1. The wires of this solenoid are connected to a battery as shown: To an observer looking down the center of the solenoid from the left of the screen, (through )current flows through it 1. clockwise.2. counter-clockwise. 3. up. 4. down. 5. right. 6. left.

2. The wires of this solenoid are connected to a battery as shown: This forms an electromagnet with its North pole at which of the above positions?

3. The wires of this solenoid are connected to a battery as shown: An aluminum ring is placed as shown at position, its center aligned with the solenoid’s. The magnetic flux through the ring is A. up.B. down.C. left D. right. E. out of the screen.F. into the screen. G. zero.

4. The wires of this solenoid are connected to a battery as shown: An aluminum ring is placed as shown at position, its center aligned with the solenoid’s. The magnetic flux through the ring is A. up.B. down.C. left D. right. E. out of the screen.F. into the screen. G. zero.

5. An aluminum ring is placed as shown at position, its axis parallel with the solenoid’s. The magnetic flux through the ring is A. up.B. down.C. left D. right. E. out of the screen.F. into the screen. G. zero.

7. An aluminum ring is placed as shown at position, centered on the solenoid’s axis, but with its own axis perpendicular to the solenoid’s (facing up). The magnetic flux through the ring is A. up.B. down.C. left D. right. E. out of the screen.F. into the screen. G. zero.

8. The wires of this solenoid are connected to a battery as shown: With steady current flowing through the solenoid, our observer left of the screen would note the loop at carries A. an induced clockwise current. B. an induced counter-clockwise current. C. zero current.

9. + - The wires of this solenoid are connected to a battery as shown: The switch is opened to shut the solenoid down. Our observer left of the screen notes the loop at momentarily carries A. an induced clockwise current. B. an induced counter-clockwise current. C. zero current.

10. The current carrying solenoid’s right end is tipped up as it is slowly rotated counter clockwise about the pivot shown: The flux through the aluminum loop A. is increasing.B. is decreasing. C. remains constant.D. is zero.

11. The current carrying solenoid’s right end is tipped up as it is slowly rotated counter clockwise about the pivot shown: The loop will experience an induced A. clockwise current. B. counter-clockwise current. C. current of zero.

12. secondary coil + - primary coil Michael Faraday (1831)

14.                                                                                                                                                                                A coil carrying a steady current I generates a uniform magnetic field within its enclosed area.                                                                                                                                                                                I B=oIB=oIBIBI We use the symbol  to represent the flux (number of magnetic field lines) through the surface defined by a loop. B

15. A wire coiled into two loops carries a current, I. It faces a single independent loop enclosing the same cross-sectional area. The flux through the secondary (red) loop must be 1) . 2) ½ . 3) 2 . 4) 3 . 5) zero. Here assume  represents the flux (number of magnetic field lines) through one loop carrying current I.

16. Wire #1 (length L) forms a 1-turn loop, and a bar magnet is dropped through. Wire #2 (length 2L) forms a 2-turn loop, the same magnet is dropped through. Compare the induced currents for both. 1)I 1 < I 2 2)I 1 = I 2 3) I 1 > I 2 #1 #2

17. To replace the loss of flux (which was pointing left) our Right-Hand-Rule (with thumb to the left) shows our fingers curling up in front, in at top. QUESTION 1 QUESTION 2 3) 2 . The two black loops reinforce one another, combining to create a magnetic field B = 2  o I. QUESTION 3 1) I 1 < I 2 QUESTION 4 QUESTION 5 QUESTION 6 QUESTION 7 QUESTION 8 QUESTION 9 QUESTION 10 QUESTION 11 2. counter-clockwise Current flows from + to – (so in at the bottom of the solenoid, up in back, out at the top, down in front. Right-hand rule: fingers coiling in the same direction current leaves your thumb pointing to the left end. C. left There’s a South pole at that end, where field lines point inward. C. left There’s a North pole at this end, with field lines pointing outward. D. right. Field lines double back along the sides of a magnet (see slide 6). G. zero. Field lines pass above and below the ring, but none pass from below the ring, through it, to above. C. zero current. Current is induced in an effort to moderate sudden changes in magnetic fields. If current through the solenoid is “steady” there are NO CHANGES to respond to. B. a counter-clockwise current. B. is decreasing. The highest concentration of field lines is straight out of either pole. Turning the pole away decreases the flux. B. counter-clockwise current. To replace the loss of flux (pointing left into the South pole!). QUESTION 12