1516 Level NS Core Physics BGT Questions Solutions Electricity

8/19/2019 1516 Level NS Core Physics BGT Questions Solutions Electricity

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Physics Core Course

Basic Questions

Table of Contents

I. ELECTRICITY:

1. Electric Charge And Electric Field 1

2. Gauss’s Law 5

3. Electric Potential 10

4. Capacitance and Dielectrics 13

5. Current, Resistance and E.M.F. 17

6. Direct-Current Circuits 20

7. Magnetic Fields and Magnetic Forces 24

8. Sources of Magnetic Fields 27

II. RELATIVITY: 31

III. Mechanics Part 1 35

Mechanics Part 2 38

Mechanics Part 3 41

Mechanics Part 4 50

Mechanics Part 5 53

Mechanics Part 6 57

Mechanics Part 7 60

Mechanics Part 8 62

IV. Chapter 21 (Student Kinematics) 65

V. WAVES:

Chapter 15 (Mechanical Waves) 71

Chapter 16 (Sound and Hearing) 76


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Chapter 21: Electric Charge and Electric Field

B 1. (a) State Coulomb’s law and give its mathematical expression.

[chap 21-page 717]

Solution: The magnitude of the electric force between two point charges is directly

proportional to the product of the charges and is inversely proportional to the

square of the distance between them.

1 2

2

k q q F

r

(b) Two point charges, q1 = -3.0 nC and q2 = +5.0 nC are separated by a distance of 2.0

cm. Find the magnitude of the electric force that q1 exerts on q2.

[chap 21-page 720]

Solution:

9 91 2 4

22

2

3.0 10 5.0 10

3.37 10 N

m2.0cm 10

cm

k k q q

F r

(c) Define electric field.

[chap 21-page 722]

Solution: The electric field is defined as the electric force F experienced by a test

charge q0 placed in the field divided by the charge q0 (providing q0 is small

enough so as not to disturb the charges generating the field).

B 2. (a) Find the magnitude of the electric field at a field point 4.0 m from a point chargeq = 2.0 nC.

[chap 21-page 724]

Solution: 9

2 2

2.0 101.12 N/C

(4.0)

k k q E E

r

(b) The electric field caused by a certain charge at a distance 2.0 cm from the charge is

4.5 x 105 NC-1 away from the charge. What is the charge?

[chap 21-page 725]


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Solution: 2 2 2 5

8

2

(2.0cm 10 m/cm) 4.5 102.0 10 C

E r k q E E q

r k k

Since the field is away from the charge, the charge is positive8

2.0 10 Cq

B 3. (a) Find the distance from a charge of 16 nC where the electric field due to the charge hasa magnitude of 4.0 x 104 NC-1.

Solution: 9

9

2 2

16 104.0 10

k k q E

r r

92

4

.(16 10 )6.00 10 m

4.0 10

k r

(b) A point charge q = 6.00 nC is located at the origin. What is the magnitude of the

electric field vector at the point of coordinates (- 15.0 cm, - 20.0 cm)?

[chap21-page 725]

Solution: Since q is positive, the electric field is away from it.

9

222 2

2

6.00 10

(0.15) (0.20)

8.63 10 N/C

k k q E

r

E

B 4. Three charges lie along the x-axis as shown in the diagram. The positive charge q1 = 4.00

C is at x = 4.00 m, and the positive charge q2 = 8.00 C is at the origin. A negative chargeq3 is placed on the x-axis such that the resultant force on it is zero. What is the x-coordinate

of q3?

[chap 21-page 720]

Solution: The changes q1 and q2 exert forces on q3: F13 and F23 since the resultant forceis zero.

q2

O

q1 q3

Q = 6.00 nC

+ y

+ x P

– 15

– 20


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13 23

1 3 2 3

2 2

13 23

0

. .

( ) ( )

x F F F

k q q k q q

r r

6 61 2

2 2 2 2

13 23

8 10 4 10

( ) ( ) (4 )

q q

r r x x

22 2

2

2

2 2

2 32 16 216 8

16 32 0

4 ( 16) 4(1)(32) 128

16 12813.66 or 2.34

2 2

x x x x

x x

x x

b ac

b x x x x

a

x is a distance so it must be positive, x is also between 0 and 4 as given in the

figure:

16 128

2.34m2

x

BG 5.

Consider three point charges located at the corners of a triangle, as shown in the diagram,

where q1 = q3 = 3.00 C, q2 = - 8.00 C, and a = 0.120 m.

(a) Find the magnitude of the force F1 of q1 on q3.

(b) Find the magnitude of the force F2 of q2 on q3.

(c) What is the x-component of the resultant force on q3?

(d) What is the y-component of the resultant force on q3?

G (e) Find the magnitude of the resultant force on q3.

(f) Calculate the angle the resultant force makes with the positive x-axis.

[chap 21-page 721]

Solution: Since q1 and q3 are both positive, the force F1 by q1 on q3 is a repulsive force.

Since q2 and q3 are of opposite sign, the force F2 by q2 on q3 is an attractiveforce.

x

ay

q1

q2 q3

+

+-

a


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1 1 0tan tan 1 45

a

a

(a)

6 6

1 3

1 222 213

3.00 10 3.00 102.81N

( )

k k q q F

r a a

(b)

6 6

2 3

2 2 2

23

8.00 10 3.00 1015 N

( )

k k q q F

r a

(c) 1 2 1 2cos45 2.81cos 45 15 13.01 N x x x x R F F F i F i i i i

(d) 1 2 1 sin 45 0 1.99 N y y y R F F F i i

(e) 2 2 13.16 N x y R R R

(f) 1 1 01.99

tan tan 8.7 with the positive -axis13.01

y

x

R x

R

B 6.

A charge q1 = 8.00 C is located at the origin, and a second charge q2 = -3.00 C is located

on the x-axis 0.800 m from the origin, as shown in the diagram. Point P has coordinates

(0, 0.600) m.

x

P

+ -

q1 q2


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(a) Find the magnitude of the electric field at P due to q1.

(b) Find the magnitude of the electric field at P due to q2.

(c) What is the x-component of the resultant electric field at P?

(d) What is the y-component of the resultant electric field at P?

(e) Find the magnitude of the resultant electric field at point P.

(f) Calculate the angle the resultant electric field makes with the positive x-axis.

[chap 21-page 728]

Solution: Since q1 is positive, the electric field at P due to q1 is away from q1.

Since q2 is negative, the electric field at P due to q2 is towards q2.

1 1 00.6 3tan tan 36.87

0.8 4

(a)

6

1 5

1 22

1

8.00 10

1.99 10 N/C( ) 0.6

k k q

E r

(b)

6

2 4

2 22 2

2

3.00 102.70 10 N/C

( ) 0.6 0.8

k k q E

r

(c) 41 2 2

0 cos 2.16 10 N/C x x x

E E E E i i

(d) 51 2 2 sin ( ) 1.83 10 N/C y y y i E E E E j E j j

(e) 2 2 51.84 10 N/C x y E E E


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(f) 1 1 0tan tan 83.3 above the +ve -axis y

x

E y x

x E

BG 7.

An electron of mass 9.11 x 10 -31 kg enters the region of a uniform electric field as shown

in the diagram, with v0 = 2.00 x 106 m/s and E = 300 N/C. The width of the plates is l =

0.250 m.

G (a) Find the acceleration of the electron while in the electric field.

G (b) Find the time it takes the electron to travel through the region of the electric field.

(c) What is the vertical displacement y of the electron while it is in the electric field?

G (d) Calculate the speed of the electron as it emerges from the electric field.

[chap 21-page 726]

Solution: Since the electron e – is negatively charged, the electric force on it is in adirection opposite to the direction of the electric field.

