일반물리2 중간고사 범위 레시테이션 자료

UNIST

20131241 박용현

21. 전하

전기력𝑟

+𝑞1 +𝑞2

𝐹 𝐹

𝐹=𝑞1𝑞24 𝜋 𝜀0𝑟

2

22. 전기장

전기장+𝑞

+𝑄

𝐹

+𝑞+𝑞 𝐹𝐹

점전하에 의한 전기장

𝑄

𝐸=𝑄

4𝜋 𝜀0𝑟2

𝑞𝐹=𝑞𝑄

4 𝜋 𝜀0𝑅2𝑅

쌍극자에 의한 전기장

+𝑄−𝑄0𝑑

• 쌍극자 : 부호가 서로 반대이고 같은 크기의 전하를 가진 두 입자가 매우 가까운 거리를 유지하는 구조

쌍극자에 의한 전기장

𝑑

+𝑄

−𝑄

𝑟+¿¿

𝑟−

𝑧

𝐸=𝐸+ ¿+𝐸−=

𝑄4 𝜋 𝜀 0𝑟+¿

2 − 𝑄4 𝜋 𝜀0𝑟 −

2 ¿¿

점전하의 전하량과 같은 역할

𝐸=𝑄

4𝜋 𝜀0𝑟2

선전하에 의한 전기장

𝑑𝑠

𝑑𝑞=𝜆𝑑𝑠

• 점전하에 의한 전기장을 모두 더하는 방식

31. the three particles are fixed in place and have charges and . Distance What are the (a) magnitude and (b) direction of the net electric field at point P due to the particles?

+5𝑒

+5𝑒+2𝑒𝑎

𝑎

𝑞1

𝑞3 𝑞2

𝐸1𝐸2𝐸3

(𝑎 )𝐸𝑛𝑒𝑡=𝐸1+𝐸2+𝐸3=𝐸3

𝑃

33. two charged particles fixed in place on an x axis with separation .The ratio of their charge magnitudesis 4.00. the component of their net electric field along the x axis just to the right of particle 2.The x axis scale is set by 30.0 cm. (a) At what value of is maximum? (b) If particle 2 has charge , what is the value of that maximum?

−𝑞2+𝑞1𝐿 𝑥𝑠

𝐸𝑛𝑒𝑡 .𝑥

0

20𝑐𝑚

짜 전기장이 0 인 위치에서 찾는다 =

𝑥

23. 가우스 법칙

플럭스

𝐴 ′𝐴𝑣

’

𝐴𝑣+ 𝐴′ 𝑣 ′=0어떤 단면적에 지나가는 유체의 양

플럭스

𝐴𝜃𝐸

Φ=𝐸 ∙ �⃗�=𝐸𝐴cos𝜃

플럭스Φ=∑ �⃗� ∙ �⃗�

𝐸

𝑑𝐴𝜃𝐸 𝐸

𝑑𝐴

Φ=∮𝐸 ∙𝑑 �⃗�

가우스 법칙

𝑞𝑞𝑞

=

전하 밀도• 선전하밀도•

• 면전하밀도

• 부피전하밀도

가우스 법칙 적용 ( 점전하 )• 가우스 곡면 설정• 값 구하기• 계산

• 가우스 법칙 적용

𝑄𝑞𝑒𝑛𝑐=𝑄

𝜀0∮ �⃗� ∙𝑑 �⃗�=𝑞𝑒𝑛𝑐

점전하를 중심으로 한 구𝑑 𝐴1

𝑑 𝐴2

𝑑 𝐴3

Charged isolated conductor

conductor

𝐸=0

Charged isolated conductor

conductor

+𝑄

−𝑄

𝜀0∮ �⃗� ∙𝑑 �⃗�=𝑞𝑒𝑛𝑐

가우스 법칙 적용 ( 선전하 )

• 가우스 곡면 설정• 값 구하기• 계산• 가우스 법칙 적용

선전하를 중심으로 한 원기둥

h𝑞𝑒𝑛𝑐=𝜆h

가우스 법칙 적용 ( 면전하 )

• 가우스 곡면 설정• 값 구하기• 계산• 가우스 법칙 적용

면전하와 평행한 밑면을 가진 원기둥

𝑞𝑒𝑛𝑐=𝜎 𝐴𝐴

가우스 법칙 적용 ( 도체 구 )

