1. Magnetism Ferromagnetism: Substances that exhibit strong magnetic properties ex. iron, nickel, cobalt neodymiumNd 2 Fe 14 B Paramagnetism: form of magnetism that occurs only in the presence of an externally applied magnetic field; weakly attractive Check this out examples:platinum, aluminum, chromium, oxygen, tungsten

2. Diamagnetism : property of an object which causes it to create a magnetic field in opposition to an externally applied magnetic field Check this out examples:nitrogen, gold, silver, mercury, diamond

3. Magnetic Field Around a Straight Wire B =  o I 2πr2πr  o = 4  x 10 -7 T m/A I = current r = distance from wire r Check this out

4. ex. A current of 1.5 A is traveling through a wire from left to right, as shown below. Determine the magnitude and direction of the field at a location 0.45 m above the wire. 0.45 m 1.5 A B =  o I 2πr2πr  π x 10 -7 Tm/A )( 1.5 A ) 2π ( 0.45 m ) = B = 6.7 x 10 -7 T

5. Forces on Current-Carrying Wires Consider a wire that is hanging in a magnetic field created by a horseshoe magnet: NS B Pass current up through the wire: Direction of force can be found by a Right-Hand-Rule I

6. L Magnitude can be found from: F = B I L where L = length of wire in field NS B I

7. N S B I L What if the wire is at an angle to the field? F = L ( I x B ) magnitude : F = BIL sin x direction : right-hand fingers in direction of I ; curl to B; thumb points in direction of F direction of force is into the paper: L sin x

8. B = 2.0 T I = 0.75 A 60 o ex. A wire 0.40 m long is placed in a 2.0-T magnetic field at a 60 o angle to the field lines, as shown. A current of 0.75 A is sent through the wire. Find the magnitude and direction of the force on the wire. F = L ( I x B ) magnitude: F = BIL sin x = ( 2.0 T )( 0.75 A )( 0.40 m ) sin 60 F = 0.52 N direction: Fingers with I, curl to B : F = 0.52 N

9. Forces on Charges Moving in Magnetic Fields A positive charge is directed up through the magnetic field at left. B p+p+ Magnitude of force on particle: F = q v B Direction can be found from a RHR More general: F = q ( v x B )

10. F = q ( v x B ) B p+p+ F = qvB sin x in the direction determined as shown at right

11. ex. A proton is sent into a 0.50-T magnetic field with a speed of 2.5 x 10 6 m/s. (a) Find the magnitude and direction of the force exerted on the proton. B x x x x x x x x x x x x x x x p+p+ F = q ( v x B ) direction: proton is deflected to the left magnitude: F = qvB sin x x = 90 o = ( 1.6021 x 10 -19 C )( 2.5 x 10 6 m/s )( 0.50 T ) F = 2.0 x 10 -13 N

12. (b) The proton is deflected into a circular path of radius r. Find r. B x x x x x x x x x x x x x x x p+p+ r r = 0.052 m mv 2 F cent = F mag r = qvB mv 2 = qvB r mv = q B r r = mv q B = ( 1.67 x 10 -27 kg )( 2.5 x 10 6 m/s ) ( 1.6021 x 10 -19 C )( 0.50 T )

13. Mass Spectrometer:

14. Bubble Chamber:

15. Force Between Two Parallel Current-Carrying Wires Consider two parallel wires: Pass current through each wire in the same direction; I I determine the direction of magnetic field generated by each wire above and below the wires; do the wires attract or repel? A video clip

16. Magnetic Flux B A The flow of magnetic field lines through a surface area Φ = B A Magnetic flux

17. If field lines are at an angle to the surface, take the perpendicular component Φ = A ( B cos x ) Use angle with the normal to the surface More generally: Φ = A B = A B cos x dot product

18. B = 1.5 T normal to loop Φ = A B = A B cos x = A B cos 90 Φ = 0 ex. A loop of wire of radius 4.5 cm is placed in a magnetic field of strength 1.5 T. (a) If the plane of the loop is parallel to the field lines, find the magnetic flux through the loop. 90 o

19. ex. A loop of wire of radius 4.5 cm is placed in a magnetic field of strength 1.5 T. (a) If the plane of the loop is perpendicular to the field lines, find the magnetic flux through the loop. normal to loop Φ = A B = A B cos x = A B cos 0 B = 1.5 T = π r 2 B = π ( 0.045 m ) 2 ( 1.5 T ) Φ = 0.0095 T m 2 = 0.0095 webers = 0.0095 Wb = A B Wilhelm Weber ( 1804 – 1891 )

20. ex. A loop of wire of radius 4.5 cm is placed in a magnetic field of strength 1.5 T. (c) If the plane of the loop is at a 50 o angle to the field lines, find the magnetic flux through the loop. = π ( 0.045 m ) 2 ( 1.5 T )( 0.766 ) Φ = 0.0073 T m 2 = 0.0073 webers = 0.0073 Wb Φ = A B = A B cos x = A B cos 40 = π r 2 B cos 40 40 o B = 1.5 T