(a) 19 17

0

. (300 )( 1.60 10 ) 4.8 10 Nee

F E F E e j j

q

13 2

13 2

5.27 10 m/s

5.27 10 m/s vertically downwards

e y e

F mg F ma F mg ma a j

m

a

(b) 0 0 0cos cos0 X X V t V t V t V t

yx

- - - - - - - - - - - - - - - - - - - -

+ +| + + + + + + + + + + +

l

-v0

E


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7

6

0

0.2501.25 10 s

2.00 10

X t t

V

(c) 2 21 1

02 2

oy y at V t at

13 7 21 ( 5.27 10 )(1.25 10 ) 0.4122

y j m j

(d) 13 7 6( 5.27 10 )(1.25 10 ) 0 6.59 10 m/s y oV at V j

62.00 10 m/s

x oxV V i

2 2 6 1 0

6 0

6.88 10 m/s; tan 73.1

6.88 10 m/s at an angle of 73.1 below the +ve x-axis

y

x y

x

V V V V

V

V

B 8. (a) In a thundercloud there may be an electric charge of + 30.0 C near the top and – 30.0C near the bottom. These charges are separated by approximately 3.00 km. Find the

magnitude of the electric force between them.

[chap 21-page 720]

Solution: 1 2 52 3 2

30 308.99 10 N

(3 10 )

k q q k F

r

(b) What are the magnitude and direction of the electric field that will balance the weight

of three protons, of mass 1.67 x 10-27 kg each?

[chap 21-page 725]

Solution: 0 y F

0

27

7

19

.

3 1.67 10 9.83. .1.02 10 N/C

3 3 1.60 10

e

proton

proton

F mg

E q m g

m g E

q

Since the protons are positively charged, for e F to be vertically upwards, E must also be vertically upwards:

71.02 10 N/C E j

B 9. A 45.0-g insulating sphere carries a charge Q = - 40.0 C and is suspended by a silk threadfrom a fixed point. An external electric field which is uniform and vertical downward is

applied.

(a) The applied electric field has a magnitude of 2.00 x 103 N/C. What is the tension in the

thread?

Solution: Since E is vertically downwards and Q is negative, Fe is vertically upwards:

Fe

+y

mg


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03 6

0

(0.045)(9.8)

(0.045)(9.8) ( 2.00 10 )( 40.0 10 )0.361N

y e

e

F T F mg

T mg F

T E q

T T

(b) The applied electric field holds the sphere in place above the fixed point of suspension,

and the tension in the thread is 0.450 N. Find the magnitude of the electric field.

[chap 21-page 722]

Solution: Since Q is negative and Fe is vertically upwards, E must be verticallydownwards:

0

6

4

4

0 0 . ( ) (0.450)(9.8)(0.045) (0.450)

40.0 10

2.23 10 N/C

2.23 10 N/C

y e F F T E q j mg j j j j

E

E j

E j

T

+y

Fe

Q = – 40.0 C

T

+y

Fe

Q = – 40.0 CE


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Chapter 22:Gauss’s Law

BG 1.G (a) State Gauss’s law in words and give its mathematical expression.

[chap 22-page 759]

Solution: The total electric flux through any closed surface is directly proportional to

the net electric charge enclosed within the surface.

.

E E dA

(b) When excess charge is placed on a solid conductor, where does it reside?

[chap 22-page 761]

Solution: Entirely on the surface.

G (c) Write the three forms of the general definition of electric flux.

[chap 22-page755]

Solution: . . cos E

E dA E dA E dA

G (d) Give the expression of the magnitude of the electric field outside a charged conducting

sphere.

[chap 22- page 762]

Solution: Let the Gaussian surface be a sphere with radius R larger than that of the

conducting sphere.

E is radially outward and perpendicular to the Gaussian surface at every

point.

0 0 0

2

2

0 0 0

. . cos

0; cos 1 . .44

enc enc enc E

enc enc enc

q q q E dA E dA E dA

q q q E dA E R E

R


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BG 2.G (a) What Gaussian surface is used to find the electric field outside an infinitely long

charged wire?

[chap 22-page 763]

Solution: A cylinder with the wire as its axis and with ends perpendicular to the wire.

The radius of the cylinder is R and the length is L.

G (b) What is the magnitude of the field of an infinite sheet of charge?

[chap 22-page 764]

Solution: Take as Gaussian surface a cylinder bisected by the sheet of charge.

E = E

since E points perpendicularly away from the sheet at every point.

1 2 3 3

1

2

1 2

0

0 0

.

but ; 0 since 0

. .

& . .

2

.2

2

E

E

enc E

E dA

E

E dA E dA E dA EA

E dA E dA E dA EA

q EA

A EA E

Surfaces 1 & 2 are the ends of the cylinder; surface

3 represents the lateral area of the cylinder.

Surface charge density =q

A

G (c) What is the magnitude of the field between oppositely charged parallel conducting

plates?

[chap 22-page 765]

Solution: 0

E

(d)

The shaded conductor shown in the figure carries a total charge of Q = +7.0 nC. The

charge within the cavity, insulated from the conductor, is q = -2.0 nC. What is the

charge on the inner surface of the conductor?

qQ

+

+

+

+

+

+

+

+

+

+

+

2

3

1

E

E E

E


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[chap 22-page 767]

Solution: Induced charge on inner surface = +2.0 nC

B 3.

Four closed surfaces S1 through S4, together with the charges q1, q2, q3, q4 and q5 are

sketched in the figure. q1 = -4Q, q2 = +Q, q3 = -2Q, q4 = -3Q, q5 = -Q

(a) Find the electric flux through surface S1.

(b) Find the electric flux through surface S2.

(c) Find the electric flux through surface S3.

(d) Find the electric flux through surface S4.

[chap 22-page 759]

Solution: (a) 11

0 0 0

4encq q Q b

(b) 1 2 3 52

0 0 0 0

4 2 6encq q q q q Q Q Q Q Q b

(c) 3 4 53

0 0 0 0

2 3 6encq q q q Q Q Q Q b

(d) 54

0 0 0

encq q Q

b

B 4. (a) A point charge Q = 7.00 C is located at the centre of a cube of side L = 0.150 m. Sixother point charges, each carrying a charge q = -1.50 C, are positioned symmetricallyaround Q, inside the cube. What is the electric flux through one face of the cube?

(permittivity of free space = 8.85 x 10-12)

Solution: Total E through all 6 faces:4

0 0

6

22.6 10

enc

E

q Q q

b

E through one face =46 3.77 10

E b

S!

S4

S2 S3

q1

q2

q3

q4

q5


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(b) The total electric flux through a closed surface in the shape of a cylinder is

4.80 x 104 Wb. What is the net charge within the cylinder?

(permittivity of free space = 8.85 x 10-12)

[chap 22-page 759]

Solution: 4 70 00

. 4.8 10 4.25 10enc E enc E

q q C

B 5. A thin square conducting plate 35.0 cm on a side lies in the xy plane. A total charge of1.96 x 10-8 C is placed on the plate. (permittivity of free space = 8.85 x 10-12).

(a) Find the charge density on the plate.

(b) What is the magnitude of the electric field just above the plate?

[chap 22-page 764]

Solution: Since the plate is thin, we may model it as a sheet.

(a)8

7 21.96 10 1.6 10 C/m(0.35)(0.35)

q

A

(b) for a thin plate (sheet): 7

3

0 0

1.6 109.04 10 N/C

2 2 E

BG 6. A conducting spherical shell of radius 13.0 cm carries a net charge of 5.07 C uniformlydistributed on its surface.

G (a) Find the magnitude of the electric field just outside the shell.

(b) Find the magnitude of the electric field inside the shell.

(c) What is the magnitude of the electric field at a point very far away from the shell?

[chap 22-page 762]

Solution: (a)6

6

2 2

0 0

5.07 102.7 10 N/C

4 4 (0.13) .

encq

E r

(b) E = 0 since qenc = 0

(c)2

0

;as , 04

encq E r E r

B 7. A conducting spherical shell having an inner radius of 5.00 cm and outer radius of 6.00 cm

carries a net charge of 9.00 C. If a +4.00 C point charge is placed at the centre of theshell, determine the surface charge density on:


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(a) the inner surface.

(b) the outer surface.

