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Sources of Magnetic Field Chapter 28 • Study the magnetic field generated by a moving charge • Consider magnetic field of a current-carrying conductor • Examine the magnetic field of a long, straight, current-carrying conductor • Study the magnetic force between currentcarrying conductors • Consider the magnetic field of a current loop • Examine and use Ampere’s Law 1 The magnetic field of a moving charge • A moving charge will generate a magnetic field relative to the velocity of the charge. 2 Magnetic Field of a Moving Charge) Magnitude of B Permeability of free space 0 q v sin B 4 r2 0 4x10 7 (28-1) Direction of B determined by v xr 0 10 7 4 The vector form: 3 Force between two moving protons • • • • Two protons moving at the same velocity (much less than speed of light) in opposite directions. The electric force FE is repulsive. The right-hand rule indicates the magnetic force FM is repulsive. (i x k=-j) Find the ratio of the magnitude of the forces. FB 0 4 FB v2 2 FE c q2 v2 r2 The ratio of the two forces. Where c=speed of light. Therefore: FE>>FB Magnetic Field of a Current Element Total magnetic field of several moving charges = vector sum of fields caused by individual charges Let dQ = charge in wire segment dl Let A = cross section area of wire segment dl Let n = charge density in wire segment dl dQ = nqAdl I = nqvdA Figure 28-3 dQvd sin 0 nqvd Adl sin dB 0 2 4 r 4 r2 0 Idl sin dB (28-5) Biot-Savart Law 2 4 r Direction of dB determined by Idl xr Vector form of Biot-Savart Law 5 Magnetic field of a straight current-carrying conductor • Biot and Savart contributed to finding the magnetic field produced by a single current-carrying conductor. 6 Magnetic Field of a Current-Carrying Conductor 0 Idl sin dB 4 r2 sin sin( ) x x2 y2 dl dy 0 Idy 0 Ix x dy 4 ( x 2 y 2 ) x 2 y 2 4 ( x 2 y 2 )3 / 2 0 Ix a dy B 4 a x 2 y 2 3 / 2 dB Figure 28-5 Based upon symmetry around the y-axis the field will be a circle 0 Ix 1 B 4 x 2 If a x a 0 I 2a y 2 2 2 2 4 x x y a x a 0 I 2a 0 I B 2 4 x a 2 x 7 Magnetic Field of a Current-Carrying Conductor Figure 28-6 where r = perpendicular distance from the current-carrying wire. 8 Force between Parallel Conductors Only field due to I shown Each conductor lies in the field set up by the other conductor ' F I L B Note: If I and I’ are in the same direction, the wires attract. If I and I’ are in opposite directions, the wires repel. 0 I F I LB sin I LB I L 2 r 0 II ' Substitute for B FL 2 r See Example 28.5 Page 966 F 0 II ' L 2 r ' ' ' 9 Magnetic Field of a Circular Current Loop 0 Idl sin dB 4 r2 sin sin 90 1 r 2 x2 a2 0 I dl cos 4 ( x 2 a 2 ) 0 I dl a dBx dB cos 4 ( x 2 a 2 ) ( x 2 a 2 )1/2 dB Figure 28-12 0 I 0 Ia a Bx dl (2a) 2 2 3/ 2 2 2 3/ 2 4 ( x a ) 4 ( x a ) By= 0 10 Magnetic Field of a Circular Current Loop Bx 0 NIa 2 2( x 2 a 2 )3 / 2 (on the axis of N circular loops) (x=0) x Figure 28-13 Figure 28-14 11 Ampere’s Law I—specific then general Similar to electric fields if symmetry exists it is easier to use Gauss’s law 12 Ampere’s Law II • The line integral equals the total enclosed current • The integral is the sum of the tangential B to line path Ampere’s Law (Chapter 28, Sec 6) Figure 28-15 For Figure 28-15a B dl B cos dl Bdl B dl I B 0 2 r dl 2r 0 I B dl B(2r ) 2r (2r ) 0 I For Figure 28-15b B dl 0 I For Figure 28-15c b c d a a b c d B dl B dl B dl B dl B dl b c d a c d B dl B dl 0 dl B dl 0 dl 1 2 a b 0 I 0 I B dl 2r1 (r1 ) 0 2r2 (r2 ) 0 B dl 0 14 Ampere’s Law B dl B cos dl cos dl rd 0 I B 2 r 0 I 0 I B dl 2r (rd ) 2 d Figure 28-16 d 2 I B dl (2 ) I 2 0 0 d 0 B dl 0 15 Applications of Ampere’s Law Example 28-9 Field of a Solenoid (magnetic field is concentrated in side the coil) n = turns/meter Figure 28-20 B dl 0 I encl 0 nLI b Figure 28-21 c d a B dl B dl B dl B dl B dl a b b c c d N n l d a B dl B dl 0 dl 0 dl 0 dl BL a BL 0 nLI b c where N = total coil turns l = total coil length d N B 0 nI 0 I l turns/meter (28-23) 16 Applications of Ampere’s Law Example 28-9 Field of a Solenoid N n l turns/meter where N = total coil turns l = 4a = total coil length Figure 28-22 17 Applications of Ampere’s Law Example 28-10 Field of a Toroidal Solenoid – (field is inside the toroid) N turns Path 1 B dl I 0 encl 0 B0 No current enclosed Path 3 B dl I 0 encl 0 B0 Figure 28-23 Current cancels Path 2 B dl I NI B dl B dl B(2r ) 0 encl B(2r ) 0 NI 0 NI B 0 2r (28-24) 18 Magnetic materials • The Bohr magneton will determine how to classify material. • Ferromagnetic – can be magnetized and retain magnetism • Paramagnetic – will have a weak response to an external magnetic field and will not retain any magnetism • Diamagnetic – shows a weak repulsion to an external magnetic field Bohr Magneton- In atoms electron spin creates current a loop, which produce magnetic their own field 19 Ferromagnetism and Hysteresis loops • • • The larger the loops the more energy that is lost magnetizing and demagnetizing. Soft iron produce small loops and are used for transformers, electromagnets, motors, and generators Material that produces large loops are used for permanent magnet applications M B 0