𝑄• 가우스 곡면 설정• 값 구하기• 계산• 가우스 법칙 적용

도체 구와 중심이 같은 구

𝑞𝑒𝑛𝑐=𝜌𝑉=𝑄

43𝜋 𝑅3

× 43𝜋𝑟 3= 𝑟

3

𝑅3𝑄

∮𝐸 ∙𝑑 �⃗�=𝐸 ∙𝑑 𝐴1+𝐸 ∙𝑑 𝐴2+𝐸 ∙𝑑 𝐴3+⋯

𝜀0∮ �⃗� ∙𝑑 �⃗�=𝑞𝑒𝑛𝑐⇒𝜀0𝐸 ∙ (4𝜋 𝑟2 )= 𝑟3

𝑅3𝑄⇒𝐸=

𝑟𝑄4𝜋 𝜀0𝑅

3

3 An unknown charge sits on a conducting solid sphere ofradius 10 cm. If the electric field 15 cm from the cen-ter of the sphere has the magnitude N/C and is di-rected radially outward,what is the net charge on the sphere?

10𝑐𝑚15𝑐𝑚

𝜀0∮ �⃗� ∙𝑑 �⃗�=𝑞𝑒𝑛𝑐

4. A charge of uniform linear density 2.0 nC/m is dis-tributed along a long, thin, nonconducting rod.The rod is coaxial with a long conducting cylindrical shell (in-ner radius 5.0 cm, outer radius 10 cm). The net charge on the shell is zero. (a) What is the magnitude of the electric field 15 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?

+𝜆h

− 𝜆h+𝜆h

7. two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have ex-cess surface charge densities of opposite signs and magnitude 7.00 C/. In unit-vector notation, what is the electric field at points (a) to the left of the plates, (b) to the right ofthem, and (c) between them?

) ) )

11. The volume charge density of a solid nonconduct-ing sphere of radius R 5.60 cm varies with radial dis-tance r as given by r (14.1 pC/)(a) What is the sphere’s total charge? What is the field magnitude E at (b) , (c) , and (d) ? (e) Graph E versus r.

𝑄𝑅

(𝑑 )𝜀0𝐸 ∙ (4𝜋𝑟2 )=∫

0

𝑅

𝜌 𝑑𝑉

𝑑𝑉=4𝜋𝑟 2𝑑𝑟

11. The volume charge density of a solid nonconduct-ing sphere of radius R 5.60 cm varies with radial dis-tance r as given by r (14.1 pC/m3). (a) What is the sphere’s total charge? What is the field magnitude E at(b) , (c) , and (d) ? (e) Graph E versus r.

𝑄𝑅 (𝑒 )𝜀0𝐸 ∙ (4𝜋 𝑟2 )=𝑞𝑒𝑛𝑐=∫𝑑𝑞𝑒𝑛𝑐=∫

0

𝑟

𝜌𝑑𝑉𝐸

𝑟

17. proton is a distanced/2 directly above the center of a square of side d. What is the magnitude of the electric fluxthrough the square? (Hint:Think of the square as one face of a cube with edge d.)

𝑑

+𝑒𝑑2

25. Two charged concentric spherical shells have radii 10.0 cm and 15.0 cm. The charge on the inner shell is C, and that on the outer shell is C. Find the electric field (a) at cm and (b) at cm

𝑟 𝑖𝑛

𝑟𝑜𝑢𝑡

𝑞𝑖𝑛𝑞𝑜𝑢𝑡

25. Two long, charged, thin-walled, concentric cylin-drical shells have radii of 3.0 and 6.0 cm. The charge per unit length is 5.0C/m on the inner shell and 7.0C/m on the outer shell.What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance cm? What are (c) E and (d) the direction at cm?

=

𝑞𝑖𝑛

𝑞𝑜𝑢𝑡

24. 전기 포텐셜

포텐셜 vs 포텐셜 에너지

𝑀𝑚

𝑟중력포텐셜에너지

포텐셜 vs 포텐셜 에너지

𝑄

𝐸=𝑄

4𝜋 𝜀0𝑟2

𝑞𝐹=𝑞𝑄

4 𝜋 𝜀0𝑅2𝑅

포텐셜 vs 포텐셜 에너지

𝑄

𝑉

𝑞𝑈=𝑞𝑉𝑅

포텐셜 vs 포텐셜 에너지

𝑚

𝑅

h (𝑟 )𝑈=𝑚 h𝑔 (𝑅)

0

포텐셜 vs 포텐셜 에너지

𝑞

𝑅

𝑉 (𝑟 )𝑈=𝑞𝑉 (𝑅)

