BQ 1112 Level NS Core Physics BGT- Electricity

Physics Core Course

Basic Questions

Table of Contents

I. ELECTRICITY: 1. Electric Charge And Electric Field 1 2. Gausss 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

SABIS Proprietary 1

Chapter 21: Electric Charge and Electric Field

B 1. (a) State Coulombs 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 22k q q

Fr

= (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

222

3.0 10 5.0 103.37 10 N

m2.0cm 10cm

kk q qF

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 charge

q = 2.0 nC. [chap 21-page 724]

Solution: 9

2 2

2.0 101.12 N/C

(4.0)kk q

E Er

= = = =G

(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]

SABIS Proprietary 2

Solution: ( )2 2 2 5 82 (2.0cm 10 m/cm) 4.5 10 2.0 10 CE rk qE E qr k k

= = = = = GG

82.0 10 Cq

Since the field is away from the charge, the charge is positive = +

B 3. (a) Find the distance from a charge of 16 nC where the electric field due to the charge has

a magnitude of 4.0 x 104 NC-1.

Solution: 9

92 2

16 104.0 10

kk qE

r r

= = = G

9

24

.(16 10 ) 6.00 10 m4.0 10

kr

= =

(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.

Q = 6.00 nC

+y

+x P

15 20

( )9

222 2

2

6.00 10

(0.15) (0.20)

8.63 10 N/C

kk qE

r

E

= =+

=

G

G

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 charge q3 is placed on the x-axis such that the resultant force on it is zero. What is the x-coordinate of q3?

O

q3 q1

[chap 21-page 720] q2

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

is zero.

SABIS Proprietary 3

13 23

1 3 2 32 2

13 23

0. .( ) ( )

xF F Fk q q k q q

r r

= = =

6 6

1 22 2 2 2

13 23

8 10 4 10( ) ( ) (4 )

q qr r x x

= =

22 2

2

2

2 2

2 32 16 216 816 32 0

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

16 128 13.66 or 2.342 2

x x x xx x

x xb ac

bx x x xa

= + = + + = = = =

= = = =

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.

x

a y

q1

q2 q3

+

+ -

a

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 attractive force.

SABIS Proprietary 4

1 1 0tan tan 1 45aa

= = =

(a) ( )6 6

1 31 22

2 213

3.00 10 3.00 102.81N

( )kk q q

Fr a a

= = =+

(b) 6 6

2 32 2 2

23

8.00 10 3.00 1015 N

( )kk q q

Fr a

= = =

( )

( ) (c) ( )1 2 1 2cos 45 2.81cos 45 15 13.01 Nx x x xR F F F i F i i i i= + = + = + = G G G G

( ) (d) 1 2 1 sin 45 0 1.99 Ny y yR F F F i i= + = + =G G G (e) 2 2 13.16 Nx yR R= + =

GR

(f) 1 1 01.99tan tan 8.7 with the positive -axis13.01

y

x

Rx

R = = =

B 6. y P

x -+ q2 q1

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.

SABIS Proprietary 5

(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 36.870.8 4

= = =

(a) ( )6

1 51 22

1

8.00 101.99 10 N/C

( ) 0.6

kk qE

r

= = = G

(b) ( )6

2 42 2 2 2

2

3.00 102.70 10 N/C

( ) 0.6 0.8

kk qE

r = = =

+G

(c) 41 2 20 cos 2.16 10 N/Cx x xE E E E i i= + = + =

G G G

(d) 51 2 2 sin ( ) 1.83 10 N/Cy y y iE E E E j E j j= + = + =

G G G

(e) 2 2 51.84 10 N/Cx yE E E= + =

G

SABIS Proprietary 6

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

Ey xx E

= = = BG 7. y 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 a direction opposite to the direction of the electric field.

