294' Amperes Law

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29-4. Amperes Law
Sign of the current?
ccw up positive
2
29-4. Amperes Law
Same current (parallel or antiparallel) Rank
the loops according to the magnitude of the
integral
d gt a c gtb
3
29-4. Amperes Law
What is the magnitude of the integral ?

• ccw -i1 i2

(2) cw, i1 up, ccw i1 down, cw i2 up
-i1 -i1 i2
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29-4. Amperes Law
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29-4. Amperes Law
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29-5. Solenoids
Solenoid 50 cm long, diameter 5 cm Current3
A B-field inside 7 mT What is the length of
the wire?
B n (1.26 x 10-6 ) (3) 7 x 10-3 T
n N/(50 x 10-2 )
L N (2 p) (2.5 x 10-2 )
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29-5. Solenoids

• No B-Field outside
• Uniform Inside

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29-6. Current-Carrying Coil as a Magnetic Dipole
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29-6. Current-Carrying Coil as a Magnetic Dipole
Solenoid 200 turns, diameter 5 cm Current3
A (1) What is the magnitude of the magnetic
moment? (2) What is the magnitude of the
B-field at axial distance z 80 cm?
m (200) (3) p (2.5 x 10-2 )2 A/m2
B (1.26 x 10-6 )/(2p) x m /(80 x 10-2 )3 T
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30-3. Faradays Law
Magnetic Flux
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30-3. Faradays Law
B(t) given. Rank the regions according to the
magnitude of the emf induced in a conducting
loop.
b gt d e gt a c
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30-3. Faradays Law
a

• 3 loops are shown.
• B 0 everywhere except in the circular region
  where B is uniform, pointing out of the page and
  is increasing at a steady rate.
• Rank the 3 loops according to the magnitude of
  the induced EMF.

b
c
a b gt c
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30-4. Lenz s Law
If the magnetic flux increases as time goes, what
is the direction of the induced current?
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30-4. Lenz s Law

• 100-turn coil of radius 5cm and 5W.
• Solenoid of diameter 3cm with 200 turns/cm.
• The solenoid current drops from 2A to zero in
  Dt2ms.
• What is the current induced in the coil during
  Dt?

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30-5. Induction and Energy Transfer

• Flux change in time current

• Force on a current by the B-field

• Power dissipation?

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30-5. Induction and Energy Transfer

• Induced current in a solid conducting plate
• Mechanical energy is transferred to thermal energy

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30-6. Induced E-Fields
Current
E-field
Potential has no meaning if the E-field is
produced by induction
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30-6. Induced E-Fields

• A long solenoid has a circular cross-section of
  radius R.
• The current through the solenoid is increasing at
  a steady rate di/dt.
• Compute the E-field as a function of the distance
  r from the axis of the solenoid.

r
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30-6. Induced E-Fields
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30-6. Induced E-Fields

• Two versions of Faradays law
• A varying magnetic flux produces an EMF
• A varying magnetic flux produces an E-field

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30-7. Inductors and Inductance
i
Inductor Generate B-flux by current
Inductance L Generate B-flux by current
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30-7. Inductors and Inductance
Inductance of a solenoid
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30-8. Self-Induction
Current appears to oppose the increase
B
i
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30-8. Self-Induction

• The current in a 10 H inductor is decreasing at a
  steady rate of 5 A/s.
• If the current is as shown at some instant in
  time, what is the magnitude and direction of the
  induced EMF?

• Magnitude (10 H)(5 A/s) 50 V
• Current is decreasing
• Induced emf must be in a direction that OPPOSES
  this change.
• So, induced emf must be in same direction as
  current

29
30-9. RL Circuits

• Set up a single loop series circuit with a
  battery, a resistor, a solenoid and a switch.
• Describe what happens when the switch is closed.
• Key processes to understand
• What happens JUST AFTER the switch is closed?
• What happens a LONG TIME after switch has been
  closed?
• What happens in between?

At t0, a capacitor acts like a wire an inductor
acts like a broken wire. After a long time, a
capacitor acts like a broken wire, and inductor
acts like a wire.
30
30-9. RL Circuits

• Immediately after the switch is closed, what is
  the potential difference across the inductor?
• (a) 0 V
• (b) 9 V
• (c) 0.9 V

• Immediately after the switch, current in circuit
  0.
• So, potential difference across the resistor 0
• So, the potential difference across the inductor
  E 9 V

31
30-5. RL Circuits

• Immediately after the switch is closed, what is
  the current i through the 10 W resistor?
• (a) 0.375 A
• (b) 0.3 A
• (c) 0

• Immediately after switch is closed, current
  through inductor 0.
• Hence, current trhough battery and through 10 W
  resistor is i (3
  V)/(10W) 0.3 A

• Long after the switch has been closed, what is
  the current in the 40W resistor?
• (a) 0.375 A
• (b) 0.3 A
• (c) 0.075 A

• Long after switch is closed, potential across
  inductor 0.
• Hence, current through 40W resistor (3
  V)/(40W) 0.075 A

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30-9. RL Circuits

• How does the current in the circuit change with
  time?

i
i(t)
Small L/R
Large L/R
Time constant of RL circuit L/R
t
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30-10. Energy Stored in a B-Field
30-11. Energy Density of a B-Field
35
30-10. Energy Stored in a B-Field

• The switch has been in position a for a long
  time.
• It is now moved to position b without breaking
  the circuit.
• What is the total energy dissipated by the
  resistor until the circuit reaches equilibrium?

• When switch has been in position a for long
  time, current through inductor (9V)/(10W)
  0.9A.
• Energy stored in inductor (0.5)(10H)(0.9A)2
  4.05 J
• When inductor discharges through the resistor,
  all this stored energy is dissipated as heat
  4.05 J.