𝑄

0

전기 포텐셜𝐹=

𝑞𝑄4 𝜋 𝜀0𝑟 2

𝐸=𝑄

4𝜋 𝜀0𝑟2𝑉=

𝑄4𝜋 𝜀0𝑟

𝑈=𝑞𝑄4𝜋 𝜀0𝑟

𝐸=𝐹𝑞 𝑉=

𝑈𝑞 =−𝑊𝑞

전기 포텐셜 계산

∆𝑉=−∫𝐸 ∙𝑑 �⃗�

점전하에서의 전기 포텐셜

𝑄𝑉=

𝑄4𝜋 𝜀0𝑅

𝑞𝑅

여러 점전하에서의 전기 포텐셜

𝑄1𝑄3

𝑄2

𝑄5𝑄4

𝑄6 𝑄7𝑄8

𝑞𝑉=∑

𝑖𝑉 𝑖=∑

𝑖

𝑄𝑖4𝜋 𝜀0𝑟 𝑖

쌍극자에서의 전기 포텐셜

+𝑄

−𝑄𝑑

𝜃

𝜃

𝑟+¿¿

𝑟−

𝑟−−𝑟+¿ ¿

𝑃

𝑟

7 particles with the charges =+15e and 5e are fixed in place with a separation of d 24.0 cm. With electric po-tential defined to be V 0 at infinity, what are the finite (a) positive and (b) negative values of x at which the net electric potential on the x axis is zero?

+15𝑒 −5𝑒𝑑

𝑥𝑥

18 seven charged particles are fixed in place to form a square with an edge length of 4.0 cm. How much work must we do to bring a particle of charge 6e initially atrest from an infinite distance to the center of the square?

−3 𝑒−2𝑒−𝑒

+2𝑒

+3𝑒 +3𝑒 +𝑒𝑉 0=∑

𝑖=1

7

𝑉 𝑖(0 )

+6 𝑒

36. Sphere 1 with radius has positive charge . Sphere 2 with ra-dius 3.00 is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to re-tain only negligible charge, (a) is potential of sphere 1 greater than,less than, or equal to potential of sphere 2? What fraction of ends up on (b) sphere 1 and (c) sphere 2? (d) What is the ratio / of the surface charge densities of the spheres?

𝑅1𝑅2

𝑉 1 𝑉 2𝑞2𝑞1𝑞

36. Sphere 1 with radius has positive charge . Sphere 2 with ra-dius 3.00 is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to re-tain only negligible charge, (a) is potential of sphere 1 greater than,less than, or equal to potential of sphere 2? What fraction of ends up on (b) sphere 1 and (c) sphere 2? (d) What is the ratio / of the surface charge densities of the spheres?

𝑞1

𝑅1𝑅2

𝑞2𝐸1=

𝑞14𝜋 𝜀0𝑟

2 (𝑅1<𝑟 )

))

𝐸2=𝑞2

4𝜋 𝜀0𝑟2 (𝑅2<𝑟 )

36. Sphere 1 with radius has positive charge . Sphere 2 with ra-dius 3.00 is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to re-tain only negligible charge, (a) is potential of sphere 1 greater than,less than, or equal to potential of sphere 2? What fraction of ends up on (b) sphere 1 and (c) sphere 2? (d) What is the ratio / of the surface charge densities of the spheres?

𝑞1

𝑅1𝑅2

𝑞2𝑉 ∞−𝑉 𝑅1=−∫𝑅1

∞ 𝑞14𝜋 𝜀0𝑟

2 𝑟 ∙𝑑 �⃗�𝑉 ∞−𝑉 𝑅2=−∫

𝑅2

∞ 𝑞24𝜋 𝜀0𝑟

2 �̂� ∙𝑑𝑟

𝑉 1=𝑉 𝑅1=𝑞1

4𝜋 𝜀0𝑅1

𝑉 2=𝑉 𝑅2=𝑞2

4𝜋 𝜀0𝑅2

𝑉 𝑅1−𝑉 0=−∫0

𝑅1

0 ∙𝑑𝑟=0𝑉 𝑅2

−𝑉 0=−∫0

𝑅2

0 ∙𝑑 �⃗�=0

𝑞=𝑞1+𝑞2

𝑞14𝜋 𝜀0𝑅1

=𝑞2

4𝜋 𝜀0𝑅2

𝜎 1=𝑞1

4𝜋 𝑅12

𝜎 2=𝑞2

4 𝜋 𝑅22

45. A metal sphere of radius 15 cm has a net charge of(a) What is the electric field at the sphere’s surface? (b) If at infinity, what is the electric potential at the sphere’s surface? (c) At what distance from the sphere’s surface has the electric potential decreased by 500 V?