[chap 22-page 767]

Solution: Since Q = +4.00 C in the cavity, a charge of – 4.00 C is induced on the

inner surface of the shell. Consequently the charge on the outer surface of theshell is +13.00 mC (since 13 + ( – 4) = 9)

(a)

6 64 2

inner inner 2 2

inner

4.00 10 4.00 101.27 10 C/m4 4 (0.05)

q

A r

(b)6 6

4 2outer outer 2 2

outer

13.00 10 13.00 102.87 10 C/m

4 4 (0.06)

q

A r

B 8.

Two infinite, nonconducting sheets of charge are parallel to each other as shown in the

figure.

(a) Both sheets have positive uniform charge densities 1 = 2 = 5.31 x 10-4 C/m2. What is

the electric field at point A?

Solution: Diagram for parts (a), (b) & (c):

A B C

1 2

+

+

+

+

+

+

+

+

+

++

+

+

+

+

+

+

+

+

+

++

Plate 1 Plate 2

1 2

+x

E1 E2 E2 E1 A CBE2 E1


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* Note: 1) the electric field caused by a positively charged plate always

points away from it.

2) the electric field caused by a negatively charged plate always

points towards it.

(a) At A we must consider the two electric fields 1 2and E E

caused by plates 1 and 2 respectively:

47

total 1 2

0 0 0 0

(5.31 10 )6.0 10 N/C

2 2 E E E

76.0 10 N/Ci A E

(b) Both sheets have positive uniform charge densities 1 = 2 = 5.31 x 10-4 C/m2. What is

the electric field at point B?

Solution: At B, the two electric fields 1 2and E E are equal in magnitude and opposite

in direction; they cancel so: 0N/C B

E .

(c) Both sheets have positive uniform charge densities 1 = 2 = 5.31 x 10-4 C/m2. What is

the electric field at point C?

Solution: At C, we must consider the two electric fields1 2and E E caused by plates 1

and 2 respectively:4

7

total 1 2

0 0 0 0

5.31 106.0 10 N/C

2 2

E E E

76.0 10 N/CiC E [thus could also be done using symmetry from part (a)]

(d) The sheet on the left has a uniform surface density 1 = 5.31 x 10-4 C/m2, and the one

on the right has a uniform charge density 2 = -5.31 x 10-4 C/m2. What is the electric

field at point A?

Solution: Diagram for parts (d), (e) & (f):

++

+

+

+

+

+

+

+

+

+

– –

–

–

–

–

–

–

–

–

–

Plate 1 Plate 21 2

+x

E2 E1 B CE2 E1 AE1 E2


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(d) At A, the two electric fields1 2and E E (caused by plates 1 & 2

respectively) are equal in magnitude0

2

but opposite in

direction; they cancel so: 0N/C A

E

(e) The sheet on the left has a uniform surface density 1 = 5.31 x 10-4 C/m2, and the one


field at point B?

Solution: At B, we must consider the two electric fields1 2and E E caused by plates 1

and 2 respectively:4

8

total 1 2

0 0 0 0

5.31 101.2 10 N/C

2 2 E E E

81.2 10 N/C i B E

(f) The sheet on the left has a uniform surface density 1 = 5.31 x 10-4 C/m2, and the one


field at point C?

[chap 22-page 765]

Solution: By a similar argument to that in (d), we can show that:

0 N/CC E

B 9.

A non-uniform electric field is directed along the x-axis at all points in space. The

magnitude of the field varies with x, but not with respect to y and z. The axis of a cylindrical

surface, 1.00 m long and 0.500 m in diameter, is aligned parallel to the x-axis. The electric

fields E1 and E2, at the ends of the cylindrical surface, have magnitudes 5.00 x 103 N/C and

1.50 x 103 N/C respectively, and are directed as shown in the diagram.

(a) Find the electric flux entering the cylindrical surface at the left end.

(b) Find the electric flux leaving the cylindrical surface at the right end.

(c) Find the net electric flux through the cylindrical surface.

[chap 22-page 755]

(d) What is the net charge enclosed by the cylindrical surface?

[chap 22-page 760]

Solution: (a) 1 11

1. . .

left E E dA E dA E dA E A

zx

y

E1 E2


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23 2 3

5 10 5 10 0.25r

2( ) 312.5 Nm /C 312.5 E

left b

This flux is into the surface, therefore: 312.5left E

b

(b) 2 222. . .right E E dA E dA E dA E A

23 2 3

2

1.5 10 1.5 10 0.25

( ) 93.75 Nm /C 93.75 E

r

right b

This flux is out of the surface, therefore: 93.75right E

b

(c) 312.5 93.75 218.75total left right E E E

b

(d) 90 0

0

. ( 218.75 )( ) 6.08 10 Ctotal total

enc E enc E

qq

6.08nCenc

q

BG 10.

Three hollow, concentric spherical conductors are charged as follows: The inner sphere

carries -2Q, the middle sphere carries +3Q and the outer sphere carries -Q.

(a) Find the charge on the outer surface of the middle sphere.

G (b) Find the charge on the outer surface of the outer sphere.

[chap 22-page 767]

Solution:

Since there is no charge in the innermost cavity, the charge of – 2Q rests

entirely on the outer surface of the innermost sphere. This induces a charge

of +2Q on the inner surface of the middle sphere. Since the total charge on

the middle sphere is +3Q, we conclude that the charge on the outer surface of


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the middle sphere is +Q. This induces a charge of – Q on the inner surface of

the outermost sphere. The net charge on the outermost sphere is – Q, so the

charge on the outer surface of the outermost sphere is 0.

(a) +Q

(b) 0


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Chapter 23:Electric Potential

BG 1. G (a) What is the electric potential energy of two point charges q and q0 a distance r apart?

[chap 23-page 784]

Solution: 0

0

1.

4

qqU

r

(b) A pair of charged metal plates (the top is negative and the bottom is positive) sets up

a uniform electric field with magnitude E. What are the direction and magnitude of

the force exerted on a positive charge q placed in that field?

[chap 23-page 782]

Solution: magnitude of force: F = q.E

direction: vertically upwards

(c) The potential at a certain distance from a point charge is 720 V, and the electric field

is 180 NC-1. What is the charge?

[chap 23-page 788]

Solution: . 4 mV E x x

73.20 10kq rV xV V q C r k k

BG 2. G (a) The potential difference between two oppositely charged parallel metal plates is 120

V and the distance between them is 3.0 cm. Find the magnitude of the electric field

between the plates.

[chap 23-page 796]

Solution: 3120

4.0 10 N/C0.02

V E

d


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(b) The potential difference between two oppositely charged parallel metal plates is 450

V. Find the work done on a charge of 40 mC as it moves from the higher potential

plate to the lower.

[chap 23-page 788]

Solution: 30( ) (40 10 )(450 0) 18Ja b a b a bW U U q V V

(c) When a charge of 20 C is moved between two points M and N, in a uniform

electric field, 240 J of work is done. What is the potential difference between M

and N?

[chap 23-page 788]

Solution: 6

6

0

240 1012.0V

20 10

MN

MN

W V

q

BG 3.G (a) A positive charge of 4.0 C is moved through an electric field to a point where the

potential is +250 V higher than before. Find the increase in the potential energy of

the system.

[chap 23-page 788]

Solution: 6 30. 4.0 10 (250) 1.0 10 JU q V

A proton, of mass 1.67 x 10 -27 kg, is released from rest in a uniform electric field of

magnitude 7.50 x 104 V/m directed along the positive x-axis. The proton undergoes a

displacement of 0.600 m in the direction of E.

G (b) What is the change in the electric potential?

[chap 23-page 796]

G (c) Find the change in potential energy of the proton for this displacement.

[chap 23-page 781]

(d) What is the speed of the proton after it has moved 0.600 m, starting from rest?

[chap 23-page 783]

Solution: (b) 4 4. 7.50 10 (0.6) 4.5 10 VV E x

(c) 19 4 150. 1.602 10 ( 4.5 10 ) 7.21 10 JU q V

(d) . .k e q V

2 2 4

0

1 1( 4.5 10 )

2 2mV mV q

27 2 19 41 (1.67 10 ) 0 (1.602 10 )( 4.5 10 )2

V


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62.94 10 m/sV

BG 4. A 3.00-C point charge is located at the origin, and a second point charge of – 6.00 C islocated on the y-axis at the position (0,2.00) m.