21. Electromagnetic Induction Consider a loop of wire: If an EMF (voltage) is induced in the loop, current will flow. + _ I Faraday: Anytime the magnetic flux that is passing through the loop changes, an EMF is induced

22. Faraday's Law of Induction: N = number of coils negative sign refers to Lenz's Law EMF = - N ΔΦ ΔtΔt EMF = - N Δ (BA) ΔtΔt

23. ex. A loop of wire of radius 0.36 m is in a 4.5-T magnetic field. If the loop is taken out of the field in 0.50 s, what is the magnitude of the induced EMF in the loop? EMF = - N Δ (BA) ΔtΔt N = 1Δt = 0.50 s A = πr 2 = π (0.36 m) 2 = 0.40715 m 2 ΔB = 4.5 T A ΔB ΔtΔt = = ( 0.40715 m 2 )( 4.5 T ) 0.50 s EMF = 3.6 V

24. Use Lenz's Law: If a bar magnet is brought near a loop of wire, in what direction will current be induced? S N N S I

25. Consider loop of wire in magnetic field B x x x x x x x x x x x x x x x Pull loop out of field; in what direction will current be induced in the loop?

26. Current induced to try to keep the field in the loop x x x x x x x x x x x x x x x Consider loop of wire in magnetic field B Pull loop out of field; in what direction will current be induced in the loop? I x x x x x Induced current will be clockwise around loop

27. ex. A loop of wire in the shape of a square with sides of 12.0 cm is in a magnetic field of strength 24.5 T, as shown at right. The wire has a resistance of 1.5 Ω. x x x x x x x x x x B = 24.5 T 12.0 cm

28. ex. A loop of wire in the shape of a square with sides of 12.0 cm is in a magnetic field of strength 24.5 T, as shown at right. The wire has a resistance of 1.5 Ω. (a) If the loop is removed from the field in 0.30 s, find the magnitude and direction of the induced current in the loop. x x x x x x x x x x B = 24.5 T 12.0 cm EMF = - N Δ (BA) ΔtΔt A ΔB ΔtΔt = = ( 0.120 m ) 2 ( 24.5 T ) 0.30 s EMF = 1.2 V = 1.176 V

29. ex. A loop of wire in the shape of a square with sides of 12.0 cm is in a magnetic field of strength 24.5 T, as shown at right. The wire has a resistance of 1.5 Ω. x x x x x x x x x x B = 24.5 T (a) magnitude of EMF = 1.2 V direction of current in loop = ? Field inside loop is decreasing; how will loop respond? x x x Current is induced to keep field in loop I Direction of current is clockwise around loop

30. x x x x x x x x x x B = 24.5 T x x x I (b)Find the force necessary to pull the wire from the field. EMF = 1.176 V force to pull wire = force that field exerts on the wire R = 1.5 Ω F = B I L I = ? I = V R = 1.176 V 1.5 Ω = 0.784 A F = B I L = ( 24.5 T )( 0.784 A )( 0.12 m ) F = 2.3 N

31. x x x x x x L ex. A rod of zero resistance and length L = 0.45 m slides to the right on two zero-resistance wires. The wires are joined by a 12.5 Ω resistor, and the system is placed in a 4.8-T magnetic field. (a) Find the speed v necessary to produce a current in the resistor of 0.36 A. B = 4.8 T v L = 0.45 m R = 12.5 Ω I = 0.36 A B = 4.8 T Can find voltage in loop V = I R = ( 0.36 A )( 12.5 Ω ) = 4.5 V EMF = 4.5 V

32. x x x x x x x x x x x L B = 4.8 T v EMF = 4.5 V L = 0.45 m EMF = - N Δ (BA) ΔtΔt ΔtΔt = EMF = B ΔA ΔtΔt d = B Δ( L d ) ΔtΔt EMF = B L Δd ΔtΔt = B L v EMF = B L v v = B L EMF = 4.5 V ( 4.8 T )( 0.45 m ) v = 2.1 m/s EMF induced by moving a wire of length L at a speed v through a B field

33. x x x x x x (b) Find the direction of the current through the resistor ? Use Lenz’s Law

34. x x x x x x ? As rod moves to the right, field through loop (into the page) increases How does loop respond? Field will be set up to oppose the increase; field will be generated that is out of the page (b) Find the direction of the current through the resistor Use Lenz’s Law

35. x x x x x x I As rod moves to the right, field through loop (into the page) increases How does loop respond? Field will be set up to oppose the increase; field will be generated that is out of the page Current is induced that flows down through the resistor (b) Find the direction of the current through the resistor Use Lenz’s Law

36. x x x x x x L B = 4.8 T v (c) How would your answers to (a) and (b) change if the rod were moved to the left instead of to the right? Answer to (a) would be unchanged, since direction of v did not play a role ? Would direction of induced current also be the same?

37. x x x x x x B = 4.8 T (c) How would your answers to (a) and (b) change if the rod were moved to the left instead of to the right? x Current is induced that moves up through the resistor I Answer to (a) would be unchanged, since direction of v did not play a role Would direction of induced current also be the same? As rod is moved to the left, the field inside loop decreases; loop responds by generating field into page to keep it