+3.0×108𝐶𝑥

𝑉 𝑥 (𝑐 )𝑉 𝑥−𝑉 𝑅=−∫𝑅

𝑥 𝑞14 𝜋𝜀0𝑟

2 𝑟 ∙𝑑�⃗�=−500

500𝑉

(𝑏)𝑉 ∞−𝑉 𝑅=−∫𝑅

∞ 𝑞14𝜋 𝜀0𝑟

2 𝑟 ∙𝑑𝑟

𝐸=𝑞1

4𝜋 𝜀0𝑟2

48. Two isolated, concentric, conducting spherical shells have radii 0.500 m and1.00 m, uniform charges 2.00 mC and 21.00 mC, and negligible thicknesses.What is the magnitude of the electric field at radial distance (a)4.00 m, (b) 0.700 m, and (c) 0.200 m? With 0 at infinity, what is V at (d)4.00 m, (e) 1.00 m, (f) 0.700 m, (g)0.500 m,(h)0.200 m, and (i) 0? ( j) Sketch and

𝑞1

𝑞2

𝑅1

𝑅2

𝐸=𝑞1+𝑞24𝜋 𝜀0𝑟

2 (𝑅2≤𝑟 )

𝐸=𝑞1

4𝜋 𝜀0𝑟2 (𝑅1≤ 𝑟 ≤𝑅2)

(

48. Two isolated, concentric, conducting spherical shells have radii 0.500 m and1.00 m, uniform charges 2.00 mC and 21.00 mC, and negligible thicknesses.What is the magnitude of the electric field at radial distance (a)4.00 m, (b) 0.700 m, and (c) 0.200 m? With 0 at infinity, what is V at (d)4.00 m, (e) 1.00 m, (f) 0.700 m, (g)0.500 m,(h)0.200 m, and (i) 0? ( j) Sketch and

𝑞1

𝑞2

𝑅1

𝑅2

𝑉 ∞−𝑉 𝑅1=𝑉 ∞−𝑉 𝑅2+ (𝑉 𝑅2−𝑉 𝑅1 )(𝑅1≤𝑟 ≤𝑅2)

(

55. A nonconducting sphere has radius R 2.31 cm and uniformly distributed charge =3.50 fC.Take the electric potential at the sphere’s center to be . What is V at radial distance (a)1.45 cm and (b) . (c) If, instead, at infinity, what is V at

𝑄𝐸=

𝑟𝑄4𝜋 𝜀0𝑅3

𝑉 0−𝑉 𝑟=−∫𝑟

0 𝑟𝑄4 𝜋 𝜀0𝑅

3 �̂� ∙𝑑𝑟

𝑉 𝑟=−𝑟2𝑄

8𝜋 𝜀0𝑅3

𝑉 0−𝑉 𝑅=−∫𝑅

0 𝑟𝑄4𝜋 𝜀0𝑅

3 𝑟 ∙𝑑 �⃗�

𝑉 𝑅=−𝑄

8𝜋 𝜀0𝑅𝑉 ∞−𝑉 𝑅=−∫

𝑅

∞ 𝑄4𝜋 𝜀0𝑟

2 𝑟 ∙𝑑�⃗�

𝑉 𝑅=𝑄

4𝜋 𝜀0𝑅

𝐸=𝑄

4𝜋 𝜀0𝑟2

𝑄

55. A nonconducting sphere has radius R 2.31 cm and uniformly distributed charge q3.50 fC.Take the electric potential at the sphere’s center to be V0 0. What is V at radial distance (a) r 1.45 cm and (b) . (c) If, instead, at infinity, what is V at r=R

58.Two large parallel metal plates are 1.5 cm apart and havecharges of equal magnitudes but opposite signs on their facing surfaces.Take the potential of the negative plate to be zero. If the potential halfway between the plates is then 10.0 V, what is the electric field in the region between the plates??

1.5 cm

∆𝑉=∫ �⃗� ∙𝑑 �⃗�

𝑉=0 𝑉=10𝑉𝑉=20𝑉

25. 축전기

축전기+𝑞

−𝑞

전기용량+𝑞

−𝑞

𝑄=𝐶𝑉

𝐶𝑉

전기용량

𝑞=𝐶𝑉

−𝑞

+𝑞

𝜀0∮ �⃗� ∙𝑑 �⃗�=𝑞𝐸

𝑉= ∫−

+¿ �⃗� ∙𝑑�⃗�¿

¿

𝑑

𝐶=𝜀0𝐴𝑑

𝐴

실린더형 축전기

𝑟h

𝑉= ∫−

+¿ �⃗� ∙𝑑�⃗�¿

¿

𝑏𝑎

𝑞=𝐶𝑉

구형 축전기

𝑉= ∫−

+¿ �⃗� ∙𝑑�⃗�¿

¿

𝑞=𝐶𝑉

회로에서의 축전기 ( 병렬연결 )