(a) 19 170

. (300 )( 1.60 10 ) 4.8 10 Ne eFE F E e j jq

= = = =

( )13 2

13 2

5.27 10 m/s

5.27 10 m/s vertically downwards

ey e

F mgF ma F mg ma a jm

a

= = = = =

(b) ( )0 0 0cos cos0XX V t V t V t V t= = = =

x l

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

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

v0 E -

SABIS Proprietary 7

760

0.250 1.25 10 s2.00 10

Xt tV

= = =

(c) 2 21 1 02 2oy

y at V t at = + = +G G G

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/sy oV at V j

= + = + = G G 62.00 10 m/sx oxV V i= =

G G

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

yx y

x

VV V V

V

V

= + = = = =

G

B 8. (a) In a thundercloud there may be an electric charge of + 30.0 C near the top and 30.0

C 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 3 230 30

8.99 10 N(3 10 )

k q q kF

r= = =

G

(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]

Fe +y

mg

Solution: 0yF =

( )( )( )

0

277

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 gE

q

= =

= = =

GG

G

eF

GEG

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

71.02 10 N/CE j = G B 9. A 45.0-g insulating sphere carries a charge Q = - 40.0 C and is suspended by a silk thread

from 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?

EG

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

SABIS Proprietary 8

( )( )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 mgT mg F

T E q

TT

= + = = = = = =

G

T +y

Fe Q = 40.0 C E

(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]

EG

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

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 eF F T E q j mg j j

j jE

E j

E j

= = = ++ =

= =

G

GGG

T

+y

Fe

Q = 40.0 C

SABIS Proprietary 9

Chapter 22: Gausss Law

BG 1. G (a) State Gausss 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 = GG

(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]

. . cosE E dA E dA E dA = = = GGSolution:

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. EG

is radially outward and perpendicular to the Gaussian surface at every point.

0 0 02

20 0 0

. . cos

0;cos 1 . .44

enc enc encE

enc enc enc

q q qE dA E dA E dA

q q qE dA E R ER

= = = =

= = = = =

GG

G

SABIS Proprietary 10

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 = since E E

G points perpendicularly away from the sheet at every point.

1 2 3 3

1

2

1 20

0 0

.

but ; 0 since 0

. .

& . .

2

.22

E

E

encE

E dA

E

E dA E dA E dA EA

E dA E dA E dA EA

qEA

AEA E

== + + = =

= = = == = = =

= + = =

= =

GG

GGGG

+ + + + + + + + + + +

2

3

1

EG

EGEG

EG

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

3 represents the lateral area of the cylinder.

Surface charge density = qA

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

plates? [chap 22-page 765]

Solution: 0

E =

(d) Q q


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

[chap 22-page 767]

Solution:

charge on the inner surface of the conductor?

Induced charge on inner surface = +2.0 nC

3.

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.

he tric f [chap 22-page 759]

B

Four closed surfaces S1 through S4, together

(c) Find the electric flux through surface S3.

(d) Find t elec .

lux through surface S4

Solution: (a) 10 0 0

4encq q Q b1 = = =

(b) 1 2 3 520 0 0 0

4 2encq q q q q Q Q Q Q 6Q b + + + + = = = =

(c) 3 4 50 0

encq q q3

0 0

2 3 6q Q Q Q Q b + + = = = =

(d) 540 0 0

encq q Q b = = =

A point charge Q = 7.00 C is located at t B 4. (a) he centre of a cube of side L = 0.150 m. Six

other point charges, each carrying a charge q = -1.50 C, are positioned symmetrically f the cube?

ittiaround Q, inside the cube. What is t o(perm vity of free space = 8.85 x 10-1

he electric flux through one face 2)

S!

S4

S2 3 S

q2 q4 q1 q3

q5


Solution: Total E through all 6 faces: 40 0

22.6 10encE b6q Q q

+= = = E through one face = 46 3.77 10E b =

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

x ittiv

[chap 22-page 759]

S

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

ity of free space = 8.85 x 10-12) 4.80 (perm

olution: ( )4 70 0. 4.8 10 4.25 10encE enc Eq q C = = = = ( )0

5. A thin square conducting plate 35.0 cm on a side lies in the xy plane. A total charge of 10-12).

te?

[chap 22-page 764]

Solution:

B1.96 x 10-8 C is placed on the plate. (permittivity of free space = 8.85 x

(a) Find the charge density on the plate.

