GenPhyExercises 뜻(meaning)



1. ch26

1.1. p1


$I=\frac{\Delta Q}{\Delta t}=qnv_dA$
$v_d=\frac{I}{qnA}$


$J=qnv_d$
$v_d=\frac{J}{qn}$
v_d=drift velocity
J=전류 밀도
q=전자 전하
n=개수 밀도
1/((7.4e-7)*(8.4e28)*(1.6e-19))=0.0001


1.2. p2

A는 r=0.5, B는 r=1(밖), r=0.5(안)
면적은 r^2에 비례하므로
0.5^2:(1^2-0.5^2)=0.25:0.75=1:3

1.3. p3

5.0mm=5e-3m
2.0cm=2e-2m
P=1.0W=1.0J/s, P=I²R, P=V²/R
비저항 ρ=3.5e-5
저항R = ρ × (길이) / (단면적) = (3.5e-5)*(2e-2)/(pi*(5e-3)^2) = 0.00891268
I²=P/R=112.2, I=10.59, J=I/A=10.59/(pi*(5e-3)^2)=1.34836 × 10^5=1.3e5 A/m^2 (a)
V²=PR=0.0089, V=0.0943=9.4e-2 (b)

=p64

1.4. p4

=p77
기체 온도가 변하지 않으려면,
$P_{\rm R}=i^2R$
과 피스톤의 에너지 증가 비율
$P_{\rm m}=mg(dh/dt)=mgv$
이 같아야 함
$i^2R=mgv$
$v=\frac{i^2R}{mg}=\frac{(0.24A)^2(550\Omega)}{(12kg)(9.8m/s^2)}=0.27m/s$


2. ch 27


2.1. p1

반지름 d/2이고 길이 9L인 도선의 저항 = rho * (9L) / pi*(d/2)^2 = 36*rho*L / pi*d^2
반지름 x이고 길이 L인 도선의 저항 = rho * (L) / pi*(x^2) = rho*L / pi*x^2
36:d^2=1:x^2
d^2=36*x^2
x^2=d^2/36, x=d/6
구하는 2x는 d/3
C

2.2. p2

전체 emf = E1-E2
전체 내부 저항 = r+r=2r

전류: I = (total emf)/(total resistance) -= (E1-E2)/(2r)
Current is the same everywhere as it is a series circuit.

Power in smaller cell (cell 2) P = VI = (emf) x I = E2 x (E1-E2)/(2r) = (E1-E2)E2/2r
전력 = P = VI = (emf)*

This is answer D

If you connect a 2V emf cell and a 1.5V emf cell the 'wrong way round', the overall emf is 2-1.5=0.5V, because the 'driving force' of one cell is partly cancelled by the opposite 'driving force' of the other.

So the overall emf = E1- E2

Internal resistance of cells can be thought of simply as series resistors. So the total resistance is r+r=2r

The circuit is equivalent to a cell of emf (E2-E1), connected to a resistance of 2r.

Current: I = (total emf)/(total resistance) -= (E1-E2)/(2r)
Current is the same everywhere as it is a series circuit.

Power in smaller cell (cell 2) P = VI = (emf) x I = E2 x (E1-E2)/(2r) = (E1-E2)E2/2r

This is answer D.

2.3. p3

= p23
(a) 아래쪽에 고리 법칙 적용하면
$\mathcal{E}_2-i_1R_1=0$
$i_1=\frac{\mathcal{E}_2}{R_1}=\frac{5.0V}{100\Omega}=0.05A$

(b) 위쪽에 적용하면
$E_1-E_2-E_3-i_2R_2=0$
$i_2=\frac{E_1-E_2-E_3}{R_2}=\frac{6.0V-5.0V-4.0V}{50\Omega}=-0.060A$

(c) If $V_b$ is the potential at point b, then the potential at point a is
$V_a=V_b+E_3+E_2$
so
$V_a-V_b=E_3+E_2=4.0V+5.0V=9.0V$


2.4. p4

= p79
전하
$q(t)=EC(1-e^{-t/RC})$
충전전류
$i(t)=dq/dt=(E/R)e^{-t/RC}$
emf 장치 에너지 공급율
$P=Eidt$

(a)
emf가 주는 에너지는
$U=\int_0^{\infty}Pdt=\int Eidt=\frac{E^2}{R}\int e^{-t/RC}dt=CE^2=2U_C$
capacitor에 저장된 에너지는
$U_C=\frac12CE^2$

(b)
$U_R=\int_0^{\infty}i^2Rdt=\frac{E^2}{R}\int e^{-2t/RC}dt=\frac12CE^2$


3. ch 28

3.1. p1

E=5e5
B=0.8
a=0 : qE=qvB
v=E/B= 625000
D

3.2. p2

$\mu=IA=5.0e-4$
$B=0.50T$
$W=U_f-U_i=-\mu B\cos\theta_f+\mu B\cos\theta_i$
$=0+5\times 10^{-4}\times 0.5=2.5\times 10^{-4}J$

B

3.3. p3

=p15

(a)
$|\vec{E}|=v|\vec{B}|=(20.0m/s)(30.0T)=600V/m$
$\vec{E}=-(0.600V/m)\hat{k}$
F는 상쇄됨(vanish)

(b)
V=Ed=(600V/m)(2.00m)=1200V

3.4. p4

=p51