𝑉𝐶1 𝐶2 𝐶3

𝑉=𝑉 1=𝑉 2=𝑉 3

회로에서의 축전기 ( 직렬연결 )

𝑉

𝐶1 𝐶2 𝐶3

𝑞=𝑞1=𝑞2=𝑞3

축전기에 저장된 에너지

−𝑞

+𝑞𝑑

𝑉𝐸𝑉

부도체가 들어간 축전기

−𝑞

+𝑞𝑑

𝑉

𝑄=𝐶𝑉

𝐸

부도체가 들어간 축전기

−𝑞

+𝑞

𝐸

부도체가 들어간 축전기

−𝑞

+𝑞

𝑑 𝐸 𝐸 ′

+𝑞 ′

−𝑞 ′

𝑉= ∫−

+¿ �⃗� ∙𝑑�⃗�¿

¿

𝑞=𝐶𝑉

𝐶=𝜅 𝜀0𝐴𝑑

17. The parallel plates in a capacitor, with a plate area of 8.50 cm2 and an air-filled separation of 8.00 mm, are charged by a 16.00 V battery. They are then dis-connected from the battery and pushed together (without discharge) to a separation of 3.00 mm. Ne-glecting fringing, find (a) the potential difference be-tween the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the (negative)work in pushing them together.

𝐶

18. a parallel-plate capacitor with a plate area A 5.56 cm2 and separation d 5.56 mm. The left half of the gap is filled with material of dielectric constant = 7.00; the right halfis filled with material of dielectric constant =12.0.What is the capacitance?

𝜅1𝜅2

𝐶=𝐶1+𝐶2

𝐶=𝜅 𝜀0𝐴𝑑

19. 10.0 mF, 20.0 mF, and25.0mF. If no capacitor can withstand a potential difference of more than 100 V without failure,what are (a) the magnitude of the max-imum potential difference that can exist between points A and B and (b) the maximum energy that can be stored in the three-capacitor arrangement?

𝑎 𝑏𝐶1 𝐶2 𝐶3𝑞=𝑞1=𝑞2=𝑞3

100𝑉125𝑉250𝑉

19. 10.0 mF, 20.0 mF, and25.0mF. If no capacitor can withstand a potential difference of more than 100 V without failure,what are (a) the magnitude of the max-imum potential difference that can exist between points A and B and (b) the maximum energy that can be stored in the three-capacitor arrangement?

𝑎 𝑏𝐶1 𝐶2 𝐶3𝑞=𝑞1=𝑞2=𝑞3

40𝑉50𝑉100𝑉

𝑈=12 𝐶𝑉

2

𝜅

25. A certain parallel-plate capacitor is filled with a di-electricfor which =5.5. The area of each plate is 0.034 m2, and theplates are separated by 2.0 mm.The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 200 kN/C.What is the maximum energy that can be stored in the capacitor?.

−𝑞

+𝑞

𝐸𝑑

𝐴

𝑉= ∫−

+¿ �⃗� ∙𝑑�⃗�¿

¿

𝑈=12 𝐶𝑉

2

53. A 100 pF capacitor is charged to a potential differ-ence of 50 V, and the charging battery is disconnect-ed.The capacitor is then connected in parallel with a second (initially uncharged) capacitor. If the potential difference across the first capacitor drops to 35 V, what is the capacitance of this second capacitor?

𝐶1

𝑉𝐶2

𝑞1=𝐶1𝑉 1

’==

26. 전류와 저항

전류• 도선 내의 양전하의 흐름

𝑖

전류밀도• 도선 내의 양전하가 얼마나 많이 흐르나 ?-> 전류• 주어진 단면적을 기준으로 어떤 방향으로 움직이는 양전하가 얼마나 많이 있는가 ?-> 전류밀도