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

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

(a) 81.96 10q

= = = 7 21.6 10 C/m (0.35)(0.35)A

(b) for a thin plate (sheet): ( )7 31.6 10 9.04 10 N/CE = = =

0 02 2

G 6. A conducting spherical shell of radius 13.0 cm carries a net charge of 5.07 C uniformly

G

at is the magnitu e of the eld at a point very far away from the shell?

[chap 22-page 762]

Solution:

B

distributed on its surface. (a) Find the magnitude of the electric field just outside the shell.

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

(c) Wh d electric fi

6

62 2

5.07q (a) 0 0r

10 2.7 10 N/C4 4 (0.13) .

encE

= = =

(b) E = 0 since qenc = 0

(c) 20

;as , 04

encqE r Er =


B A conducting spherical 7. 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 the

(a) the inner surface. (b) the outer

[chap 22-page 767]

Solution:

shell, determine the surface charge density on:

surface.

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 the shell is +13.00 mC (since 13 + (4) = 9)

(a) 6 6

4inner2 2

4.00 10 4.00 10 1.27 10 C/4 4 (0.05)

qA r

= = = = 2 inner

inner

m

(b)

6 6

4 2outerouter 2 2

outer

13.00 10 13.00 10 2.87 10 C/m4 4 (0.06)

qA r

= = = = B 8.

Two infinite, nonconducting sheets of charge ar

(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:

A B C

1 2 e parallel to each other as shown in the

figure.

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

+++++++++++

+ + + + + + + + + + +

E1 E2 E2 E1 A C B E2 E1


Plate 1 Plate 21 2

* Note: 1) aused by a positively charged plate always

ays

(a) At A we m 2

the electric field cpoints away from it.

2) the electric field caused by a negativ ed plate alwpoints towards it.

ely charg

ust consider the two electric fields 1 aE End G G

caused by plates 1 and 2 respectively:

47

total 1 20 0 0 0

6.0 10 N/C2 2

E E E (

(5.31 10 ) = + = + = = =

6.0E = G )7 10 N/CiA she 1 2 = 5.31 x 10 C/m . What is

the elect

S t

(b) Both ets have positive uniform cha es -4 2ric field at point B?

rge densiti =

olu ion: At B, the two electric fields 1 2and E EG G

are equal in magnitude and opposite in direction; they canc 0 N/CBE =el so: .

charge densities 1 = 2 . What is the elect

Solution:

(c) Both sheets have positive uniform -4 2

ric field at point C? = 5.31 x 10 C/m

At C, we must 2E consider the two electric fields 1E and G G

2 respectively: caused by plates 1

and45.31 10 6.0 10E E E

= + = + = = = 7total 1 2 N/C2 2

0 0 0 0

( )76.0 10 N/CiCE = G [thus cou ld also be done using symmetry from part (a)]

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:

(d) The sheet on the left has a uniform surface

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

+++++++++++

Plate 1 Plate 21 2

+x

E2 E1 B C E2 E1 A E1 E2


1 2 (d) At A, the two electric fields Eand EG G

(caused by plates 1 & 2

respectively) are equal in magnitude 02

= but o pposite in

direction; they cancel so: 0 N/CAE =

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 2

(e) uniform charge density 2 = -5.31 x 10-4 C/m . What is the electric

field at point B?

Solution: der the two electric fields At B, we must consi 1 2and E EG G

caused by plates 1 and 2 respectively:

48

al 1 20 0

5.3 102

+ = = =tot0 0

1 1.2 10 N/CE E E = + =

2

( ) 81.2 10 N/CiBE = G (f)

uniform charge density 2 = -5.31 x 10-4 C/m2. What is the electric [chap 22-page 765]

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 afield at point C?

Solution: By a sim ilar argument to that in (d), we can show that: 0 N/CCE =

B 9.

s 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.

