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

Michael Faraday (1831)

=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

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

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

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.

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

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 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

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

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

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

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 

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 – – + + – – – – + + + + – – + +

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

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

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

http://micro.magnet.fsu.edu/electromag/java/varcapacitor/

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

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

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

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.