𝑖=∫ �⃗� ∙𝑑 �⃗�

전류밀도𝑖=∫ �⃗� ∙𝑑 �⃗� 𝐿

𝑞=(𝑛𝐴𝐿)𝑒𝑡= 𝐿𝑣𝑑 �⃗�=(𝑛𝑒)𝑣𝑑

저항

𝑖

𝑉

𝑅

𝑉=𝐼𝑅

전하들은 도선을 이루는 원자들의 구조와 운동에 영향을 받음 재질마다 고유한 저항값이 있음저항기장과 전류밀도의 비율로 구함

저항𝐸

𝜌=𝐸𝐽 𝜌=𝑉 /𝑑𝑖 / 𝐴 ⇒𝑅=𝜌 𝐴𝑑

전력

𝑖

𝑉

𝑅

𝑖 𝑖

𝑑𝑈=𝑉𝑑𝑞=𝑉𝑖𝑑𝑡𝑃=𝑑𝑈𝑑𝑡 =𝑉𝑖=𝑖2𝑅=

𝑉 2

𝑅

27. 회로

저항의 직렬연결𝑅1 𝑅2 𝑅3

𝑉=𝑉 1+𝑉 2+𝑉 3

저항의 병렬연결𝑉

𝑅1 𝑅2 𝑅3

𝐼=𝐼 1+𝐼 2+ 𝐼3

복잡한 회로에서의 계산• Junction rule

𝑖1 𝑖2

𝑖3

𝑖1=𝑖2+𝑖3

복잡한 회로에서의 계산• loop rule

𝑉𝑅1 𝑅2 𝑅3

−𝑖2𝑅2+𝑖1𝑅1=0

𝑖1 𝑖2

s

복잡한 회로에서의 계산𝑉

𝑅1 𝑅2 𝑅3

𝑉𝑅4

𝑖1 𝑖2 𝑖3

𝑖4𝑖2+𝑖4𝑖1+𝑖2+𝑖4

• 1. 각 저항에 흐르는 전류를 미지수로 두고 각 전류의 방향을 정한다 .• 2.junction rule 을 이용해 각 전류들의 관계를 구한다 .

• 3. 닫힌 회로를 설정하고 loop rule 을 이용하여 연립방정식을 세운다 .

RC 회로𝑅 𝐶

𝑉

𝑡

𝑞

RC 회로𝑅

𝐶

𝑡

𝑞

8. 100 , 50.0 , 75.0 , and the ideal battery has em-f6.00 V.(a) What is the equivalent resistance? What is in (b) resistance 1,(c) resistance 2, (d) resistance 3, and(e) resistance 4?

𝑉

𝑅1𝑅2 𝑅4 𝑅3

병렬연결

==

21. Switch S is closed at time t =0,to begin charging an initially uncharged capacitor of capacitance through a resistor of resistance . At what time is the potential across the capacitor equal to that across the resistor?

𝑉

𝑅

𝐶

𝑉 𝑅=𝑉 𝐶=12 𝑉

𝑞=𝐶𝑉 𝐶=𝐶𝑉 (1−𝑒− 𝑡𝑅𝐶 )

60. five 5.00 resistors. Find the equivalent resistancebetween points (a) F and H and (b) F and G. (Hint: For eachpair of points, imagine that a battery is connected across the pair.)

𝐹

𝐺

𝐻

𝑅 𝑅

𝑅𝑅

𝑅

𝑉

𝑉

𝑉

2. the ideal batteries have emfs =10.0 V and 0.500, and the resistances are each 4.00 .What is the current in (a)resistance 2 and (b) resistance 3?

𝑉 1

𝑅1𝑅3

𝑅2

𝑉 2

𝑖1 𝑖2

𝑖3

Juction rule:

𝑆

Roop rule1:Roop rule2:

69. 100 , 0 , and the ideal batteries have emfs 6.0 V, 5.0 V, and 4.0 V. Find (a) the current in resistor1, (b) the current in resistor 2,and (c) the potential

𝑉 1

𝑉 2 𝑉 3

𝑅1

𝑅2

𝑆

Roop rule1:𝑖1

𝑖2

Roop rule2:

𝑆

28. 자기장

자기장의 정의

𝐵𝐸=𝐹𝑞

𝐵=𝐹

|𝑞|𝑣

자기력

𝐹

�⃗�=𝑞𝑣× �⃗�

𝐹=𝐵𝑞𝑣 ,𝐹=𝑚𝑣2𝑟

자기장내에서 원운동하는 전하

𝐵

𝐹

도선에 작용하는 자기력

𝐹

�⃗�=𝑖 �⃗�× �⃗�𝑞=𝑖𝑡=𝑖 𝐿𝑣𝑑

, �⃗�=𝑞 �⃗�× �⃗�

𝐿

𝐵

𝐹

도선에 작용하는 토크

𝑁 𝑆𝑖

𝑖

𝐹

𝜏=𝑏2 ×𝐵𝑖𝑎+𝑏2 ×𝐵𝑖𝑎=𝐵𝑖𝑎𝑏

𝑎

𝑏

3. A particular type of fundamental particle decays by transforming into an electron and a positron . Suppose the decaying particle is at rest in a uniform magnetic field of magnitude 3.53 mT and the e and e move away from the decay point in paths lying in a plane perpendicular to . How long after the decay do the e and e collide?

