Download 國立中山大學 103 學年度普通物理(二)期末考 104.6.24 1. (15%) A

國立中山大學 103 學年度普通物理(二)期末考
104.6.24
答題須知：
一、請由答題紙第六列開始填寫，並標明題號。考試時間為 19：00~21：30，20：00 以後方可交卷。
二、請詳述相關物理觀念及計算過程，記得寫上單位，建議以M.K.S制表示。遇力學題目時，請畫出力學圖解圖，未畫者將扣分。
1. (15%) A rectangular loop of wire has dimensions 0.500 m by 0.300 m.
The loop is pivoted at the x axis and lies in the xy plane as shown in
Figure 1. A uniform magnetic field of magnitude 1.50 T is directed at an
angle of 40.0° with respect to the y axis with field lines parallel to the yz
plane. The loop carries a current of 0.900 A in the direction shown.
Fig. 1
(Ignore gravitation.) We wish to evaluate the torque on the current loop.
(a) What is the direction of the magnetic force exerted on wire segment ab? , Explain. (b) What is the
direction of the magnetic force exerted on segment cd? , Explain. (c) Can the forces examined in
parts (a) and (b) combine to cause the loop to rotate around the x axis? (d) What is the direction of the
magnetic force exerted on segment bc? (e) What is the direction of the torque associated with this
force about an axis through the origin? (f) What is the torque on segment ad about an axis through
the origin? (g) From the point of view of Figure 1, once the loop is released from rest at the position
shown, will it rotate clockwise or counterclockwise around the x axis? (h) Compute the magnitude of
the magnetic moment of the loop. (i) What is the angle between the magnetic moment vector and the
magnetic field? (j) Compute the torque on the loop using the results to parts (h) and (i).
(a) The current in segment ab is in the -y direction. According to the right-hand rule 【1%】, the
magnetic force on it is in the -x direction. 【1%】
(b) The current in segment cd is in the +y direction. According to the right-hand rule 【1%】, the
magnetic force on it is in the +x direction.【1%】
(c) No.【1%】
(d) The magnetic force is perpendicular to both the direction of the current in bc and the magnetic
field. According to the right-hand rule, the magnetic force is in the yz plane at 50o clockwise
from the +y direction. 【1%】
(e) The force acting on segment bc tends to rotate it clockwise about the x axis, so the torque is in
the -x direction. 【1%】
(f) Because the lever arm of the force on segment ad is zero, the torque is zero. 【1%】
(g) Clockwise. 【1%】
(h) 𝜇 = 𝐼𝐴𝑁 = (0.9𝐴)[(0.5𝑚)(0.3𝑚)](1) = 0.135 𝐴 ∙ 𝑚2 . 【2%】
(i) Because the current flows counterclockwise around the loop, the magnetic moment vector is
directed upward, in the positive z direction. Therefore, the angle between it and the magnetic
field is 𝜃 = 90𝑜 − 40𝑜 = 500 . 【2%】
(j) 𝜏 = 𝜇𝐵𝑠𝑖𝑛𝜃 = (0.135 𝐴 ∙ 𝑚2 )(1.5 𝑇) sin(50𝑜 ) = 0.155 𝑁 ∙ 𝑚. 【2%】
2. (10%) A long, cylindrical conductor of radius R has the uniform
current density J as shown in Figure 2a. Find an expression for the