G (a) Find the total electric potential due to these charges at the point P, whose coordinates

are (1.50,0) m.

(b) What is the work required to bring a -3.00-C point charge from infinity to the point

P?

[chap 23-page 792]

Solution:

(a)6 6

3

2 20 0

1 1 3 10 6.0 10. 3.60 10 V

4 4 1.50 1.5 2

i P

ii

qV

r

(b) 6 3 2. 3.0 10 3.6 10 1.08 10 J P W qV

B 5. (a) What is the change in potential energy a 16.0-C charge experiences when it is moved between two points for which the potential difference is 75.0 V?

[chap 23-page 790]

Solution: 6 30. 16.0 10 75 1.2 10 JU q V

(b) The gap between the electrodes in a spark plug is 0.0800 cm. To produce an electric

spark in a gasoline-air mixture, an electric field of 4.50 x 106 V/m must be achieved.

When starting the car, what is the minimum voltage that must be applied by the

ignition circuit?

[chap 23-page 789]

Solution: 6(4.50 10 )(0.08 cmV Ed 2

1m

10 cm

33.6 10 V

B 6. What is the speed of a proton, of mass 1.67 x 10-27 kg, that is accelerated from rest through a potential difference of 105 V?

[chap 23-page 788]

2.00 m

1.50 mq1 =3.00 C

q2 =-6.00 C

2 22.0 1.5

P


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Solution: . .k e q V

2 2 1901 1

1.602 10 1052 2

mV mV

5(1.42 10 )m/sV

*Note: the proton will accelerate from +ve to – ve: fi nal in iti al=V V V

since V decreases as the proton moves from +ve to – ve: V = 0 – 105 = – 105

B 7. An electron, of mass 9.11 x 10-31 kg, moving parallel to the x-axis has an initial speed of3.00 x 106 m/s at the origin O. Its speed is reduced to 2.10 x 105 m/s at point A. Find the

potential difference VAO of point A relative to the origin O.

[chap 23-page 788]

Solution: . .k e q V

2 2

0 0 0 0

31 2 2 19

0 0

0

1 1. * Note :

2 21

(9.11 10 ) (1.602 10 )( )2

25.5V

A A A A

A A

A

mV mV q V V V V V

V V V

V

BG 8.

The three charges in the figure are at the vertices of an isosceles triangle ABC, where

AB = AC = 5.00 cm, and BC = 4.0 cm. q1 = +4.00 C, q2 = q3 = -4.00 C. O is the

midpoint of BC.

(a) What is the potential at O due to q1?

(b) What is the potential at O due to q2?

G (c) Find the total electric potential at O.

G (d) A point charge q0 = 3.00 C is placed at O. Find the electric potential energy

associated with qo.

[chap 23-page 788]

q2 q3 B

q1 A

CO


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Solution:

2 2 3(0.05) (0.02) 2.1 10 AO

(a)1 1

651 1

30 0

1 1 4.00 10. . 7.85 10 V

4 4 2.1 10q q

kq qV V

r AO

(b)2 2

662 2

0 0

1 1 4.00 10. . 1.80 10 V

4 4 0.02q q

kq qV V

r OB

(c) By symmetry: 2 3

61.80 10 Vq qV V

1 2 3Total q q qV V V V

5 6 6 67.85 10 1.80 10 + 1.80 10 = 2.82 10 V

(d) 0 0 31 20

0 0 1 2 3

. .4 4

itotal

ii

q q q qq qU q V r r r r

6 6(3.00 10 )( 2.82 10 )

8.45JU

B 9. Two point charges Q1 = +7.00 nC and Q2 = -2.00 nC, are separated by 36.0 cm.

(a) What is the potential energy of the pair?

[chap 23-page 785]

(b) Find the electric potential at a point midway between the charges.

[chap 23-page 788]

Solution: (a)9 9 9

71 2

2

(9 10 )(7.00 10 )( 2.00 10 )3.5 10

(36.0 10 )

kQ QU J

r

(b)

991 2

1 2 1 2 2

2 2 (9 10 )( ) (7.00 2.00) 10

36.0 10

2 2

kQ kQ k V V V Q Q

r r r

22.50 10 250V V V


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Chapter 24: Capacitance and Dielectrics

BG 1. (a) What is the capacitance of a parallel-plate capacitor? Define all its terms.

[chap 24-page 816]

Solution: 0kA A

C d d

C : capacitance; A: area of the plates; d : distance between the platesk : relative permittivity of the dielectric dielectric constant

permittivity of the dielectric; 0: permittivity of free space

(b) Give the expression of the capacitance C of a parallel-plate capacitor with common

plate area A and plates separated by a layer of dielectric constant K and thickness d.

[chap 24-page 830]

Solution: 0kA A

C d d

G (c) A capacitor acquires a charge of 156 C when it is connected to a potential

difference of 24.0V, find its capacitance.

[chap 24-page 817]

Solution: 6

6156 106.5 10 F

24.0

QC

V

BG 2. G (a) The plates of a parallel-plate capacitor in vacuum are 3.54 mm apart and 4.00 m2 in

area. Permittivity of free space = 8.85 x 10 -12 F/m. Calculate its capacitance.

[chap 24-page 817]

Solution: 0 80

3

4.001.0 10 F

3.54 10

AC

d

G (b) Find the equivalent capacitance of the following combination.

2.0 F 4.0 F 9.0 F


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[chap 24-page 821]

Solution: 66 6 6

1 2 3

1 1 1 1 1 1 1 1.16 10 F

2.0 10 4.0 10 9.0 10 eq

eq

C C C C C

BG 3. (a) Find the equivalent capacitance of the following combination.

2.0 F

4.0 F

9.0 F

[chap 24-page 822]

Solution: 6 6 6 51 2 3 (2.0 10 ) (4.0 10 ) (9.0 10 ) 1.5 10 Feq eqC C C C C

(b) A 0.25-F capacitor is charged so that the potential difference between its plates is400 V. What is the stored energy?

[chap 24-page 824]

Solution: 2 6 2 21 1

0.25 10 (400) 2.0 10 J2 2

U CV

B 4. (a) A capacitor of capacitance 25.0 F has a charge of 12.5 C. Find the stored energy.

Solution:

2

626

6

12.5 103.125 10 J

2 2(25.0 10 )

QU

C

(b) The potential difference across a capacitor is 60 V and its charge is 3.5 C. Calculate

the stored energy.

[chap 24-page 824]

Solution: 6 41 1

3.5 10 (60) 1.05 10 J2 2

U QV

B 5. A parallel-plate capacitor has plates of dimensions 1.50 cm x 5.00 cm separated by a 1.40-

mm thickness of bakelite of relative permittivity 4.90 and dielectric strength 2.40 x 107 V/m.Permittivity of free space 8.85 x 10

-12.


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(a) Find the capacitance of the device.

[chap 24-page 830]

(b) What is the maximum energy that can be stored in the capacitor?

[chap 24-page 833]

Solution: (a) 12

110

3

4.90 8.85 10 0.015 0.052.32 10 F

1.40 10

k A AC

d d

(b) 7 3 4max max 2.40 10 1.40 10 3.36 10 VV E d

2

2 11 4 2

max max

1 1. 2.32 10 3.36 10 1.31 10 J

2 2U C V

BG 6.

(a) When a potential difference of 175 V is applied to the plates of a parallel-platecapacitor, the plates carry a surface charge density of 35.0 nC/cm2. What is the

spacing between the plates?

Solution:

60 0 0

9 40

.

(175)4.43 10 m

35.0 10 10

Q AC

A V AV V d

A V d C

d

G A small object with a mass of 300 g carries a charge of 20.0 nC and is suspended by a

thread between the vertical plates of a parallel-plate capacitor. The plates are separated by5.00 cm. The thread makes an angle of 18.0 with the vertical.

G (b) Calculate the tension in the thread.