[chap 22-page 755]

[chap 22-page 760]

z x

y

E1 E2

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 magnitude

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

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


1 111. . .

leftEE dA E dA E dA E A = = = = GG

( )( )Solution: ) (a

( )( )( )23 2 35 10 5 10 0.25r = = b

2( ) 312.5 Nm /C 312.5E left = = lefE

This flux is into the surface, therefore: t

b 312.5 =

2 222. . .

rightEE dA E dA E dA E A = = = = GG

( ) (b)

( ) ( )( )( )23 2 31.5 10 1.5 10 0.25rb

2( ) 93.75 Nm /C 93.75E right = =

= =

rightE

This flux is out of the surface, therefore: b 93.75 =

(c) b

312.5 93.75 218.75total left rightE E E

= + = + = 9

0 00

. ( 218.75 )( ) 6.08 10 C (d) total

encE enq = totalE

q = = =

G 10.

arries -Q.

.

[chap 22-page 767]

Solution:

c

6.08nCencq = B

Three hollow, concentric spherical conductors are charged as follows: The inner sphere carries -2Q, the middle sphere carries +3Q and the outer sphere c

(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.

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 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


Electric Potential

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

[chap 23-page 784]

Solution:

Chapter 23:

BG 0

00

1 q.4

qUr=

(b)

ction and magnitude of the force exerted on a positive charge q placed in that field?

[chap 23-page 782]

Solution:

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 dire

direction: vertically upwards

magnitude of force: F = q.E

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

[chap 23-page 788]

Solution:

is 180 NC-1. What is the charge?

. 4 mV E x x= = 73.20kq rV xVq = = = 10V C

r k k=

2. G (a)

etween them is 3.0 cm. Find the magnitude of the electric field

[chap 23-page 796]

olution:

BG The potential difference between two oppositely charged parallel metal plates is 120 V and the distance bbetween the plates.

S 3120V 4.0 10 N/C0.02

Ed

= = =


(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

60

240 10 12.0V20 10

MNMN

WVq

= = =

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]

( )4 4. 7.50 10 (0.6) 4.5 10 VV E x = = = JG GSolution: (b) (c) ( )19 4 150. 1.602 10 ( 4.5 10 ) 7.21 10 JU q V = = =

. .k e q V =

(d)

2 2 401 1 ( 4.5 10 )2 2

mV mV q =

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

V =


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 is

located 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:

2.00 m

1.50 m q1 =3.00 C

q2 =-6.00 C

2 22.0 1.5+

P

(a) 6 6

3

2 20 0

1 1 3 10 6.0 10. 3.60 10 V4 4 1.50 1.5 2

iP i

i

qVr

= = + = +

(b) ( )( )6 3 2. 3.0 10 3.6 10 1.08 10 JPW q V = = = + 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] ( )( )6 30. 16.0 10 75 1.2 10 JU q V = = = Solution:

(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]

2

1m10 cm

33.6 10 V= Solution: 6(4.50 10 )(0.08 cmV Ed= =

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]


Solution: . .k e q V = ( ) ( )2 2 1901 1 1.602 10 1052 2mV mV =

5(1.42 10 ) m/sV = final initial=V V V

*Note: the proton will accelerate from +ve to ve: 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 of

3.00 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 20 0 0 0

31 2 2 190 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.

qB 2 q3 OC

qA 1 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]


Solution:

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

(a) 1 1

651 1

30 0

1 1 4.00 10. . 7.85 10 V4 4 2.1 10

q qkq qV Vr AO

= = = =

(b) 2 2

662 2

0 0

1 1 4.00 10. . 1.80 10 V4 4 0.02q q

kq qV Vr OB

= = = =

2 3

61.80 10 Vq qV V= =

1 2 3Total q q qV V V V = + +

( )

(c) By symmetry:

( ) ( )5 6 6 67.85 10 1.80 10 + 1.80 10 = 2.82 10 V= + (d) 0 0 31 2 0

0 0 1 2 3

. .4 4

itotali

i

q q q qq qU q Vr 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 22

(9 10 )(7.00 10 )( 2.00 10 ) 3.5 10(36.0 10 )

kQ QU Jr

= = =

(b) 9

91 21 2 1 2 2

2 2 (9 10 )( ) (7.00 2.00) 1036.0 10

2 2

kQ kQ kV V V Q Qr r r

= + = + = + =

22.50 10 250V V V = =


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: 0kAACd d

= = C: capacitance; A: area of the plates; d: distance between the plates k: 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: 0kAACd 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 10 6.5 10 F24.0

