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Unformatted text preview: NAME PHYS1200 PHYSICS II QUIZ 1 FEBRUARY 16, 2005 SOLUTIONS
PART A.
1. D
2. B
3. C
4. B
5. D
6. A 7. E
8. C PART B.
1. a) Since the Gaussian surface is in the conducting metal, the electric field must be equal to
zero everywhere on the Gaussian surface. Therefore, ( ∫ E ⋅ dA )A = 0
¯¯¯¯¯¯¯¯¯¯¯¯
b) In this case, the integral equals the net charge enclosed by the surface divided by ε0.
Then, ( ∫ E ⋅ dA )B = q/ε0
¯¯¯¯¯¯¯¯¯¯¯¯¯¯
2. 3. Only wires 1, 3, 6, and 7 are enclosed by the path. The sign of the integral is positive for 1, 3,
and 7, and negative for 6.
a) POSITIVE There are more currents making a positive contribution than negative.
b) ∫ B ⋅ ds = µ i 0 enc = µ 0 (i1 + i3 + i6 + i7 ) = µ 0 (i0 + i0 − i0 + i0 ) 1 ∫ B ⋅ ds = 2µ i 00 NAME
PART C.
1. a) Since the light bulbs all have the same resistance, and they are all in parallel, the 50 A is
i
50 A
=
divided equally among them. Then, i1 =
i1 = 0.50 A
100 100
b) For any one bulb, P = i2R, so R = P
100 W
=
2
i
(0.50 A)2 R = 400 Ω
P 100 W
=
i 0.50 A
E = 200 V c) Since all the light bulbs are in parallel, for any one bulb, P = Ei, so E = d) INTO THE PAGE According to the right hand rule, the fields from both wires point
into the page between the wires. e) The magnetic field due to one of the wires is given by, B1 =
the two wires add , so B = 2 B1 = 2. a) THE FIELD IS INCREASING µ 0i
. The contributions of
2π r µ0i ( 4π × 10 −7 H/m)(50 A)
=
πr
π (0.05 m) B = 4.0 ×104 T The current produces a field opposite to the existing
field. According to Lenz’s law, the induced current
must oppose the change, so the field is increasing.. dΦ B d ( BA)
dB
2 dB
=
=A
. Since A = π r², this becomes, E = π r
. Then,
dt
dt
dt
dt
E
dB
0.020 V
=
=
2
dt π r
π (0.10 m)2
dB
= 0.64 T/s
dt
¯¯¯¯¯¯¯¯¯¯¯ b) E = 2 ...
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This note was uploaded on 10/17/2011 for the course PHYS 1010 taught by Professor Schroeder during the Spring '07 term at Rensselaer Polytechnic Institute.
 Spring '07
 Schroeder
 Physics

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