magnetic field magnitude B (a) at a distance r1 < R and (b) at
distance r2 > R (c) now, this conductor has one cylindrical cavities
with radius R/2 through its entire length as shown in Figure 2b, find
the magnitude of the magnetic field at point P.
Fig. 2
(a) For r1 < R, take a circle of radius r1 to apply Ampere’s law
⃑ ∙ 𝑑𝑠 = 𝜇0 𝐼 = 𝜇0 𝐽𝐴 【1%】
∮𝐵
2𝜋𝑟1 𝐵 = 𝜇0 𝐽𝜋𝑟1 2 , 𝐵 =
𝜇0 𝐽𝑟1
2
【2%】
(b) For r2 > R, take a circle of radius r2 to apply Ampere’s law
⃑ ∙ 𝑑𝑠 = 𝜇0 𝐼 = 𝜇0 𝐽𝐴 【1%】
∮𝐵
2𝜋𝑟2 𝐵 = 𝜇0 𝐽𝜋𝑅 2 , 𝐵 =
𝜇0 𝐽𝑅 2
2𝑟2
【2%】
(c) By the superposition principle, we can consider the system as an imaginary superposition of
two parts. One is the current in the direction of the actual current, distributed over the entire
cylinder of radius R and the other is the current in the opposite direction flowing over the
removed region.
For the cavity:
take a circle of radius (r2-R/2) to apply Ampere’s law 【1%】
⃑ ∙ 𝑑𝑠 = 𝜇0 𝐼 = 𝜇0 𝐽𝐴
∮𝐵
𝑅
𝑅 2
2𝜋(𝑟2 − 2 )𝐵 = 𝜇0 𝐽𝜋 ( 2 )
, 𝐵=
𝜇0 𝐽𝑅 2
【2%】
𝑅
2
8(𝑟2 − )
1
so the resultant magnetic field is 𝐵 = 𝜇0 𝐽𝑅 2 (2𝑟 −
2
1
𝑅
2
) 【1%】
8(𝑟2 − )
3. (10%) Consider a circular wire loop of radius a located in the
yz plane and carrying a steady current I in Figure 3a.
Calculate the magnetic field (a) at an axial point P a distance
from the center of the loop and (b) at the center of the loop (c)
if now there is a conductor consist of a circular loop and the
two long, straight sections as shown in the Figure 3b. Find the
magnetic field at the center of the loop.
(a) Use Biot-Savart law to find the magnitude of
element
Fig. 3
due to the current in any length
:
【1%】
Find the x component of the field element:
【1%】
Integrate over the entire loop:
From the geometry, evaluate cosθ:
Substitute cos θ into the integral:
【2%】
Integrate around the loop:
(b)
【1%】
【2%】
(c) By the superposition principle, we can consider the system as an imaginary
superposition of two parts. One is the long straight wire, having magnitude
directed into the page.
𝜇0 𝐼
2𝜋𝑎
【1%】
The other is the circular loop, having
𝜇0 𝐼
2𝑎
and directed into the page.
⃑ = (1 + 1 ) 𝜇0 𝐼
The resultant magnetic field is: 𝐵
𝜋 2𝑎
【1%】
【1%】
and
4. (15%)A long solenoid of radius R and length  has n turns of wire per
unit length and carries a time-varying current that varies sinusoidally as
I = Imax cos ωt, where Imax is the maximum current and ω is the
angular frequency of the alternating current source (Fig.4).
(a) Determine the magnitude of the induced electric field outside the
solenoid at a distance r > R from its long central axis.
(b) What is the magnitude of the induced electric field inside the
solenoid, a distance r from its axis?
(c) Find the self-inductance of the solenoid.
dB
d
dB
   B R 2    R 2
dt
dt
dt
(a) 
&
(1)
B = µ0nI = µ0nImax cos ωt
【1%】
(2)
Sub. (2) into (1):
dB
d
  R 2 0 nI max  cos t    R 2  nI max sin t
dt
dt
 