QCV

= = =

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 34.00 1.0 10 F3.54 10AC

d

= = =

G (b) Find the equivalent capacitance of the following combination. 2.0 F 4.0 F 9.0 F


[chap 24-page 821]

Solution: 66 6 61 2 3

1 1 1 1 1 1 1 1.16 10 F2.0 10 4.0 10 9.0 10 eqeq

CC 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 is 400 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: ( )262 6612.5 10 3.125 10 J2 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.


(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 10k AAC

d d

= = = =

( )( )7 3 4max max 2.40 10 1.40 10 3.36 10 VV E d = = = (b) ( )( )22 11 4 2max max1 1. 2.32 10 3.36 10 1.31 10 J2 2U C V = = = BG 6. (a) When a potential difference of 175 V is applied to the plates of a parallel-plate

capacitor, 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 m35.0 10 10

Q AC A VAV V dA V dC

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 by 5.00 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)


0 cos18 (0.3)(9.8) 3.09 Ny yF T mg T T = = = =

0 sin18x e xF F T T = = = (c)

790

sin18 4.78 10 N/C20.0 10

eF TE Eq

= = = = G

( )

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

C3

C2 C1

V

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: 646 6

4 1 2

1 1 1 1 1 2.4 10 F4.00 10 6.00 10

CC 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 = =


( ) ( )( )6 61 14.00 10 6.00 10 15V V = 1 9.0VV =

(d) V3 = 15 V

6 533 3 3 33

(3.60 10 )(9) 5.4 10 CQC Q C VV

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

supply 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 then

reconnected 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 61 2

1 1 1 1 14.00 10 16.0 10eqC C C

= + = + 63.2 10 FeqC

= 6 4

1 2 . (3.2 10 )(50.0) 1.6 10 Ceq eqQ Q Q C V = = = = =

(b) In series (c)

Qtotal = Q1 + Q2 (charge is conserved)

* since C1 & C2 have two points in common, they are in parallel and V1 = V2 = V


( ) ( )4 6 60 1 2. 3.2 10 4 10 16.00 10eqC V C V C V V = + = + 16.0VV =

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

( )( )6 52 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 2U C V = = = (b) 0

0

. 84.7k A d cC kd A

= = =

(c) 6maxmax max max 3400( ) 1.6 10 V/m

0.25 10VV E d E

d = = = =


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 through

the 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 AI 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 0 112 4 10

l lRA 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 = + = + = . . . 72 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

IR r= = = =+ +. .. . .

3 3 2 9 60 VV IR= = =. .


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 ItR 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 IRR r = = = = = = = =+ +. . .. . .

6 4 24 WP 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]


Solution: (a) 14 10 4 0 40 A1 5 3 3 1 2 4 10

IR

e = = = = + + + .. . . (b) 2 20 4 10 1 6 WP I R= = =. . (c) 10 0 4 4 0 WP Ie= = =. . (d) 14 0 4 5 6 WP I= = =. . B 6. A resistance thermometer, which measures temperature by measuring the change in

resistance 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 01R R T T= + 0 3

0

1 85 11 1 31 327 C40 3 8 10

RT TR

= + = + = .D

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 2150 2250 W 2 25 kW

10VPR

= = = = .

(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 ll A A R R

A A = = = =

' ' 1 1 1 20' 1 25 ' 4 4 16 16 16

R l A RRR l A

= = = = = = . B 8. (a) A 2.10-V potential difference is maintained across a 2.00-m length of tungsten wire of

resistivity 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 10 9 375 9 38 A5 6 10 2

V V VAI lR lA

= = = = = . . . ..


(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 10R R T T R T T = + = = + = + = . D

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

[chap 25-page 853]

Solution: '' ' ' 1 5'

A lV V lA l AA l

= = = = .