𝑟=𝑚𝑣𝐵𝑞

𝑇=2𝜋𝑟𝑣 =

2𝜋𝑚𝐵𝑞

𝑡=𝑇2

8. a charged particle moves into a region of uniformmagnetic field , goes through half a circle, and then exits that region.The particle is either a proton or an electron (you must decide which). It spends 130 ns in the region.(a) What is the magnitude of ? (b) If the particle is sent back through the magnetic field (along the same initial path) but with 2.00 times its previous kinetic energy, how much time does it spend in the field during this trip?

�⃗�

?𝑇=

2𝜋𝑟𝑣 =

2𝜋𝑚𝐵𝑞

32. a metal wire of mass m 24.1 mg can slide with negligible friction on two horizontal parallel rails sepa-ratedby distance d 2.56 cm. The track lies in a vertical uni-form magnetic field of magnitude 56.3 mT. At time t 0, device G is connected to the rails, producing a con-stant current i 9.13 mA in the wire and rails (even as the wire moves). At , what are the wire’s (a) speed and (b) direction of motion (left or right)?

𝐺

𝑖

𝑖 𝑖

�⃗�

�⃗�=𝑖 �⃗�× �⃗�=𝑚𝑎𝑚�⃗�

59. A mass spectrometer is used to separate uranium ions of mass kg and charge 3.20C from related species.The ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.00 m. After traveling through 180° and passing through a slit of width 1.00 mm and height 1.00 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 100 mg of material per hour, calculate (b)the current of the desired ions in the machine and (c) the thermal energy produced in the cup in 1.00 h.

𝑟

𝑉

�⃗�에너지 보존 :등속원운동 :

𝑀𝑚 =𝑁 ,𝑖=𝑞𝑁 ,𝐸=𝑞𝑉=𝑉𝑖𝑡

𝑀 : h𝑤𝑒𝑖𝑔 𝑡 𝑜𝑓 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑝𝑒𝑟 h𝑜𝑢𝑟

29. 전류에 의한 자기장

전류에 의한 자기장 계산

𝑖 𝑑𝑠𝑟

𝑑𝐵=𝜇04𝜋𝑖𝑑 �⃗�× �̂�𝑟 2

𝑑𝐵

𝑖𝑑 �⃗�=𝑑𝑞𝑑𝑡 𝑑 �⃗�=𝑑𝑞𝑑 �⃗�𝑑𝑡 =𝑑𝑞𝑣

무한직선도선에서의 자기장

𝑅

𝑠 𝑟

𝑖

𝑑𝑠 𝜃

오른손 법칙𝑖

원형 도선에 의한 자기장𝑑𝑠

𝑅

도선 사이의 자기력𝑑𝑖1

𝑖2𝐹

𝐵1

𝐵1=𝜇0𝑖12𝜋 𝑑

앙페르 법칙

∮ �⃗� ∙𝑑 �⃗�=𝜇0𝑖𝑒𝑛𝑐

𝑖1

𝑖2

무한 직선도선에서의 앙페르 법칙

∮ �⃗� ∙𝑑 �⃗�=𝜇0𝑖𝑒𝑛𝑐𝑖

𝑟

도선안에서의 앙페르 법칙𝑖

𝑟

𝑅

솔레노이드

𝑖

솔레노이드h

𝑎 𝑏

𝑐𝑑

h𝐵 0 0 0

8. four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a =7.00 cm. Each wire carries 15A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3. In unit vector notation, what is the net magnetic force per meter of wire length on wire 1?

1 2

34

�⃗�2

�⃗� 2�⃗� 3�⃗� 4

𝐵=𝜇0 𝑖  2𝜋𝑟

�⃗�=𝑖 �⃗�× �⃗�

27. A long solenoid with 10.0 turns/cm and a radius of 7.00 cm carries a current of 20.0 mA. A current of 6.00 A exists in a straight conductor located along the cen-tral axis of the solenoid. (a) At what radial distance from the axis will the direction of the resulting mag-netic field be at 45.0° to the axial direction? (b) What is the magnitude of the magnetic field there?

