Download TEST 3 over chapters 29

COLLEGE PHYSICS PH 222-3A (MIROV)
Exam 3 (04/05/10)
STUDENT NAME: ________KEY____________STUDENT id #: ______________________
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WORK ONLY 5 QUESTIONS
ALL QUESTIONS ARE WORTH 30 POINTS
-----------------------------------------------------------------------------------------------------------------------------------------NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)
Important Formulas:
1.
2.
3.
4.
5.
6.
7.
8.
9.
Ch.28. Magnetic fields



Magnetic force exerted on a point charge by a magnetic field: F=
qv × B
B
Bi
The number density n of charge carriers (Hall Effect): n =
Vle
mv
Circular orbit in magnetic field: r =
q B



Magnetic force on a straight current carrying wire of length L: FB
= iL × B



The force acting on a current element in a magnetic field: dF
=
idL × B
B
Magnetic dipole moment of current loop: µ= [current] x [area]


Torque on a current carrying coil: τ= µ × B
 
Orientation energy of a magnetic dipole: U (θ ) =
−µ ⋅ B
The work done on the dipole by the agent is: Wa =
∆U =
U f − Ui
10. Permeability constant: µo = 1.26x10-6 N⋅s2/C2 = 1.26x10-6 H/m;
11. Permittivity constant: εo = 8.85x10-12 C2/(N⋅m2) = 8.85x10-12 F/m;
1.
2.
3.
4.
5.
6.
7.
Ch. 29 Magnetic fields due to currents


µo ids × rˆ
The Bio-Savart Law: dB =
4π
r2
µ0I
Magnetic field of a long straight wire current: B =
{Unit 1 tesla = 1T = 1 N/(C⋅m/s)}
2πr
µoiφ
Magnetic field of a circular arc: B =
4π R
µo Lia ib
Forces between parallel currents:
Fba
ib=
LBa sin 90
=
2π d


Ampere’s Law: 
B
⋅
ds
=
µ
i
o enc
∫
Magnetic field of an ideal solenoid: B = µoin
µoiN 1
Magnetic field of a toroid: B =
2π
r
1.


µo µ
Field of a magnetic dipole: B ( z ) =
2π z 3
Ch. 30. Induction and Inductance


∫ B ⋅ dA
1.
Magnetic flux: Φ B =
2.
Faraday’s Law of Induction: ε = −
3.
4.
5.
6.
7.
8.
9.
{Unit 1weber = 1Wb = 1 T m2}
dΦB
dt
Lenz’ Law: Induced emf opposes change that produced it

dΦB

Emf and the Induced Electric Field: ε =
−
∫ E ⋅ ds =

dt
NΦB
Inductance: L =
{SI unit henry(H), where 1 H=1Tm2/A)
i
L
The inductance per unit length of solenoid:
= µo n 2 A
l
di
Self-induction: ε L = − L
dt
Φ B2
Φ B1
di1
di2
Mutual induction: M =
=
−M
−M
; ε2 =
; ε1 =
i1
i2
dt
dt
=
i
Series RL Circuits:
ε
R
(1 − e
−t
τL
) (rise of current)
−t
i = io e τ L (decay of current)
1
UB =
Li 2 (magnetic energy)
2
10. Magnetic Energy:
B2
(magnetic energy density)
uB =
2 µo
Ch. 31. Electromagnetic Oscillations and Alternating Current
q2
Li 2
, UB =
, U = U E + U B = const
2C
2
1.
LC Energy transfer: U E =
2.
Emf of an electromagnetic generator:
3.
1
LC Charge and Current Oscillations: q =
−ωQ sin(ωt + φ )
Q cos(ωt + φ ); ω = ; i =
LC
ε= - ∆Φ /∆t=NABω sin(ω t)
4. Damped Oscillations: q =
Qe − Rt 2 L cos(ω ' t + φ ), where ω ' =
ω2 − ( R 2L)
5. Alternating Currents; Forced Oscillations:
A series RLC circuit may be set into forced oscillation at a driving angular
frequency ωd by an external alternating emf ε =ε m sin
=
ωd t ; i
I sin(ωd t − φ )
2
εm
ωd
=
when
ω. Then
=
XC
=
X
L, φ
1.
Resonance:
=
I
2.
Capacitive reactance: VC=IXC; X C =
3.
π/2 rad)
Inductive reactance: VL=IXL; X L = ωd L ; the current lags behind the voltage by π/2
4.
R
0
1
; the current leads voltage by π/2 radians (φ=ωd C
radians(φ=π/2 rad)
Series RLC Circuits
Relation between emf and current: εm=IZ
Impedance: Z =
R 2 + (X L − X C )
2
Phase angle between current and voltage: tan φ =
XL − XC
R
Average Power dissipated:
2
=
Pavg
I=
ε rms I rms cos φ , =
I rms
rms R
ε rms
2; =
I/
Vrms V /
2; =
ε/
2
Power factor of the circuit: cosφ
5.
Transformers:
Vs = V p
Ns
(transformation of voltage)
Np
Is = I p
Np
Ns
(transformation of currents)
The equivalent resistance of the secondary circuit, as seen by the generator
2
Req
N 
=  p  R, where R is the resistive load of the secondary circuit
 Ns 
Ch. 32. Maxwell’s Equations. Magnetism of matter
1.
2.
3.


B
⋅
dA
= 0
∫


dΦ
Maxwell’s Extension of Ampere’s Law: 
B
=
⋅
dS
+ µ0ienc
µ 0ε 0
∫
dt
dΦE
Displacement current: id = ε 0
dt
Gauss’ Law for Magnetic Field: Φ B=
1.
4
2.
3.
6
4.
7
5. A capacitor in an LC oscillator has a maximum potential difference of 15V and a
maximum energy of 360 μJ. At a certain instant the energy in the capacitor is 40 μJ. At
that instant what is the emf induced in the inductor?
1) U E ,max
2
2U E ,max 2 ⋅ 360 ×10−6
CVmax
=
; ⇒ C=
=
= 3.2 × 10−6 F = 3.2 µ F
2
2
2
15
Vmax
CV12
2) At a certain instant t=
; ⇒ potential difference across the capacitor
1 U E1
2
=
V1
2U E1
=
C
2 ⋅ 40 ×10−6
=
5.0V
−6
3.2 ×10
Li 2 q 2
3) At any instant in an oscillating LC circuit U = U B + U E =
+
2 2C
Since U remains constant with time
dU d  Li 2 q 2 
di q dq
di
=
+
=
= Li + Vi= 0;

 Li +
dt dt  2 2C 
dt C dt
dt
di
V1 =
0; ⇒ at t1 ε L =
5.0V
⇒ L +V =
−ε L + V =
dt
6.
I
VR
VL
θ=-28
VC-VL
VC
φ = arccos 0.88 = −28deg
Vo
A
) is connected to an emf that is increasing uniformly with time at a
d
rate of 100V/s. What is the displacement current between the plates of the capacitor?
7. A 1-μF capacitor ( C = ε o
V
d ( A)
dΦE
d ( EA)
A dV
dV
d
id ε=
ε 0 = ε 0 = ε 0= C
0
dt
dt
dt
d dt
dt
V
−6
(1 10 F ) ⋅ (100 ) =
1× 10−4 A
⇒ id =×
s
8.3.
Immediately after switch S in the circuit shown is closed, what is the current
through the battery?
The inductor prevents a fast build-up of the current through it,
so immediately after the switch is closes, the current in the inductor is zero.
V0
It follows that i =
R1 + R2