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

l l R l AR R RA A R l A = = = = = = . . . .

B 10.

E2 r2

E1

R

r1

A B

C D

In the above circuit : E1 = 14.0 V, r1 = 3.00 , E2 = 10.0 V, r2 = 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

IR += = = = + + .

(b) 1 1 14 2 4 3 6 80 VABV Ir= = =. .


(c) 2 2 10 2 4 2 5 20 VDCV Ir= = =. . (d) 2 22 4 10 57 6 WP I R= = =. .


Chapter 26: Direct-Current Circuits

B 1. (a) Three resistors of resistances R1 = 1.00 , R2 = 3.00 and R3 = 5.00 are connected

in 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 R1 = 1.00 , R2 = 3.00 and R3 = 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 15 0 652 1 3 5 15 23

RR 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 baI I= + = = . . .


(b) Applying Kirchhoffs 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 40 45 2 0 505 4 1 5 0 1 946 1 95 1 5

x x + + = = = . .. . . . ..

B 3. (a) A galvanometer has coil resistance RC and f.s.d. current Ifs. If it is to be converted to an ammeter 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 RI I R I R RI I

= =

(b) A galvanometer has coil resistance RC 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

VV I R R R RI

= + =

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

[chap 26-page 892]

Solution: fs Csha fs

3 48 5 1 5 in parallel100 3

I RRI I

= = = . .


(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 3fs

16 20 7980 2 10

VR RI

= = =

3s 7 98 10 in seriesR = .

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) 01t t

RC RCFq Q e i I e

= =

(ii) 0 0t t

RC RCq 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 sRC = = = .

(c) 012 1 00 A12

IR

= = = . (d) 2 12 24 0 CFQ C = = = . B 6. 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.

(a) Calculate the current at one time constant. (b) Calculate the charge at one time constant.

[chap 26-page 897]


Solution: (a) 0 01212 2 24 0 s 1 A12

tRCRC i I e I

R = = = = = = =.

11 0 368 Ai e = = .

(b) 1 2 12 24 Ct

RCF Fq Q e Q C = = = = (

) 124 1 15 2 Cq e = = . B 7.

R1 R2

R3

b c

a R4 Four resistors are connected as shown. R1 = 3.00 , R2 = 4.00 , R3 = 24.0 , R4 = 12.0 .

Vac = 22.5 V (a) What is the equivalent resistance between a and c ? (b) Find the current through R3.

[chap 26-page 882]

Solution: (a) 5 1 2 66 3 4

1 1 1 1 1 13 4 7 8 24 12 8

R R R RR R R

= + = + = = + = + = = eq 5 6 7 8 15 0 R R R= + = + = .

(b) eq

22 5 1 5 A15

ACVIR

= = =. .

66 3 3 33

1 5 8 0 500 A24bc

IRV IR I R IR

= = = = =. . B 8. (a) The currrent in a loop circuit that has a resistance R1 is 4.0 A. The current is reduced to

3.0 A when an additional resistor R2 = 2.5 is added in series with R1. What is the value of R1?

( ) ( )Solution: 2 21 1 2 1 2 2 1 2 2 1 1 2 2 2 11 2

I RI R I R R I R I R R I I I R RI I

= + = + = =

13 2 5 7 5 4 3

R = = . .


(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 2 2 5

I R rr R I = = = = + . ..

2 22 5 5 2 32 5 WP I R= = =. . . B 9. The current in a circuit is quadrupled by connecting a 600- resistor in parallel with the

resistance R of the circuit. Calculate the resistance R. [chap 26-page 883]

Solution: eqeq

1 1 1 600 600600 600 600

R RRR R R R

+= + = = +

I

4R I= eqR R 6004 R= 2400 600 1800 1 80 k600 R RR = + = = + .JJJJJJJJM BG 10. Three resistors each having resistance R = 3.0 are connected as in the figure. Each can

dissipate a maximum power of 48 W without being excessively heated. Find the maximum power the network can dissipate.

