# Physics 1304 - Summer 1998

# Homework Assignment #6

**Due:** Tuesday 14 July 1998

### READING

Chapters 33 and 34.
### QUESTIONS

Chapter 33 - 1, 3, 5, 8, 12, 17, 19, 20, 22, 23, 28.

Chapter 34 - 1, 12, 13, 14, 19.
### PROBLEMS

Chapter 33 - 8, 12, 18, 19A, 19, 24, 25, 29, 31, 37, 45, 59,
Review Problem.

Chapter 34 - 1, 2, 6, 12, 19, 25.
### ANSWERS

These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to
copy these and turn them in as homework. You must show your work.
**Q 33-1)** The voltage across the inductor peaks first, then one quarter
of a cycle later the current through the inductor reaches its maximum value.

**Q 33-3)** The reactance of a capacitor 1/(wC) is frequency dependent.
At high frequencies (large w) the capacitive reactance is small and the
cap offers no impedance to current flow, behaving like a short (a piece of
wire). At low frequencies (small w) the capacitive reactance is large
and the cap offers much impedance to current flow behaving like an open
(a break in the wire).

**Q 33-5)** The voltages across the resistor, capacitor, and inductor
do not all peak at the same time. Kirchhoff's voltage law, which is just
energy conservation, is satisfied at ANY ONE TIME.

**Q 33-8)** R is not frequency dependent;

X_{L} doubles;

X_{C} is halved.

**Q 33-12)** At resonance, Z (the impedance) equals R (the resistance).

**Q 33-17)** No, transformers are strictly AC devices. The changing
current in one coil causes a changing magnetic field, which causes a
changing magnetic flux in another coil, which induces an emf by Faraday's
law (Lenz' law) in the second coil, which pushes a current in the second
coil.

**Q 33-19)** The time average is zero; the graph spends as much time
above zero level as below. The root-mean-square voltage is
V_{max}/sqrt(2).

**Q 33-20)** The time average is V_{max}/2. The root-mean-square
voltage is V_{max}/sqrt(2).

**Q 33-22)** No. If the RLC circuit is capacitive, the current leads
the applied voltage; if the RLC circuit is inductive, the current lags
the applied voltage.

**Q 33-23)** No, power is only dissipated in resistors.

**Q 33-28)** The iron has a high permeability and concentrates the lines
of magnetic field to the inside of the coils, instead of letting B lines
spread all over space. They are only useful to the transformer if the
B lines pass through the coils.

**Q 34-1)** The momentum given to the reflecting surface by light photons
is twice as large as the momentum given to the absorbing surface. The
radiation pressure is the sum of all the microscopic momentum transfers.
The same effect would be seen if bullets were fired at two walls, one
reflecting the bullets and one absorbing the bullets.

**Q 34-12)** Oscillating electric fields push electrons in the metal
antenna up and down, creating a current.

**Q 34-13)** A changing magnetic flux through the conducting loop antenna
induces a current around the loop.

**Q 34-14)** The voltage induced in a VHF dipole antenna by the electric
field in the signal is V = Ed.

The voltage (emf) induced in a UHF loop antenna by the magnetic field in
the signal is V = -d/dt(magnetic flux) = (Area of loop) (cos theta) dB/dt.
Because the oscillating magnetic field is B(t)=B_{0}sin(kx-wt),
the derivative dB/dt will be proportional to the angular frequency of
oscillation (w).

**Q 34-19)** There is no signal directly over (or under) a dipole antenna.

**P 33-8)** 0.413 Wb

**P 33-12)** 3.14 A

**P 33-18)**
- 141 A
- 235 A

**P 33-19A)** C V_{max}

**P 33-19)** 2.77 nC

**P 33-24)** 2792 Hz

**P 33-25)**
- 78.5 ohms
- 1590 ohms
- 1520 ohms
- 0.138 A
- -84.3 degrees

**P 33-29)** 1.88 V

**P 33-31)**
- 146 V
- 213 V
- 179 V
- 33.3 V

**P 33-37)** 8 W

**P 33-45)** 1.82 pF

**P 33-59)** 687 V

**P 33-Review Problem)**
- V
_{max}/R sin(wt)
- V
_{max}^{2}/2R
- V
_{max}/sqrt[R^{2} + (wL)^{2}]
cos[wt + arctan(wL/R)]
- 1/(w
^{2}L) at the resonant frequency W
- Z=R at the resonant frequency W
- V
_{max}^{2}L/2R^{2}
- (same as above)
- arctan[3/2R sqrt(L/C)]
- w/sqrt(2)

**P 34-1)** 680 years from now

**P 34-2)** 7.33 x 10^{-7} T

**P 34-6)** 3c/4 = 2.25 x 10^{8} m/s

**P 34-12)** 3.33 x 10^{-6} J/m^{3}

**P 34-19)**
- 3.32 x 10
^{5} W/m^{2}
- E
_{surface} = 1880 V/m

B_{surface} = 2.22 x 10^{-4} T

**P 34-25)** 8.33 x 10^{-8} Pa (1 Pa = 1 N/m^{2})

Please report any corrections to
Professor Scalise.

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