𝐵=𝜇0 𝑖𝑛=𝜇0 𝑖 ′  2𝜋𝑟

𝐵𝑡𝑜𝑡=√2𝐵

35. The current density inside a long, solid, cylindrical wireof radius a 3.1 mm is in the direction of the central axis, and its magnitude varies linearly with radial dis-tance r from the axis according to where = 310 A/m2. Find the magnitude of the magnetic field at

∮ �⃗� ∙𝑑 �⃗�=𝜇0𝑖𝑒𝑛𝑐∮ �⃗� ∙𝑑 �⃗�=𝜇0𝑖𝑒𝑛𝑐

30. Induction and induc-tance

Magnetic flux

Φ𝐵=∫ �⃗� ∙ �⃗�𝐴

패러데이의 법칙 , 렌츠의 법칙

𝑖𝑖

𝑖𝑖

𝜀=−𝑁𝑑Φ𝐵𝑑𝑡

(𝑒𝑚𝑓 ) ,𝑁 : number  of   turns   in   the   coil

Inductor and inductance

𝑖

𝑁

𝐿=𝑁Φ𝐵𝑖

Self induction

𝑖

𝑁

𝑖 ′

𝜀𝐿=−𝑁𝑑Φ𝐵𝑑𝑡

𝐿=𝑁Φ𝐵𝑖

RL 회로𝑅 𝐿

𝑉

𝑖 𝑖 ′

𝜀𝐿=− 𝐿𝑑𝑖𝑑𝑡

𝑖=𝑉𝑅 (1−𝑒−𝑅𝑡𝐿 )

Inductor 에 저장된 에너지

=

26. a right angle. A conducting bar in contact with the rails starts at the vertex at time t 0 and moves with a constant velocity of 8.90 m/s along them. A magnetic field withB 0.350 T is directed out of the page. Calcu-late (a) the flux through the triangle formed by the rails and bar at and (b) the emf around the triangle at that time. (c) If the emf is at n, where a and n are con-stants,what is the value of n?

𝜀=−𝑁𝑑Φ𝐵𝑑𝑡 =

𝑑 (𝐵𝐴 )𝑑𝑡 =𝐵 𝑑𝐴𝑑𝑡

Φ𝐵=∫ �⃗� ∙𝑑 �⃗�=𝐵∫𝑑 �⃗�=𝐵𝐴=𝐵 12 (𝑣𝑡 )(2𝑣𝑡 )𝑣𝑡𝑣𝑡𝑣𝑡

𝑑𝐴

55. rod of length L 10.0 cm that is forced to move at constant speed v 5.00 m/s along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.400 ; the rest of the loop has negligible resistance. A current i 100 A through the long straight wire at distance a 10.0 mm from the loop sets up a (nonuni-form) magnetic field through the loop. Find the (a) emf and (b) current induced in the loop. (c) At what rate is thermal energy generated in the rod? (d) What is the magnitude of the force that must be applied to the rod to make it move at constant speed? (e) At what rate does this force do work on the rod?𝑖

𝑎

𝐿

𝜀=−𝑁𝑑Φ𝐵𝑑𝑡

Φ𝐵=∫𝑎

𝑎+𝐿

𝑑Φ𝐵=∫𝑎

𝑎+𝐿

𝐵𝑑𝐴𝑣𝑡

𝑑𝑟

55. rod of length L 10.0 cm that is forced to move at constant speed v 5.00 m/s along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.400 ; the rest of the loop has negligible resistance. A current i 100 A through the long straight wire at distance a 10.0 mm from the loop sets up a (nonuni-form) magnetic field through the loop. Find the (a) emf and (b) current induced in the loop. (c) At what rate is thermal energy generated in the rod? (d) What is the magnitude of the force that must be applied to the rod to make it move at constant speed? (e) At what rate does this force do work on the rod?𝑖

𝑎

𝐿

(b)

(c) ’

(𝑑) �⃗�=𝑖 ′ �⃗�× �⃗�=𝑖 ′ 𝐿𝐵𝑃=𝐹𝑣𝑑𝐴𝑑𝑟

𝑣𝑡

40. a rectangular loop of wire with length a 2.2 cm, width b 0.80 cm, and resistance R 0.40 m is placed near an infinitely long wire carrying current i 4.7 A. The loop is then moved away from the wire at con-stant speed v 3.2 mm/s.When the center of the loop is at distancer 1.5b, what are (a) the magnitude of the magnetic flux through the loop and (b) the current in-duced in the loop?

𝑏

𝑟

𝑖

𝑎

Φ𝐵= ∫𝑟 −𝑏 /2

𝑟+𝑏 /2

𝑑Φ𝐵= ∫𝑟−𝑏/2

𝑟 +𝑏 /2

𝐵𝑑𝐴

𝑖𝑖𝑛𝑑=𝜀 /𝑅

𝑣

𝑑𝑟𝑑𝐴