R

R

R

[chap 26-page 883]

Solution: R

R

R 1

2

3 I

I /2

I /2 48 W

3 1 2148 W 48 12 W4

P P P= = = =

1 2 3 12 12 48 72 WTP P P P= + + = + + =


BG 11. Initially, for the circuit shown, the switch S is open and the capacitor, of capacitance

20.0 F, has a voltage of 120 V, and R = 3.00 M. The switch is closed at time t = 0.

C R

S 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 CR CqV V iR q iRCC

= = = = = = .

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 40 0 120 120 34 4 V

t t tRC RC RCQqq Q e e v V e e e

C C = = = = = .


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 x

axis. 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:

37

A

B

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 N4 8 10 N,( )ve direction.

F qvBz

= = = =

. . .F .

D

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:

37

A

B

2cos cosBA B r = =

( )

2

2 34 1 5 10 cos53 1 7 10 Wb = = . .D


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 of 2.50 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

319

1 67 10 2 5 10 3 26 10 m 3 26 mm1 6 10 0 8

mv mvqvB rr 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 BF F q= E q=6

6 65 10 6 25 10 6 3 10 m/s0 8

EvB vB

= = = . ..

BG 3. (a) A wire carries a steady current of 2.5 A. A straight section of the wire is 0.60 m long

and 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:

x

y

I B

( ) ( ) ( ) 2 5 0 6 2 4 3 6 NI= = = 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:

60

A

B

90 60 30 = =D D D


4sin 60 5 36 10 4 sin30 6 78584 6 8 NmNIAB = = = . .D BG 4. (a) The Earths magnetic field at a certain place is 3.2 x 10-5 T at an angle of 30 with the

horizontal. Calculate the flux through the horizontal ceiling of a rectangular room 6.0 m by 8.0 m.

[chap 27-page 924]

Solution:

30

A B

5 4 4

90 30 60cos 3 2 10 48 cos60 7 68 10 7 7 10 WbBA

= == = = = . . .

D D D

D

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 at

an angle of 37 with respect to a magnetic field of magnitude 3.5 x 10-3 T. What is the speed of the electron ?

[chap 27-page 919]

Solution: 15

719 3

6 7 10sin 1 988 10sin 1 6 10 3 5 10 sin 37FF qvB v

qB

= = = = . .

. . D

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 the

magnetic flux through this circle due to a uniform magnetic field of magnitude 0.60 T in the +z direction.

[chap 27-page 924]

Solution:

y

z

A

B

x


( ) 22 3cos 0 6 5 10 1 4 7 10 WbBA = = = . . 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

25 255

0 5 1 6 10 0 52 1 07 10 1 1 10 kg3 9 10

mv rqBr mqB 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: sinF IlB = 0 21 30 A0 35 0 02

FIlB

= = =.

. .

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

[chap 27-page 933]

Solution:

East B

F

mg

I

West

3

2

0 6 10 9 80 6 g/cm10 3

0 196 T, West

m mg m gF mg IlB mg Bl Il l I

= = = = = = =

. ..

B .

B 8. Alpha particles (charge =+2e, mass = 6.68 x 10-27 kg) are accelerated in a cyclotron to a

final 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 final

orbit? [chap 27-page 926]


Solution: (a) 2 2r mv rqB rT r v Tv qB m = = = =

r2 mqBqB

m

=

277

19

2 6 68 10 2 186 10 s 0 2186 s 0 219 s2 1 6 10 0 6

T

= = = . . . .

. .

(b) ( )2 2

2 14 14 222 7

4 4 0 8 6 609 10 6 61 10 m/s2 186 10

a r rT

= = = =

. . .

.

B 9. A uniform magnetic field of magnitude 0.60 T in the positive z-direction is present in a

region 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:

i

j

v0

k

B

FB

FE

E

( )Right-hand rule 0 along + along 0 along 0y 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 a

rectangular 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 the magnitude of the magnetic force acting on this segment of wire.

[chap 27-page 933]

Solution:

240

180 l

2 2240 180 300 cml = + =


sinI 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. It

is 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:

( )2290 37 53

sin 250 4 5 1 25 10 0 4 sin 53

0 176 Nm

NIAB

= == = =

. . .

