Goal: To understand induction

1
Goal To understand induction

• Objectives
• Motional EMF
• Magnetic Flux
• Faradays Law

2
Exam 1

• Average 81
• Only 1 score below 63
• Lots of good scores!

3
Concepts questions

• NOTE You may have slightly different questions
  than the ones I list here.
• 1) Two charged objects attract each other with an
  electric force F. Suppose that one charge is
  halved and the distance between them is doubled.
  The new electric force equals
• (A) 2F (B) ½F (C) 4F
• (D) 8F (E) 1/8 F
• 2) A lower Capacitance tells you that
• (A) you collecting charge more efficiently
• (B) you can collect more charge
• (C) you are collecting charge less efficiently
• (D) you are using a dielectric with a higher
  dielectric constant
• (E) none of the above
• 3) Electric field lines ALWAYS
• (A) point toward decreasing values of potential.
  
• (B) point in the same direction as the electric
  force.
• (C) are parallel to equipotential surfaces.

4
continuing

• 4) A proton, released at rest in an electric
  field (with an assortment of charges), ALWAYS
  moves
• (A) along an equipotential surface. (C) in a
  straight line directly to the nearest electron
• (B) nowhere.
• (D) down an electric field line (high to low
  potential).
• (E) up an electric field line (low to high
  potential).
• 5) Electric Potential is in units of
• (A) Volts (B) Joules (C) Newton / Coulomb
• (D) Coulombs (E) Watts
• 6) A resistance is a measure of a devices
• (A) ability to store charge (B) ability to limit
  flow of charge
• (C) ability to resist heating (D) ability to
  produce force
• (E) ability to resist corrosion
• 7) A kilowatt-hour is a unit of
• (A) charge (B) energy (C) power

5
Last concepts

• 8) When current passes through a resistor
• (A) the current decreases
• (B) the potential decreases
• (C) the potential increases
• (D) the voltage increases
• (E) the current increases
• 9) Electric companies bill their customers for
• (A) current (B) voltage (C) energy
• (D) power (E) force
• 10) In general, when two resistors of unequal
  resistance are connected in parallel,
• (A) the heat generated by each resistor is the
  same.
• (B) the current through each resistor is the
  same.
• (C) the voltage across each resistor is the same.
• (D) the equivalent resistance is larger than the
  bigger resistance.
• (E) more current flows out of the resistors than
  flows into the resistors

6
Short answer 1a

• For the charge configuration shown, compute
• (A) find the x and y components of the vector of
  the electric force on the 5.0-nC charge due to
  the 4.0-nC charge.
• F -k q1 q2 / r2 and r 10
• F - 8.99 109 5 10-9 4 10-9 / 100
• F -1.8 10-9 N
• Fx F (x/r) -1.8 10-9 N (-8 / 10)
• Fx 1.44 10-9 N
• Fy F (y/r) -1.8 10-9 N (6 / 10)
• Fy -1.08 10-9 N

7
Short answer 1b

• For the charge configuration shown, compute
• (B) find the x and y components of the vector of
  the electric force on the 5.0-nC charge due to
  the -4.0-nC charge.
• F -k q1 q2 / r2 and r 10
• F - 8.99 109 5 10-9 -4 10-9 / 100
• F 1.8 10-9 N
• Fx F (x/r) 1.8 10-9 N (-8 / 10)
• Fx -1.44 10-9 N
• Fy F (y/r) 1.8 10-9 N (-6 / 10)
• Fy -1.08 10-9 N

8
Short answer 1c

• (C) find the magnitude of the net electric force
  on the 5.0-nC charge
• F F1 F2
• 1.44 10-9 N (x) -1.08 10-9 N (y) -1.44
  10-9 N (x) -1.08 10-9 N (y)
• The two X forces cancel, so you get
• F - 2.16 10-9 N (y)
• So, the magnitude is 2.16 10-9 N
• For D you needed to draw a line in the y
  direction

9
Short answer 2A

• 2) The average house would use about 3107 J of
  energy each night. The biggest area you want to
  use is about 0.3 square meters. For the cost,
  water is the best dielectric with a dielectric
  constant of 80. Since this system should power
  your household it should be set to the same
  voltage as your wall sockets which is 120 Volts.
• calculate the needed initial
• (A) the charge stored
• U ½ QV, so Q 2U / V 6107 J /120 V
• Q 5105 C

10
Short answer 2B

• 2) The average house would use about 3107 J of
  energy each night. The biggest area you want to
  use is about 0.3 square meters. For the cost,
  water is the best dielectric with a dielectric
  constant of 80. Since this system should power
  your household it should be set to the same
  voltage as your wall sockets which is 120 Volts.
• (B) the magnitude of the electric field between
  the plates
• E Q / (? e0 A) 5105 C / (80 8.8510-12
  0.3)
• E 2.351015 N/C

11
Short answer 2C

• 2) The average house would use about 3107 J of
  energy each night. The biggest area you want to
  use is about 0.3 square meters. For the cost,
  water is the best dielectric with a dielectric
  constant of 80. Since this system should power
  your household it should be set to the same
  voltage as your wall sockets which is 120 Volts.
• (C) the capacitance of the capacitor
• Q CV so C Q/V
• C 5105 C / 120 V 4.17 103 F

12
Short answer 2D

• 2) 3107 J of energy. Area 0.3 square meters.
  Dielectric constant of 80. 120 Volts.
• (D) the separation distance between the plates
• E V / D, so D V / E
• D 120 V / 2.351015 N/C
• D 5.110-14 m
• (E) Considering your answer to part D is running
  your house off of a capacitor a realistic goal?
• No, 5.110-14 m is way too small a separation
  distance (atomic sized) to be practical or
  realistic.