G (c) What is the magnitude of the electric field between the plates?

G (d) Find the potential difference between the plates.

[chap 24-page 817]

Solution: (b)


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0 cos18 (0.3)(9.8) 3.09 N y y F T mg T T

(c) 0 sin18 x e x

F F T T

7

9

0

sin184.78 10 N/C

20.0 10

e F T

E E q

(d) 7 2 64.78 10 5.0 10 = 2.39 10 VV Ed

B 7.

In the capacitor network shown in the figure: C1 = 4.00 F, C2 = 6.00 F, C3 = 3.60 F, V =

15.0V.

(a) Find the equivalent capacitance C4 of C1 and C2.

[chap 24-page 823]

(b) Find the equivalent capacitance Ceq of C3 and C4.

[chap 24-page 823]

(c) What is the potential difference across C1?

[chap 24-page 823]

(d) Find the charge on C3.

[chap 24-page 817]

Solution: (a) C1 & C2 are in series:

6

46 6

4 1 2

1 1 1 1 12.4 10 F

4.00 10 6.00 10C

C C C

(b) C3 & C4 are in parallel:

6 6 63 4 3.60 10 2.4 10 6.0 10 FeqC C C

(c) Voltage in each branch is the same = 15 V

C1 & C2 are in series Q1 = Q2

1 1 2 2 1 1 2 115C V C V C V C V

C1 C2

C3

V


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6 61 14.00 10 6.00 10 15V V

1 9.0VV

(d) V3 = 15 V

6 533 3 3 3

3

(3.60 10 )(9) 5.4 10 CQ

C Q C V

V

BG 8. A 4.00-F capacitor and a 16.0-F capacitor are connected in series across a 50.0-Vsupply line.

(a) What is the equivalent capacitance?

G (b) Find the charge on each capacitor.

[chap 24-page 821]

G. The charged capacitors are disconnected from the line and from each other and thenreconnected to each other, with terminals of like sign together.

(c) What is the final voltage across each?

[chap 24-page 821]

G (d) Find the final charges on C1 and C2 respectively.

[chap 24-page 817]

Solution: (a) In series 6 6

1 2

1 1 1 1 1

4.00 10 16.0 10eqC C C

63.2 10 FeqC

(b) In series 6 41 2 . (3.2 10 )(50.0) 1.6 10 Ceq eqQ Q Q C V

(c)

Qtotal = Q1 + Q2 (charge is conserved)

* since C 1 & C 2 have two points in common, they are in parallel and V 1

= V 2 = V


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4 6 60 1 2. 3.2 10 4 10 16.00 10eqC V C V C V V 16.0VV

(d) 6 51 1 4.00 10 16.0 6.4 10 CQ C V

6 5

2 2 16.00 10 16.0 2.56 10 CQ C V

BG 9. The square plates of a 7.50-nF capacitor measure 50.0 mm by 50.0 mm and are separated by a dielectric which is 0.250 mm thick. The voltage rating of the capacitor is 400 V.

Permittivity of free space = 8.85 x 10-12.

G (a) What is the maximum energy that can be stored in the capacitor?

[chap 24-page 832]

(b) Find the dielectric constant of the dielectric.

[chap 24-page 830]

(c) Calculate the dielectric strength of the dielectric.

[chap 24-page 833]

Solution: (a) 22 9 14max max1 1

. 7.50 10 400 6.0 10 J2 2

U C V

(b) 0

0

.84.7

k A d cC k

d A

(c) 6maxmax max max 3

400( ) 1.6 10 V/m

0.25 10

V V E d E

d


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Chapter 25:Current, Resistance and E.M.F.

B 1. (a) A current of 3.20 A flows through a bulb. How many coulombs of charge flow throughthe bulb in 7.50 h?

Solution: 43 2 7 5 3600 8 64 10 CQ It . . .

(b) A wire of cross-sectional area 1.40 mm2 has 8.20 x 1027 free electrons per cubic metre.

The electrons, each of charge 1.60 x 10-19 C, drift at an average speed of 1.10 mm/s.

Find the current in the wire.[chap 25-page 849]

Solution: 27 6 3 198 2 10 1 4 10 1 1 10 1 6 10 2 02 A I nAve . . . . .

BG 2. G (a) A wire of resistivity 9.42 x 10-8 m is 15.0 m long and has a radius of 2.00 mm. What

is its resistance?

[chap 25-page 853]

Solution: 8

2 6

9 42 10 15 0 1124 10

l l R A r

.

.

(b) The resistivity of a certain metal is 2.30 x 10-7 m at 20.0C. Its temperature

coefficient is 0.00600 (C)-1. Calculate its resistivity at 35.0 C.

[chap 25- page 852]

Solution: 7 70 01 2 3 10 1 0 006 35 20 2 507 10T T . . .

7

2 51 10 m

.

BG 3. G (a) A resistance of 3.20 is connected to the terminals of a cell of e.m.f. 10.8 V and

internal resistance 0.400 . What is the p.d across the 3.20- resistor?

Solution: 10 8 10 8

3 A3 2 0 4 3 6

I R r

. .

. . .

3 3 2 9 60 VV IR . .


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G (b) A cell of e.m.f. 12.0 V and internal resistance 1.50 is connected across a 2.50-

resistor. Find the charge which passes any point in the circuit in three minutes.

[chap 25-page 858]

Solution:

12 12

3 A 3 3 60 540 C2 5 1 5 4 I Q It R r

. .

BG4.G (a) The resistance of a certain length of copper wire is 3.00 at 20.0 C. If the

temperature coefficient of copper is 4.00 x 10-3 K -1, what is its resistance at 65.0 C ?

[chap 25-page 854]

Solution: 30 01 3 1 4 10 65 20 3 54 R R T T .

(b) A cell of e.m.f. 6.4 V and internal resistance 0.10 is connected across a 1.5-

resistor. Find the net output of the battery.

[chap 25-page 864]

Solution: 6 4 6 4

4 A 4 1 5 6 V1 5 0 1 1 6

I V Ir IR R r

. ..

. . .

6 4 24 W P VI

BG 5.1.5 ,14 V

3.3 4.0

1.2 ,10 V

(a) Calculate the current in the circuit.

[chap 25-page 862]

G (b) Find the total rate of dissipation of energy.

G (c) Find the rate of conversion of electrical energy to chemical energy.

G (d) Find the rate of conversion of chemical energy to electrical energy.

[chap 25-page 865]


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Solution: (a)14 10 4

0 40 A1 5 3 3 1 2 4 10

I R

e

.

. . .

(b) 2 20 4 10 1 6 W P I R . .

(c) 10 0 4 4 0 W P I e . .

(d) 14 0 4 5 6 W P I . .

B 6. A resistance thermometer, which measures temperature by measuring the change inresistance of a conductor, is made from silver, of temperature coefficient of resistivity 3.80 x

10-3 (C)-1, and has a resistance of 40.0 at 31.0C. When immersed in a vessel containing

melting lead, its resistance increases to 85.0 . Calculate the melting point of lead.

[chap 25-page 854]

Solution: 0 01 R R T T

0 3

0

1 85 11 1 31 327 C

40 3 8 10

RT T

R

.

B 7. (a) An electric heater is constructed by applying a potential difference of 150 V to a Nichrome wire of total resistance 10.0 . Find the power rating of the heater, in kW.

[chap 25-page 864]

Solution: 2 2

1502250 W 2 25 kW

10

V P

R .

(b) A 20.0- metal wire is cut into four pieces that are then connected side by side to form

a new wire the length of which is equal to one-fourth the original wire. What is the

resistance of this new wire?

[chap 25-page 853]

Solution:

'' ' 4 '

4 '

l l l l A A R R

A A

' ' 1 1 1 20' 1 25

' 4 4 16 16 16

R l A R R

R l A .

B 8. (a) A 2.10-V potential difference is maintained across a 2.00-m length of tungsten wire ofresistivity 5.60 x 10-8 m, that has a cross-sectional area of 0.500 mm2. Find the

current in the wire.