.

D D D

D

37

A

B

(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 6

3 5 10 4 10 0 0373 T7 5 10 5 10

mv mvr BqB rq

= = = = . ..


Chapter 28: Sources of Magnetic Field

BG 1. (a) What is the constant current flowing in a straight wire required to give a magnetic field

of 2.0 x 10-5 T at 2.5 cm from the wire? [chap 28-page 963]

Solution: 2 5

07

0

2 2 2 5 10 2 10 2 5 A2 4 10

I rBB Ir

= = = =. .

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:

B

7

30 200 4 10 2 5 2 5 10 T, page and out of the page.2 2 0 04

N IBa

= = = . .

.

B 2.

S

N

(1) (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

IIFl r

= = = ..


(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:

S N B

7

300

4 10 800 6 8 0 10 T, from south to north.0 75

NIB nIl

= = = = ..

B 3. (a) What is the magnitude of the magnetic field produced by a straight long wire carrying

a 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

IBr

= = = = .. .

(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

507

0

' 2 2 3 5 10' 6 102 4 10 2 1

IIF r FIl 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 at

its centre a magnetic field of magnitude 3.8 x 10-3 T. What is its diameter?

[chap 28-page 968]

Solution: 7

0 03

600 4 10 2 52 0 496 m 50 cm2 3 8 10

N I N IB a Da 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 70

0 0603 2999 turns/m4 10 16

BB nI nI

= = = = .


B 5. (a) y

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

Ia OP Ba

= = = . D

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 2 107 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 12B B B B= + =

x I

P

O

x x

y

O I I

d d

P

2d


7

501 2

2 4 10 7 3 5 102 2 0 08

IB B Bd

= = = = ..( )

B1 53 5 10 T= B . j

( )

B2

(b) 0 0 02 1 1 22 3 6 2I I IB B B B Bd d d

= = = =

B2

B1

0 1 212 3

IBd

= = 0

3 2I 0

75

34 10 7 1 17 10 T

3 0 08

IBdd

B

= = = ..

( )51 17 10 T= B . j B 7.

I2 A C

P

I1

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:

P

B

A

B2 B1

B


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

(a) 7

70 11

1

4 10 9 7 50 10 T, direction 2 2 2 4

IB CPr

= = = ..

(b) 7

60 22

2

4 10 16 1 00 10 T, direction A2 2 3 2

IB Pr

= = = ..

(c) ( ) ( )2 22 2 7 6 61 2 7 5 10 1 10 1 25 10 TB B B = + = + = . . (d)

71 1 21

62

7 5 10tan tan tan 0 75 36 91 10

BB

= = = =. . . D

BG 8. I1

I2

P

x

y

z

A coil consisting of 40.0 turns has a radius of 12.6 cm and carries a current I1 = 4.00 A as

shown in the figure. A long, straight conductor, which is in the plane of the coil, carries a current 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 4 7 98 10 T2 2 0 126

N IBa

= = = .. 47 98 10 T, in the positive directionz= 1B .

(b) 7

40 22

2

4 10 350 4 10 T2 2 0.175

IBr

= = =

44 00 10 T, in the negative directionz= 2B . ( ) 41 2et 7 98 4 10B B B = = . (c) N


4Net 3 98 10 T, in the positive directionz= B . ( ) 4 31 2Net 7 98 4 10 1 20 10 TB B B = + = + . . (d)

3Net 1 20 10 T, in the positive directionz= B . B 9.

x Bext

F

v e-

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

4net 19 5

1 6 10 2 50 10 T1 6 10 4 10

FF qvB Bqv

= = = =

. ..

Right-hand rule

B (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 external4 4 4current 2 5 10 1 5 10 4 00 10 T

B B B B B BB

= = + = + =

. . .

(c) 2 4

0 currentcurrent 7

0

2 2 6 28 10 4 10 40 0 A2 4 10

I a BIa

= = = =. .B


Capacitance and DielectricsDirect-Current Circuits

Sources of Magnetic Field

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BQ 1112 Level NS Core Physics BGT- Electricity