Physics 9 Friday, April 4, 2014 FYI: final exam is Friday, May 9th, at 9am, in DRL A2. Turn in HW10 today. I ll post HW11 tomorrow. For Monday: read concepts half of Ch31 (electric circuits); read equations half for Wednesday. Annotated as usual. We re trying to put together a way for you to be able to tinker in class with some batteries, light bulbs, multimeters, etc.
We learned on Monday that a changing magnetic field causes an emf (a voltage), which tends to induce electric currents to oppose the change in Φ B. E induced = dφ B What s really pushing the charged particles around the loop of wire (i.e. what s making the induced current flow) is an electric field. We can rewrite the emf E as the line integral of electric field: E induced = E d l = dφ B
In this figure (29.31), B is out of the page and increasing in magnitude. So E loops around clockwise. That electric field is what causes the induced current in the metal ring on the left. E induced = E d l = dφ B This equation is saying that the changing magnetic field causes electric field lines to circle around the changing B field lines. The minus sign means they loop around in the opposing direction.
So a changing magnetic flux causes an encircling electric field. E d l = dφ B You now see in Chapter 30 that analogously, a changing electric flux causes an encircling magnetic field. B d dφ E l = + µ 0 ɛ 0 Notice that this effect has a plus sign, not a minus sign.
Changing E field causes an encircling B field. (This one has the plus sign.) Changing B field causes an encircling E field. (This one has the minus sign: opposes change.)
Suppose the current shown here is discharging the capacitor. If the charge on each plate is decreasing in magnitude, which plate has the positive charge (and is becoming less positive)? (A) Left plate has the + charge (and is becoming less positive). (B) Right plate has the + charge (and is becoming less positive).
The current is discharging the capacitor. The right plate is positive (and becoming less positive). The left plate is negative (and becoming less negative). Which way does the electric field E point between the two plates? (A) E points to the right (B) E points to the left (C) E circles clockwise when looking from the left (D) E circles counter-clockwise when looking from the left
The current is discharging the capacitor. The electric field E between the plates points left and is decreasing in magnitude. Which way does the electric field s rate-of-change d E/ point between the two plates? (A) d E (B) d E (C) d E (D) d E points to the right points to the left circles clockwise when looking from the left circles counter-clockwise when looking from the left
The current is discharging the capacitor. Between the plates, the electric field s rate-of-change d E/ points right. Between the plates, which way does the magnetic field B point (which is caused by d E/)? (A) B points to the right (B) B points to the left (C) B circles clockwise when looking from the left (D) B circles counter-clockwise when looking from the left
The current is discharging the capacitor. Which way would the magnetic field B point if we simply deleted the capacitor and let the wire go through uninterrupted? (A) B points to the right (B) B points to the left (C) B circles clockwise when looking from the left (D) B circles counter-clockwise when looking from the left
Without the capacitor present (if it were one long wire instead), B would encircle the current, looping clockwise as seen from the left. With the capacitor present, B encircles d E in the same direction dφ E as it encircles I. So ɛ 0 is called the displacement current, because it causes an encircling magnetic field in an analogous way as a current I does. You don t need to remember that, but the capacitor example is a way to remember the connection between de/ and B.
Maxwell s equations You now have some intuition for what each piece of these equations means. Φ E = E da = q encl Φ B = ɛ 0 B d A = 0 E induced = E d l = dφ B B d l = µ 0 I enc + µ 0 ɛ 0 dφ E
In empty space (no charges or currents) this becomes B d A = 0 E d A = 0 B d l = E d l = + µ 0 ɛ 0 dφ E dφ B Encircling B field is proportional to rate of change of E. Encircling E field is proportional to (minus) rate of change of B. Imagine E(x, t) sin(2πf t 2πx/λ). Waves can propagate, with c = λf = 1/ ɛ 0 µ 0. Speed of light c = 1/ ɛ 0 µ 0 pops out of Maxwell s equations!
Physics 9 Friday, April 4, 2014 FYI: final exam is Friday, May 9th, at 9am, in DRL A2. Turn in HW10 today. I ll post HW11 tomorrow. For Monday: read concepts half of Ch31 (electric circuits); read equations half for Wednesday. Annotated as usual. We re trying to put together a way for you to be able to tinker in class with some batteries, light bulbs, multimeters, etc.