[chap 25-page 853]

Solution: 6

8

2 1 0 5 109 375 9 38 A

5 6 10 2

V V VA I

l R l

A

. .

. .

.


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(b) A segment of Aluminum wire of temperature coefficient of resistivity 3.90 x 10-3 (C)-

1 is initially at 21.0C. To which temperature must the wire be heated to quadruple its

resistance?

[chap 25-page 854]

Solution: 0 0 0 0 33 31 4 21 790 C3 9 10

R R T T R T T

.

B 9. A wire with resistance 6.00 is lengthened 1.50 times its original length by pulling itthrough a small hole. Find the resistance of the wire after it is stretched.

[chap 25-page 853]

Solution: '

' ' ' 1 5'

A l V V lA l A

A l .

' ' '' 1 5 1 5 ' 2 25 6 13 5

' '

l l R l A R R R

A A R l A

. . . .

B 10.

In the above circuit : E1 = 14.0 V, r 1 = 3.00 , E2 = 10.0 V, r 2 = 2.00 , R = 5.00 .

(a) What is the current?

(b) Find the potential VAB of point A relative to point B.

(c) Find the potential VDC of point D relative to point C.

[chap 25-page 865]

(d) What is the total rate of dissipation of energy in the circuit?

[chap 25-page 865]

Solution: (a)14 10 24

2 40 A3 2 5 10

I R

.

(b) 1 1

14 2 4 3 6 80 V AB

V Ir . .

E2 r 2

E1

R

r 1

A B

C D


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(c)2 2

10 2 4 2 5 20 V DC

V Ir . .

(d) 2 22 4 10 57 6 W P I R . .


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Chapter 26: Direct-Current Circuits

B 1. (a) Three resistors of resistances R 1 = 1.00 , R 2 = 3.00 and R 3 = 5.00 are connectedin series. Find their equivalent resistance.

[chap 26-page 882]

Solution: eq 1 2 3 1 3 5 9 00 R R R R .

(b) Three resistors of resistances R 1 = 1.00 , R 2 = 3.00 and R 3 = 5.00 are connected

in parallel. Find their equivalent resistance.[chap 26-page 883]

Solution: eqeq 1 2 3

1 1 1 1 1 1 1 23 150 652

1 3 5 15 23 R

R R R R .

B 2. (a) What is the current from a to b ?

1.20 A

a

0.920 A 2.00 A

b

[chap 26-page 888]

Solution: Junction rule: 2 0 92 2 92 A 2 92 Aab ba I I . . .


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(b) Applying Kirchhoff’s loop rule for the loop abcdefa, find the resistance x.

0.450 A

2.00

b a

c f

0.505 A 4.00 x 1.50 A

d e

[chap 26-page 888]

(1 mark)

Solution: 0 45 2 0 505 4

0 45 2 0 505 4 1 5 0 1 946 1 951 5

x x

. .

. . . . .

.

B 3. (a) A galvanometer has coil resistance R C and f.s.d. current Ifs. If it is to be converted to anammeter of f.s.d. current Ia ,what is the shunt resistance required?

[chap 26-page 892]

Solution: fs Ca fs sh fs C sha fs

I R I I R I R R

I I

(b) A galvanometer has coil resistance R C and f.s.d. current Ifs. If it is to be converted to a

voltmeter of f.s.d. VV , find the resistance required.

[chap 26-page 892]

Solution: VV fs s C s Cfs

V V I R R R R I

B 4. (a) An ammeter of 48.5 resistance gives a full scale deflection for 3.0 mA. What mustthe meter have added to it to give a full scale deflection for 100 mA?

[chap 26-page 892]

Solution: fs Csha fs

3 48 51 5 in parallel

100 3

I R R

I I

.

.


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(b) A galvanometer of coil resistance 20.0 and f.s.d. current 2.00 mA is to be converted

to a voltmeter with a maximum range of 16.0 V. What is the series resistance to be

used?

[chap 26-page 892]

Solution: Vs C 3

fs

1620 7980

2 10

V R R

I

3

s7 98 10 in series R .

BG 5. (a) In the R-C circuits, give the formulae of current and charge in terms of time:

(i) while charging

(ii) while discharging

[chap 26-page 897]

Solution: (i)0

1

t t

RC RC F q Q e i I e

(ii) 0 0

t t

RC RC q Q e i I e

A resistor with resistance 12.0 M is connected in series with a capacitor with

capacitance 2.00 F and a battery with e.m.f. 12.0 V. Before the switch is closed at t = 0,

the capacitor is uncharged.

(b) Find the time constant.[chap 26-page 898]

(c) What is the current at t=0 when the switch is closed?

(d) Find the final charge on the capacitor.

[chap 26-page 897]

Solution: (b) 12 2 24 0 s RC .

(c) 012

1 00 A12 I R

.

(d) 2 12 24 0 C F Q C .

B 6. A resistor with resistance 12.0 M is connected in series with a capacitor with capacitance2.00 F and a battery with e.m.f. 12.0 V. Before the switch is closed at t = 0, the capacitor is

uncharged.

(a) Calculate the current at one time constant.

(b) Calculate the charge at one time constant.

[chap 26-page 897]


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Solution: (a) 0 012

12 2 24 0 s 1 A12

t

RC RC i I e I R

.

11 0 368 Ai e

.

(b) 1 2 12 24 C

t

RC F F q Q e Q C

124 1 15 2 Cq e .

B 7.

Four resistors are connected as shown. R 1 = 3.00 , R 2 = 4.00 , R 3 = 24.0 , R 4 = 12.0 .

Vac = 22.5 V

(a) What is the equivalent resistance between a and c ?

(b) Find the current through R 3.[chap 26-page 882]

Solution: (a)5 1 2 6

6 3 4

1 1 1 1 1 13 4 7 8

24 12 8 R R R R

R R R

eq 5 6 7 8 15 0 R R R .

(b)eq

22 51 5 A

15

AC V

I R

.

.

6

6 3 3 3

3

1 5 8

0 500 A24

bc

IRV IR I R I R

.

.

B 8. (a) The currrent in a loop circuit that has a resistance R 1 is 4.0 A. The current is reduced to3.0 A when an additional resistor R 2 = 2.5 is added in series with R 1. What is the

value of R 1?

Solution: 2 21 1 2 1 2 2 1 2 2 1 1 2 2 2 11 2

I R I R I R R I R I R R I I I R R

I I

1

3 2 57 5

4 3 R

.

.

R 4

a

R 1 R 2

R 3

b c


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(b) A battery with an e.m.f. of 15.0 V and internal resistance of 0.800 is connected

across a load resistor R. If the current in the circuit is 2.50 A, find the power dissipated

in R.

[chap 26-page 882]

Solution: 15 0 8 5 22 5

I R r r R I

. .

.

2 22 5 5 2 32 5 W P I R . . .

B 9. The current in a circuit is quadrupled by connecting a 600- resistor in parallel with theresistance R of the circuit. Calculate the resistance R.

[chap 26-page 883]

Solution: eqeq

1 1 1 600 600

600 600 600

R R R

R R R R

I 4 R I eq R R 6004

R 2400 600 1800 1 80 k

600 R R

R

.

BG 10. Three resistors each having resistance R = 3.0 are connected as in the figure. Each candissipate a maximum power of 48 W without being excessively heated. Find the

maximum power the network can dissipate.

[chap 26-page 883]

Solution:

3 1 2

148 W 48 12 W4 P P P

1 2 312 12 48 72 W

T P P P P

R

R

R

R

R

R1

2

3 I

I /2

I /2

48 W


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BG 11. Initially, for the circuit shown, the switch S is open and the capacitor, of capacitance20.0 F, has a voltage of 120 V, and R = 3.00 M. The switch is closed at time t = 0.

G (a) Calculate the charge on the capacitor, when the current in the circuit is 35.0 A.

[chap 26-page 898]

Solution: 335 3 20 2100 2 10 10 C R C q

V V iR q iRC C

.

G (b) Find the capacitor voltage when the time is equal to 75.0 s.

[chap 26-page 897]

Solution: 75 5

0 20 3 4

0 0 120 120 34 4 V

t t t

RC RC RC Qq

q Q e e v V e e eC C

.

