <d 24 Ampere's Law, cont Ampere's Law describes the creation of magnetic fields by all continuous current configurations Most useful for this course if the current configuration has a high degree of symmetry Put the thumb of your right hand in the direction of the current through the amperian loop and your figures curl in the direction you should integrate around the loop 25 Amperian Loops Each portion of the path satisfies one or more of the following conditions: The value of the magnetic field can be argued by symmetry to be constant over the portion of the path The dot product can be expressed as a simple algebraic product B ds The vectors are parallel 26 Amperian Loops, cont Conditions: The dot product is zero The vectors are perpendicular The magnetic field can be argued to be zero at all points on the portion of the path 27 Field Due to a Long Straight Wire From Ampere's Law Want to calculate the magnetic field at a distance r from the center of a wire carrying a steady current I The current is uniformly distributed through the cross section of the wire 28 Field Due to a Long Straight Wire Results From Ampere's Law Outside of the wire, r > R r r B ds = B(2 r ) = oI oI B= 2 r Inside the wire, we need I', the current inside the amperian circle r r B ds = B(2 r ) = oI ' I B = o 2 r 2 R r2 I' = 2 I R 29 Field Due to a Long Straight Wire Results Summary The field is proportional to r inside the wire The field varies as 1/r outside the wire Both equations are equal at r = R 30 Magnetic Field of a Toroid Find the field at a point at distance r from the center of the toroid The toroid has N turns of wire o NI B= 2 r 31 r r B ds = B(2 r ) = oNI Magnetic Field of a Solenoid A solenoid is a long wire wound in the form of a helix A reasonably uniform magnetic field can be produced in the space surrounded by the turns of the wire Each of the turns can be modeled as a circular loop The net magnetic field is the vector sum of all the fields due to all the turns 32 Magnetic Field of a Solenoid, Description The field lines in the interior are Approximately parallel to each other Uniformly distributed Close together This indicates the field is strong and almost uniform 33 Magnetic Field of a Tightly Wound Solenoid The field distribution is similar to that of a bar magnet As the length of the solenoid increases The interior field becomes more uniform The exterior field becomes weaker 34 Ideal Solenoid Characteristics An ideal solenoid is approached when The turns are closely spaced The length is much greater than the radius of the turns For an ideal solenoid, the field outside of solenoid is negligible The field inside is uniform 35 Ampere's Law Applied to a Solenoid Ampere's Law can be used to find the interior magnetic field of the solenoid Consider a rectangle with side l parallel to the interior field and side w perpendicular to the field The side of length l inside the solenoid contributes to the field This is path 1 in the diagram 36 Ampere's Law Applied to a Solenoid, cont Applying Ampere's Law gives r r B ds = path1 r r B ds = B path1 ds = Bl The total current through the rectangular path equals the current through each turn multiplied by the number of turns r r B ds = Bl = oNI 37 Magnetic Field of a Solenoid, final Solving Ampere's Law for the magnetic field is N B = o I = o nI l n = N / l is the number of turns per unit length This is valid only at points near the center of a very long solenoid 38 Consider a solenoid that is very long compared with the radius. Of the following choices, the most effective way to increase the magnetic field in the interior of the solenoid is to 1. 2. 3. its ra di ub le du c do re ov er w ra p e its th e le double its length, keeping the number of turns per unit length constant, reduce its radius by half, keeping the number of turns per unit length constant, or overwrap the entire solenoid with an additional layer of current-carrying wire. 33% 33% 33% ha ... ... ke en tir e ... ng th , 10 39 us by
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