1. B=moI =BA =(moI)pr2 B I                     
Recall that a coil carrying a steady current I
generates a uniform magnetic field within its
enclosed area of
B=moI
           
            
          
         
        
      
     B I
This circular loop
encloses a total
FLUX of
            
           
            
          
         
        
      
    
=BA
=(moI)pr2

2. Michael Faraday (1831)

3. =BA =(moI)pr2 =(moI)pr2. 2) ½. 3) 2. 4) 3. 5) zero. Here assume
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
=(moI)pr2.
2) ½.
3) 2.
4) 3.
5) zero.
Here assume
=BA
=(moI)pr2

5. If N2>N1 “step up” transformer If N2<N1 “step down” transformer
Joseph Henry (1838)
flux change
generated
by primary
is seen
within
secondary so
If N2>N1 “step up” transformer
If N2<N1 “step down” transformer

6. Compared to the voltage drop across coil A, B
6 V
Compared to the voltage drop across coil A,
the voltage across coil B is
1) less than 6 V
2) 6 V
3) greater than 6 V

7. A B 6 V Compared to the voltage drop across coil A, the voltage
across coil B is
1) less than 6 V
2) 6 V
3) greater than 6 V
The voltage across B is zero
Only a changing magnetic flux induces an EMF.
Batteries provide DC current.

8. What is the voltage across the light bulb? 1) 60 V 2) 120 V 3) 240 V

9. What is the voltage across the light bulb? 1) 60 V 2) 120 V 3) 240 V
What is the current through the light bulb?
1) 1/2 Amp
2) 1 Amp
3) 2 Amps

10. What is the voltage across the light bulb? 1) 60 V 2) 120 V 3) 240 V
What is the current
through the light bulb?
1) 1/2 Amp
2) 1 Amp
3) 2 Amps
What is the current
through middle circuit?
1) 1/2 Amp
2) 1 Amp
3) 2 Amps

11. 120 V 120 V 240 V What is the voltage across the light bulb? A) 60 V
B) 120 V
C) 240 V
Twice as many loops voltage doubles.
THEN: half as many loops, voltage halves
What is the current through the light bulb?
A) 1/2 Amp
B) 1 Amp
C) 2 Amp s
Power in = Power out
Voltage out = 120 V
240 V  1 Amp = 120 V  I
so I = 2 Amp

12. “Resonance”
occurs when XC=XL
(and they cancel!) Z R fo
f 

13. “Resonance”
occurs when XC=XL
(and they cancel!) Z R fo
f 

14. Z R fo f  I=V/Z fo f  I=V/R “Resonance” occurs when XC=XL
(and they cancel!) Z R fo
f 
I=V/R
I=V/Z fo
f 

15. LC Oscillations + + + + – – – – + + – – I I I I – – + + – – – –
Fully charged capacitor
Energy: all E
Current starts to flow
Energy: some E, some B
Maximum current
Energy: all B
– – – –
+ +
– – I I
Current begins to die down
induced  
C charges in opposite direction
Energy: some E, some B
Current flows
(opposite direction!)
Energy: some E, some B
Capacitor charged
Energy: all E I I
– –
+ +
– – – –
– –
+ +

16. I I I + + + + – – – – I + + – – I Fully charged capacitor Current
– – – –
Fully
charged
capacitor I I
+ +
– –
Current
starts
to flow I
Maximum
positive
current I
Maximum
current I

17. Current begins to die down Capacitor fully charged
– –
+ +
Current begins to die down I
– – – –
Capacitor fully charged I I
– –
+ +
Current flows
(opposite direction!) I

18. LC Circuits Inductor Capacitor
Capacitor: Stores energy in E field
Energy = ½ CV2
Inductor: Stores energy in B field
Energy = ½ LI2
Two ways to store energy  Oscillations 1 1
Frequency of oscillation = 2 π LC
Adjusting the capacitance of an LC circuit is the basic was a radio tuner works.
Question: What are the two ways to store energy in a pendulum?
Answer: Gravitational potential and kinetic energy

20. A radio is tuned to 141 MHz. If the tuner is to be set for 100 MHz, the capacitance of the LC circuit in the radio must be
A) quadrupled
B) doubled
C) not changed
D) halved
E) quartered

21. A radio is tuned to 141 MHz. If the tuner is to be set for 100 MHz, the capacitance of the LC circuit in the radio must be
A) quadrupled
B) doubled
C) not changed
D) halved
E) quartered 1 LC 2 π
f =
C  2C then f  (1/2) f
2 = 1.41
141 MHz / 1.41 = 100 MHz

22. produces changing B field
changing current
produces changing B field
Recording on Tape
AC current matching frequency of voice
coil
Changing current in the coil creates
a changing magnetic field in the iron
recording head (iron ring)
Guides the magnetic field created
by the current in the coil around to
to the recording tape.
direction of
tape movement N S S N N S
magnetic tape
Thin, plastic tape with millions of tiny
permanent magnets adhered to it
magnetized
areas of tape
Gap
Gap in recording head lets the
magnetic field out, so it can align
the magnetic dipoles on the tape

23. Playback changing B field induces changing current
Changing magnetic field
inside the coil induces
a current in the coil.
changing current
Tape induces a magnetic
field in the recording head.
(Just like holding a magnet near a
nail makes it temporarily magnetic)
direction of
tape movement S N N S
Of course, there are amplifiers and other stereo
components which, for clarity, are not shown.