d
By use of Faraday’s law  E  ds   B
(4)
dt

(3)
【2%】
We have E  2r  R 2  0 nI max sin t
and E 
(b) 
0 nI max R2
2r
sin t
dB
d
dB
   B r 2    r 2
dt
dt
dt
(for r  R)
(5)
【3%】
【1%】
Sub. (2) into (5):

dB
d
  r 2 0 nI max  cos t    r 2 0 nI max sin t
dt
dt
Sub. (6) into (4):
E
(c) L 
0 nI max
2
r sin t
 for r  R 
NB nl0 nIR 2

 0 n 2lR 2
I
I
【3%】
【3%】
(6)
【2%】
Fig. 4
5. (10%)A toroid having a rectangular cross section (a = 2.00 cm by b = 3.00
cm) and inner radius R = 4.00 cm consists of N = 500 turns of wire that
carry a sinusoidal current I = Imax sin ωt, with Imax = 50.0 A and a
frequency f = ω/2π = 60.0 Hz. A coil that consists of N′ = 20 turns of wire is
wrapped around one section of the toroid as shown in Figure 5. Determine
the emf induced in the coil as a function of time.
In a toroid, the field inside the toroid at a distance r from its center is
【2%】
The magnetic flux is then
【3%】
and the induced emf is
【2%】
Substituting numerical values and suppressing units,
【2%】
【1%】
Fig. 5
6. (10%)Two solenoids #1 and #2, spaced close to each other and sharing the
same cylindrical axis, have N1=400 and N2=700 turns, respectively. A
current of 3.50 A in solenoid #1 produces an average flux of 300 μWb
through each turn of #1 and a flux of 90.0μWb through each turn of #2. (a)
Calculate the mutual inductance of the two solenoids. (b) What is the
inductance of #1? (c) What emf is induced in #2 when the current in #1
changes at the rate of 1.000 A/s?
(a) M 12 
N 212 700  90  106

 0.018 H
I1
3.50
【3%：公式 1 分，代入公式之數值 1 分，最終答案 1 分】
(b) L1 
N111 400  300  106

 0.034 H
I1
3.50
【3%：公式 1 分，代入公式之數值 1 分，最終答案 1 分】
(c)   M 12
【4%】
dI1
 0.018 1  0.018 Volt  18 mV
dt
Fig. 6
7. (12%) A DC voltage source, V, connects to the circuit consisting of a
resistor, R, and a inductor, L, at time zero.
(a) Please write the differential equation of current of the circuit
according to Kirchhoff Circuit Laws;
(b) Please write down the initial condition of the current and explain
why;
(c) Please solve the equation;
(d) What is the emf of L when time approaches to infinite?
dI
(a) V

IR L  0

dt
2 分 
L
Fig. 7
(1)
2分
(b) At t = 0 I = 0 【1%】
∵ At t = 0, the battery intends to supply finite current. The change of current at this moment is
very large. According to Faraday’s Law, the inductance (L) will response seriously to stop this
change. Therefore, I at t = 0 is zero. 【解釋 1%】
dI
R
V
 I
dt
L
L
(c) Rearrange eq. (1) =>
Let u  
R
V
I
L
L
 du  
dI
L du
R

dI =>
L
dt
R dt
(3)
Inst (4) back to (2) => 
∴ V
L
R V
I
L L
u ( t t )
t
du
R
   dt
u ( t 0 ) u
0
L

(5)
(6)
(7)
V
t0

u   L
R
V
 I 
t t
L
 L

(4)
du
L du
R
  dt
 u =>
u
R dt
L
Integrate both sides of (5) =>
0 t  0
∵I  
I t  t
(2)
(8)
du
R
  t => ln
u
L

R
V
I
L
L   R t (9)
V
L
L

Take exponential on both sides of (9) =>
(d) At t , I (t  ) 
R
V
R
R
I
L
L  e  L t => I (t )  V (1  e  L t )
V
R
L
(10) 【4%】
V
dI
 0 【2%】
is a constant independent of time. ∴  L   L
R
dt
8. (12%) An RLC circuit consists of a 150-  resister, a 21.0- μF capacitor, and 460-mH inductor
connected in series with120-V, 60.0-Hz power supply. (a) What are the inductive reactance, the
capacitive reactance, the total reactance, and the total impedance of the circuit? (b) Give the
complete diagram of the circuit. (c) What is the current amplitude of the circuit, and the phase
angle between the current and the supplied voltage? Which reaches its maximum earlier, the
current or the voltage? (d) What are the voltage amplitude of the resistor, the inductor, and the
capacitor? (e) What is the resonant frequency 0 of the circuit?
【1%】
【1%】
【1%】
【1%】
【2%】
【1%】
【1%】
【1%】
【1%】
【1%】
【1%】
9. (6%) Write down the Maxwell’s equations and the Lorentz force.
【1%】
【1%】
【2%】
【1%】
【1%】