C R

S


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Chapter 27:Magnetic Field and Magnetic Forces

BG 1. (a) An electron of charge 1.6 x 10-19 C moves with a speed of 4.0 x 106 m/s along the xaxis. It enters a region where there is a magnetic field of magnitude 1.5 T, directed at

an angle of 30 to the x-axis and lying in the xy plane. Find the magnitude and

direction of the magnetic force on the electron.

[chap 27-page 919]

Solution:

Right-hand rule: force along (-)ve z .

q F v B

19 6 13

13

sin 1 6 10 4 10 1 5 sin 30 4 8 10 N

4 8 10 N,( )ve direction.

F qvB

z

. . .

F .

G (b) A disc of radius 1.5 cm is found in a region of uniform magnetic field of 4.0 T making

an angle of 37 with the plane of the disc. Calculate the magnetic flux through the

disc.

[chap 27-page 924]

Solution:

2cos cos BA B r

2

2 34 1 5 10 cos53 1 7 10 Wb

. .

37°

A

B

37°

A

B


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B 2. (a) A proton of mass 1.67 x 10-27 kg and charge 1.60 x 10-19 C moving with a speed of2.50 x 105 m/s enters a region of uniform magnetic field perpendicular to its direction

of motion and of magnitude 0.800 T. Find the radius of its trajectory.

[chap 27-page 926]

Solution: 2 27 5

3

19

1 67 10 2 5 103 26 10 m 3 26 mm

1 6 10 0 8

mv mvqvB r

r qB

. .

. .

. .

(b) In a velocity selector of crossed electric and magnetic fields, the magnitude of the

electric field is 5.0 x 106 V/m and that of the magnetic field is 0.80 T. What is the

speed of the particle?

[chap 27-page 929]

Solution: E B F F q E q6

6 65 106 25 10 6 3 10 m/s

0 8

E vB v

B

. .

.

BG 3. (a) A wire carries a steady current of 2.5 A. A straight section of the wire is 0.60 m longand lies along the x axis within a uniform magnetic field, B = (2.4 T) k . If the current

is in the + x direction, find the magnetic force on the section of the wire.

[chap 27-page 933]

Solution:

2 5 0 6 2 4 3 6 N I F l B . . i . k . j

G (b) A circular coil 6.0 cm in radius, with 60 turns of wire, carries a current of 5.0 A . The

coil is in a uniform magnetic field with magnitude 4.0 T making an angle of 60 with

the plane of the coil. Find the torque on the coil.[chap 27-page 936]

Solution:

90 60 30

x

y

I• B

60°

A

B


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4sin 60 5 36 10 4 sin30 6 78584 6 8 Nm NIAB

. .

BG 4. (a) The Earth’s magnetic field at a certain place is 3.2 x 10 -5 T at an angle of 30 with thehorizontal. Calculate the flux through the horizontal ceiling of a rectangular room 6.0

m by 8.0 m.

[chap 27-page 924]

Solution:

5 4 490 30 60

cos 3 2 10 48 cos 60 7 68 10 7 7 10 Wb BA

. . .

G (b) A particle with a charge of – 6.0 x 10-8 C is moving with instantaneous velocity of

(4.5 x 104 m/s )j Find the force exerted on the particle by a magnetic field (5.0T)k.

[chap 27-page 919]

Solution: 8 46 10 4 5 10 5 0 0135 N 0 014 Nq F v B . j k . i . i

B 5. (a) An electron experiences a magnetic force of magnitude 6.7 x 10-15 N when moving atan angle of 37 with respect to a magnetic field of magnitude 3.5 x 10-3 T. What is thespeed of the electron ?

[chap 27-page 919]

Solution: 15

7

19 3

6 7 10sin 1 988 10

sin 1 6 10 3 5 10 sin 37

F F qvB v

qB

.

.

. .

72 0 10 m/sv .

(b) A circular area with a radius of 5.0 cm lies in the xy plane. Find the magnitude of themagnetic flux through this circle due to a uniform magnetic field of magnitude 0.60 T

in the +z direction.

[chap 27-page 924]

Solution:

30°

AB

x

y

z

A

B


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2

2 3cos 0 6 5 10 1 4 7 10 Wb BA . .

BG 6.G (a) A singly charged ion moving with a speed of 3.9 x 105 m/s enters a region of uniform

magnetic field of magnitude 0.52 T, perpendicular to the direction of motion of the

ion, and describes a circular arc of radius 50 cm. What is the mass of the ion?

[chap 27-page 926]

Solution: 19

25 25

5

0 5 1 6 10 0 521 07 10 1 1 10 kg

3 9 10

mv rqBr m

qB v

. . .

. .

.

G (b) A horizontal rod 35 cm long is mounted on a balance and carries a current. At the

location of the rod a uniform horizontal magnetic field has magnitude 0.020 T and

direction perpendicular to the rod. The magnetic force on the rod is measured by the

balance and is found to be 0.21 N. Find the current.[chap 27-page 933]

Solution: sin F IlB 0 21

30 A0 35 0 02

F I

lB

.

. .

B 7. A wire having a mass per unit length of 0.600 g/cm carries a 3.00-A current horizontally tothe north. What are the magnitude and direction of the minimum magnetic field needed to

lift this wire vertically upward?

[chap 27-page 933]

Solution:

3

2

0 6 10 9 80 6 g/cm

10 3

0 196 T, West

m mg m g F mg IlB mg B

l Il l I

. ..

B .

B 8. Alpha particles (charge =+2e, mass = 6.68 x 10-27 kg) are accelerated in a cyclotron to afinal orbit radius of 0.800 m. The magnetic field in the cyclotron is 0.600 T.

(a) Find the period of circular motion of the alpha particles.

(b) What is the magnitude of the centripetal acceleration of the alpha particles in the finalorbit?

[chap 27-page 926]

• EastB

F

mg

×

I

West


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Solution: (a)2 2r mv rqB r

T r v T v qB m

r

2 m

qBqB

m

277

19

2 6 68 102 186 10 s 0 2186 s 0 219 s

2 1 6 10 0 6

T

.

. . .

. .

(b)

2 2

2 14 14 2

227

4 40 8 6 609 10 6 61 10 m/s

2 186 10

a r r T

. . .

.

B 9. A uniform magnetic field of magnitude 0.60 T in the positive z-direction is present in aregion of space. A uniform electric field is also present. An electron projected with an initial

velocity v0 = 2.5 x 104 m/s in the positive x-direction, traverses the region without

deflection. What is the electric field vector?

[chap 27-page 929]

Solution:

Right-hand rule 0 along + along

0 along 0 y E B

q

q F F F q

B EF j F j

E j E q

4 42 5 10 0 6 1 5 10

15 kV/m

vB

E vB

. . .

E j

B 10. A 12-m length of wire carrying a current of 5.6 A lies on a horizontal table with arectangular top of dimensions 180 x 240 cm2. The ends of the wire are attached to opposite

ends of a diagonal of the rectangle. A vertical magnetic field of 0.25 T is present. Find themagnitude of the magnetic force acting on this segment of wire.

[chap 27-page 933]

Solution:

2 2240 180 300 cml

i

j

v0• •

k

B•

FB

FE

E

240

180l


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sin I F IlB F l B 5 6 3 0 25 4 2 N . . .

B 11. (a) A circular coil of wire of 250 turns and diameter 2.50 cm carries a current of 4.50 A. Itis placed in a magnetic field of 0.400 T with the plane of the coil making an angle of

37.0 with the magnetic field. Find the torque on the coil.

[chap 27-page 936]

Solution:

2

2

90 37 53

sin 250 4 5 1 25 10 0 4 sin 53

0 176 Nm

NIAB

. . .

.

(b) A particle of charge -5.00 C and mass 3.50 x 10-12 kg has velocity 4.00 km/s as it

enters a region of uniform magnetic field. The particle is observed to travel in a

circular path of radius 7.50 cm. What is the magnitude of the magnetic field in the

region?

