Physics 212 Lecture 17

1. Physics 212 Lecture 17 Faraday’s Law

2. Motional EMF EMF Change Area of loop Change magnetic fieldthrough loop Change orientation of looprelative to B In each case the flux of the magnetic field through the circuit changeswithtime and an EMF is produced. Physics 212 Lecture 17, Slide 2

3. Rotate the loop,change flux, generate emf. B

4. B1 B2 v Move loop to a place wherethe B field is different, change flux, generate emf.

5. Checkpoint 1a A copper loop is placed in a uniform magnetic field as shown. You are looking from the right. Suppose the loop is moving to the right. The current induced in the loop is: A. zero B. clockwise C. counterclockwise • Motional emf is ZERO • v X B = 0 • no charge separation • no E field • no emf • The flux is NOT changing • B does not change • the area does not change • the orientation of B and A does not change

6. Checkpoint 1c O X B B dA dA X O Now suppose that the loop is spun around a vertical axis as shown, and that it makes one complete revolution every second. The current induced in the loop: A. Is zero B. Changes direction once per secondC. Changes direction twice per second Current changes direction every time the loop becomes perpendicular with the B field emf ~ dF/dt (B  dA = max)  d/dt (B  dA ) = 0

7. Faraday’s Discovery True no matter how we change the flux. In fact, the circuit may be stationary (no motional EMF) and only the B-field changes with time. An EMF is still produced. This implies that: Faraday’s Law

8. Change the B field in time so flux changes. Induce an emf nnd therefore an Electric field. This emf tries to oppose the change in flux. (Lenz’s Law) Induces an E field even if there is no circuit there! B(t) decreasing

9. Checkpoint 1b Looking from right X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X Clockwise current tries to restore B A copper loop is placed in a uniform magnetic field as shown. You are looking from the right. Checkpoint 1b Now suppose the that loop is stationary and that the magnetic field is decreasing in time. The current induced in the loop is: A. zero B. clockwise C. counterclockwise • Motional emf is ZERO • Circuit is stationary ! • HOWEVER: The flux is changing • B decreases in time • current induced to oppose the flux change • clockwise current tries to restore B that was removed

10. Checkpoint 2 F B O X B Like poles repel Ftotal < mg a < g A horizontal copper ring is dropped from rest directly above the north pole of a permanent magnet (copper is notferromagnetic) Will the acceleration a of the falling ring in the presence of the magnet be any different than it would have been under the influence of just gravity (i.e. g)? A. a > g B.a = g C. a < g This one is hard ! B field increases upward as loop falls Clockwise current (viewed from top) is induced

11. Checkpoint 2 Looking down B B I I IL X B points UP Ftotal < mg a < g A horizontal copper ring is dropped from rest directly above the north pole of a permanent magnet HOW IT WORKS (copper is notferromagnetic) Will the acceleration a of the falling ring in the presence of the magnet be any different than it would have been under the influence of just gravity (i.e. g)? A. a > g B.a = g C. a < g This one is hard ! B field increases upward as loop falls Clockwise current (viewed from top) is induced Main Field produces horizontal forces “Fringe” Field produces vertical force

12. Calculation y A rectangular loop (height = a,length = b,resistance = R,mass = m) coasts with a constant velocity v0in + x direction as shown. At t =0, the loop enters a region of constantmagnetic field Bdirected in the –z direction. What is the direction and the magnitude of the force on the loop when half of it is in the field? B x x x x 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 a v0 x • Conceptual Analysis • Once loop enters B field region, flux will be changing in time • Faraday’s Law then says emf will be induced • Strategic Analysis • Find the emf • Find the current in the loop • Find the force on the current

13. y y B B x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 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 b a a v0 v0 x x Change in Flux = dFB = BdA = Bav0dt Calculation A rectangular loop (height = a,length = b,resistance = R,mass = m) coasts with a constant velocity v0in + x direction as shown. At t =0, the loop enters a region of constantmagnetic field Bdirected in the –z direction. What is the magnitude of the emf induced in the loop just after it enters the field? (A) e = Babv02(B) e = ½ Bav0(C) e = ½ Bbv0(D) e = Bav0(E) e = Bbv0 a The area in field changes by dA =v0dt a In a time dtit moves by v0dt

14. Calculation y B x x x x 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 a v0 x What is the direction of the current induced in the loop just after it enters the field? (A) clockwise(B) counterclockwise(C) no current is induced Induced emf produces flux out of screen y A rectangular loop (height = a,length = b,resistance = R,mass = m) coasts with a constant velocity v0in + x direction as shown. At t =0, the loop enters a region of constantmagnetic field Bdirected in the –z direction. B x x x x 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 a v0 x emf is induced in direction to oppose the change in flux that produced it Flux is increasing into the screen

15. Calculation Force on a current in a magnetic field: y B x x x x x x x x x x x x x x b v0 • Force on top and bottom segments cancel (red arrows) a • Force on right segment is directed in –x direction. I x y A rectangular loop (height = a,length = b,resistance = R,mass = m) coasts with a constant velocity v0in + x direction as shown. At t =0, the loop enters a region of constantmagnetic field Bdirected in the –z direction. B x xxxxxx x xxxxxx x xxxxxx x xxxxxx b a v0 x What is the direction of the net force on the loop just after it enters the field? (A) +y(B) -y(C) +x (D) -x

16. Calculation since y B x x x x x x x x x x x x x x b v0 a F I ILB x y A rectangular loop (height = a,length = b,resistance = R,mass = m) coasts with a constant velocity v0in + x direction as shown. At t =0, the loop enters a region of constantmagnetic field Bdirected in the –z direction. B x x x x 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 a v0 x What is the magnitude of the net force on the loop just after it enters the field? e = Bav0 (A) (B) (C) (D)

17. Follow-Up Which of these plots best represents the velocity as a function of time as the loop moves form entering the field to halfway through ? (A) (B) (C) D) (E) X X X B x x x x x x x x x x x x x x b v0 a Fright I t = dt: e = Bav0 y A rectangular loop (sides = a,b, resistance = R, mass = m) coasts with a constant velocity v0 in + x direction as shown. At t =0, the loop enters a region of constant magnetic field B directed in the –z direction. What is the velocity of the loop when half of it is in the field? B x x x x 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 v0 a x This is not obvious, but we know v must decrease Why? Fright points to left Acceleration negative Speed must decrease

18. Follow-Up where Claim: The decrease in kinetic energy of loop is equal to the energy dissipated as heat in the resistor. Can you verify?? y A rectangular loop (sides = a,b, resistance = R, mass = m) coasts with a constant velocity v0 in + x direction as shown. At t =0, the loop enters a region of constant magnetic field B directed in the –z direction. What is the velocity of the loop when half of it is in the field? B x x x x 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 v0 a x e = Bav0 Which of these plots best represents the velocity as a function of time as the loop moves form entering the field to halfway through ? (A) (D) • Why (D), not (A)? • F is not constant, depends on v Challenge: Look at energy