13
Short Answer 3

• The top right 2 are parallel, so add in inverse.
• Their effective resistance is 3.08 Ohms
• This is in series with the 6 Ohm resistor so the
  resistance on top is 9.08 Ohms
• The two bottom resistors are in series so the
  effective is 5 Ohms (2 3).
• The 5 Ohms is in parallel with the 9.08 Ohms so
  the total resistance is
• R 5 9.08 / (5 9.08) 3.22 Ohms

14
EVIL QUESTION!

• Iin Iout, I1 I2 I3, I1 5 A I3
• Bottom loop 8 V 5A 10 O 7 O I3
• So, I3 58 / 7 A 8.3 A
• Since I1 5A I3 5A 8.3 A
• I1 13.3 A
• Top loop ?V I1 11 O 5A 10 O
• ?V 11 13.3 V 50V 196 V

15
Motional EMF

• When a charge moves there is a force on it.
• If used properly you can use this to generate a
  current (energy).
• A lot of electricity is generated by generators.
• To do this you move metal through a magnetic
  field.
• The charges will move perpendicular to that.
• This creates a current.

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Motional EMF

• The potential for a emf is
• V v B L where L is the length of the rod used
  and V is the motional EMF (usually denoted as a
  squiggly E)
• If you tie this rod to a resistor then the
  current is
• I V / R v B L / R

17
Magnetic Flux

• Magnetic Flux is just a measure of the amount of
  magnetic field contained within some area.
• Magnetic Flux is in units of Webers (Wb)
• 1 Wb 1 T m2
• If the plane of some wire or object is tilted at
  an angle of ? towards or away from the plane of
  magnetic field lines the the Magnetic Flux is
  just
• F B A cos(F)
• A is area and A cos(F) is the effective area
  inside the field.

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You try one

• You have a magnetic field of 0.2 T.
• You put a circular wire perpendicular to the
  field.
• If the wire is a circle with a diameter of 0.2
  meters then what is the magnetic flux through the
  wire loop?

19
You try one

• You have a magnetic field of 20 T.
• You put a circular wire perpendicular to the
  field.
• If the wire is a circle with a diameter of 0.2
  meters then what is the magnetic flux through the
  wire loop?
• F B A cos(F) BA
• 20 T 3.14 0.01 0.63 Wb

20
Another

• A wire loop with radius of 0.3 m is inserted at a
  45 degree angle to a 5 T magnetic field.
• What is the magnetic flux through the loop?

21
Another

• A wire loop with radius of 0.3 m is inserted at a
  45 degree angle to a 5 T magnetic field.
• What is the magnetic flux through the loop?
• F B A cos(F) 5T 3.14 0.09 / 1.4
• 1.0 Wb

22
Faradays law

• When you change the flux then you produce a
  current.
• The induced emf is the rate at which you change
  the magnetic flux!
• So, V - change in flux / change in time
• (- because the induced emf attempts to oppose
  that change in magnetic flux)
• V - ? F / ?t
• And ? F Ff - Fi

23
Faradays law you try

• V - ? F / ?t
• And ? F Ff Fi
• You put a 0.2 m in radius wire loop perpendicular
  to a 3 T magnetic field.
• Then something goes horribly wrong!
• To save the plant you work in you pull the loop
  from the field as quickly as you can.
• If you pull the loop free in 1.2 sec then what is
  the induced emf on that loop during that time?

24
Faradays law you try

• V - ? F / ?t
• And ? F Ff Fi
• You put a 0.2 m in radius wire loop perpendicular
  to a 3 T magnetic field.
• Then something goes horribly wrong!
• To save the plant you work in you pull the loop
  from the field as quickly as you can.
• If you pull the loop free in 1.2 sec then what is
  the induced emf on that loop during that time?
• Ff BA 0 because you have pulled it out of the
  magnetic field.
• Fi BA 3T 0.04 3.14 0.38 Wb
• ? F Ff Fi 0 0.38 Wb -0.38 Wb
• V - ? F / ?t -(-0.38 Wb) / 1.2 s 0.31 V

25
Faradays law you try 2

• V - ? F / ?t
• And ? F Ff Fi
• You put a 0.2 m in radius wire loop perpendicular
  to a 3 T magnetic field.
• Then in 1.2 sec you rotate it 30 degrees.
• What is the induced emf?

26
Faradays law you try 2

• V - ? F / ?t
• And ? F Ff Fi
• You put a 0.2 m in radius wire loop perpendicular
  to a 3 T magnetic field.
• Then in 1.2 sec you rotate it 30 degrees.
• What is the induced emf?
• Fi BA 3T 0.04 3.14 0.38 Wb
• Ff BAcos(30 degrees) BA/2 0.19 Wb
• ? F Ff Fi 0.19 0.38 Wb -0.19 Wb
• V - ? F / ?t -(-0.19 Wb) / 1.2 s 0.16 V

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Coils

• If you have a coil you do the same thing except
  that you have more area effectively.
• N loops gives you N times more area.
• So
• V - N ? F / ?t

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Conclusion

• We have seen that motions of charges in a
  magnetic field produce a current and an induced
  EMF.
• Magnetic Flux is the amount of magnetic field
  filled by an area.
• The induced EMF is just the time rate of change
  of the Magnetic Flux.
• Tomorrow direction of current, transformers,
  and LR circuits