[chap 27-page 937]

Solution: 12 3

2 63 5 10 4 10 0 0373 T7 5 10 5 10

mv mvr BqB rq

.

.

.

37°

A

B


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Chapter 28: Sources of Magnetic Field

BG 1. (a) What is the constant current flowing in a straight wire required to give a magnetic fieldof 2.0 x 10-5 T at 2.5 cm from the wire?

[chap 28-page 963]

Solution: 2 5

0

7

0

2 2 2 5 10 2 102 5 A

2 4 10

I rB B I

r

.

.

G (b) A circular coil of radius 4.0 cm and 200 turns lies in the plane of this paper and carries

a current of 2.5 A in an anticlockwise sense. Determine the field at the centre of the

coil.

[chap 28-page 968]

Solution:

730 200 4 10 2 5 2 5 10 T, page and out of the page.

2 2 0 04

N I B

a

.

.

.

B 2.

(a) Two long parallel conductors run North-South in the same horizontal plane as shown

in the diagram. (1) carries a current 20 A northward and (2) a current 20 A southward,

and they are 5.0 cm apart. Determine the force per unit length of (2) on (1).

[chap 28-page 966]

Solution: 7 30 ' 4 10 20 20 1 6 10 N/m, West.

2 2 0 05 II F

l r

.

.

S

N

(1) (2)

• B


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(b) A solenoid with 800 turns, 75 cm long and 6.0 cm in diameter, lies with its axis North-

South and carries a current of 6.0 A in a clockwise sense when viewed from the South

end. Find the magnitude and direction of the field at the centre of the solenoid.

[chap 28-page 974]

Solution:

730

0

4 10 800 68 0 10 T, from south to north.

0 75

NI B nI

l

.

.

B 3. (a) What is the magnitude of the magnetic field produced by a straight long wire carryinga current of 45 A at 18 cm from the wire?

[chap 28-page 963]

Solution: 7 7

50 4 10 45 90 10

5 0 10 T2 2 0 18 0 18

I B

r

.

. .

(b) Two long, parallel wires are separated by a distance of 3.5 cm. The force per unit

length that each wire exerts on the other is 6.0 x 10-5

N/m, and the wires attract each

other. If the current in one wire is 2.1 A, find the current in the second wire.

[chap 28-page 966]

Solution: 2

50

7

0

' 2 2 3 5 10' 6 10

2 4 10 2 1

II F r F I

l r I l

.

.

' 5 0 A, same direction as first current. I .

B 4. (a) A closely wound circular coil with 600 turns carries a current of 2.5 A and produces atits centre a magnetic field of magnitude 3.8 x 10 -3 T. What is its diameter?

[chap 28-page 968]

Solution: 7

0 0

3

600 4 10 2 52 0 496 m 50 cm

2 3 8 10

N I N I B a D

a B

.

.

.

(b) A solenoid is designed to produce a magnetic field of 0.06030 T at its centre and it

carries a current of 16.00 A. Find the number of turns per unit length?

[chap 28-page 974]

Solution: 0 7

0

0 06032999 turns/m

4 10 16

B B nI n

I

.

S NB


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B 5. (a)

A long, straight conductor lies along the x-axis and carries a current of 200 A in the

positive x-direction. Determine the magnetic field at point P in the diagram, such that

OP = 1.50 m and = 40.0:

[chap 28-page 963]

Solution: 7

0 4 10 200sin2 2 1 5sin 40

I a OP B

a

.

54 15 10 T, out of the page, i.e. in the positive direction. B z .

(b) Give the official SI definition of the ampere.

[chap 28-page 966]

Solution: The ampere is that unvarying current that, if present in each of two parallel

conductors of infinite length and one meter apart in empty space, causes each

conductor to experience a force of exactly 2 × 10 – 7

newtons per meter of

length.

BG 6.

The figure is an end view of two, long, straight, parallel wires perpendicular to the xy-plane,

each carrying a current I = 7.00 A but in opposite directions. d = 8.00 cm.

G (a) Find the magnetic field at O.

G (b) Find the magnetic field at P.

[chap 28-page 963]

Solution: (a) 1 2 12 B B B B

xI

y

P

O

x x

y

O

II

dd

P

2d


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7

50

1 2

2 4 10 73 5 10

2 2 0 08

I B B B

d

.

.

53 5 10 T B . j

(b)

0 0 0

2 1 1 2

2 3 6 2

I I I B B B B B

d d d

0 1 21

2 3

I B

d

0

3 2

I 0

7

5

3

4 10 71 17 10 T

3 0 08

I B

d d

B

.

.

51 17 10 T B . j

B 7.

Two long, straight, parallel wires are 4.00 m apart. They carry currents in opposite direction,

as shown in the figure. I1 = 9.00 A, I2 = 16.0 A, AP = 2.40 m, CP = 3.20 m.

(a) What is the magnetic field B1 at P due to I1?

(b) What is the magnetic field B2 at P due to I2?

(c) Find the magnitude of the resultant magnetic field at P.

(d) Find the acute angle the resultant magnetic field at P makes with the direction AP.

[chap 28-page 963]

Solution:

•

B2

B1

B1 B2

I2

×I1

A C

P

P

A B

•

θ B1

B2

B


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2 2 2 2 2 2 is right at because 4 2 4 3 2 ABP P AB AP PB . .

(a)7

70 11

1

4 10 97 50 10 T, direction

2 2 2 4

I B CP

r

.

.

(b)

760 2

22

4 10 161 00 10 T, direction A

2 2 3 2

I B P

r

.

.

(c) 2 2

2 2 7 6 6

1 2 7 5 10 1 10 1 25 10 T B B B

. .

(d)7

1 1 21

6

2

7 5 10tan tan tan 0 75 36 9

1 10

B

B

.

. .

BG 8.

A coil consisting of 40.0 turns has a radius of 12.6 cm and carries a current I 1 = 4.00 A as

shown in the figure. A long, straight conductor, which is in the plane of the coil, carries acurrent I2 = 350 A at a distance of 17.5 cm from the centre P of the coil.

(a) What is the magnetic field B1 at P due to I1?

G (b) What is the magnetic field B2 at P due to I2?

[chap 28-page 963]

(c) Find the net magnetic field at P.

(d) If the current I2 were 350 A in the positive y direction, what would the net field at P

be?

[chap 28-page 968]

Solution: (a)7

41 0 11

40 4 10 47 98 10 T

2 2 0 126

N I B

a

.

.

47 98 10 T, in the positive direction z 1

B .

(b)7

40 22

2

4 10 3504 10 T

2 2 0.175

I B

r

4

4 00 10 T, in the negative direction z

2

B .

(c) 41 2 Net 7 98 4 10 B B B

.

I1

I2

Px

y

z


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4 Net 3 98 10 T, in the positive direction z

B .

(d) 4 31 2 Net 7 98 4 10 1 20 10 T B B B

. .

3 Net 1 20 10 T, in the positive direction z B .

B 9.

A conducting ring of radius 6.28 cm carries a current of unknown magnitude and sense. A

uniform external magnetic field Bext = 150 T, directed into the plane, is applied as shown in

the figure. An electron is projected from the centre of the ring, with an initial velocity 4.00 x

105 m/s towards the right. The electron experiences an initial force 1.60 x 10-17 N in the upward

direction.

(a) What is the magnitude of the net magnetic field?

[chap 27-page 919]

(b) Deduce the magnitude of the magnetic field produced by the current at the centre of the

ring.

[chap 28-page 9687](c) Find the current in the ring.

[chap 28-page 968]

Solution: (a)17

4

net 19 5

1 6 102 50 10 T

1 6 10 4 10

F F qvB B

qv

.

.

.

(b) Right-hand rule B net is out of the page. But Bexternal is into the page,

therefore Bcurrent is out of the page.

net current external current net external

4 4 4

current 2 5 10 1 5 10 4 00 10 T

B B B B B B

B

. . .

(c)2 4

0 currentcurrent 7

0

2 2 6 28 10 4 1040 0 A

2 4 10

I a B B I

a

.

.

x

Bext

F

ve-

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