1. Three long, parallel conductors each carry a current of ...

a3

I2

a6

I

a4

I

a4

I BBB B

a6

I

a)3(2

I|B| B

a4

I

2

1

2a2

I

2

1|B| 45 cos |B| B

a4

I

2

1

2a2

I

2

1|B| 45 cos |B| B

:components-y

00a4

I

a4

I BBB B

0 B

a4

I

2

1

2a2

I

2

1|B| 45sin |B|- B

a4

I

2

1

2a2

I

2

1|B| 45sin |B| B

2a2

I|B||B|

:components-x

00003y2y1yy

0023y

002

o22y

001

o11y

003x2x1xx

3x

002

o22x

001

o11x

021

1. Three long, parallel conductors each carry a current of I. The figure below is an end view of

the conductors, with each current coming out of the page. Dtermine the magnitude and direction of

the magnetic field at the following points.

(a) point A

Directions of B1 (magnetic field due to current #1), B2, and B3 are determined by right hand

rule, indicated in the figure above.

B1

#2

#3

B2

#1

B3

a 2

x

y

45o 45

o

Downward :Direction

a3

I2 Magnitude 0

Downward :Direction

a4

I Magnitude

a4

I

a4

I00 BBB B

a4

I

a)2(2

I|B| B

0 B

0 B

:components-y

002a2

I

2a2

I BBB B

0 B2a2

I |B|- B

2a2

I|B| B

2a2

I|B||B|

:components-x

0

003x2x1xx

0033y

2y

1y

003x2x1xx

3x

022x

011x

021

(b) point B


rule, indicated in the figure above

B1

#2

#3B2

#1

B3

a 2

x

y

a 2

direction No :Direction

0 Magnitude

0a2

I

a4

I

a4

I BBB B

a2

I|B| B

a4

I

2

1

2a2

I

2

1|B| 45sin |B|- B

a4

I

2

1

2a2

I

2

1|B| sin45 |B|- B

:components-y

00a4

I

a4

I BBB B

0 B

a4

I

2

1

2a2

I

2

1|B| 45 cos |B|- B

a4

I

2

1

2a2

I

2

1|B| 45 cos |B| B

2a2

I|B||B|

:components-x

0003y2y1yy

023y

002

o22y

001

o11y

003x2x1xx

3x

002

o22x

001

o11x

021

(c) point C


rule, indicated in the figure above

45o 45

o

B1

#2

#3

B2

#1

B3x

y

2121

210

2102010

2y1yy

2022y

1011y

20202

10101

2x1xx

2x

1x

II if downward , II if Upward:Direction

|I-I| d

Magnitude

)I-(I dd

I

d

I BB B

d

I |B|- B

d

I |B| B

d

I

2

d2

I|B| and

d

I

2

d2

I|B|

:components-y

0BB B

0 B

0 B

:components-x

2. Two long, parallel wires carry currents of I1 and I2 in the directions indicated in the figure

below. Take the positive x direction to be to the right.

(a) Find the magnitude and direction of the magnetic field at a point midway between the wires.

Directions of B1 (magnetic field due to current #1)and B2 are determined by right hand rule,

indicated in the figure above

B1

x

y

B2

21

11-o

210

10

1-o

y

y1-o

2221

21

0

21

221

021

0221

0

10102y1yy

2y

1010o11y

2102010

2x1xx

2022x

1010o1011x

202

101

I2I

I tan 180

)I2I(d4

d4

I

tan 180 |B|

|B| tan 180

I4II4I2d4

I

I)I2I(d4

I ]I

d4[)]I2I(

d4[ Magnitude

d4

I0

d4

I BB B

0 B

d4

I

2

1

2d2

I45sin |B| B

:components-y

)I2I(d4d2

I

d4

IBB B

d2

I |B| B

d4

I

2

1

2d2

I45 cos

2d2

I|B| B

:components-x

d2

I|B| and

2d2

I|B|

d 2

21 BB B

(b) Find the magnitude and direction of the magnetic field at point P, located d above the

wire carrying current I2.

Directions of B1 (magnetic field due to current #1)and B2 are determined by right hand rule,

indicated in the figure above

B2

B1

45o

x

y

3. Two long, parallel conductors, separated by d, carry currents in the same direction. The first

wire carries a current I1, and the second carries I2 = 8.00 A. (See figure below. Assume the

conductors lie in the plane of the page.)

(a) What is the magnetic field created by I1 at the location of I2?

By right hand rule, the magnetic field created by I1 at the location of I2 is in the + z direction.

k d 2

I B 10

(b) What is the force per unit length exerted by I1

on I2?

According to the right hand rule, the force is in the

‐y direction (see direction).

j d 2

II

L

F

j d 2

II L-

j 90sin d 2

I LI BLI F

210B

210

o102B

(c) What is the magnetic field created by I2 at the location of I1?

By right hand rule, the magnetic field created by I1 at the location of I2 is in the - z direction.

k d 2

I - B 20

(d) What is the force per length exerted by I2 on I1?

According to the right hand rule, the force is in the

+y direction (see direction).

j d 2

II

L

F

j d 2

II L

j 90sin d 2

I LI BLI F

210B

210

o201B

y

xz I1

I2 B

FB

(for part (a) and (b))

(for part (c) and (d))

y

xz I1

I2

B

FB

4. In the figure below, the current in the long, straight wire is I1 and the wire lies in the plane of the

rectangular loop, which carries a current I2 . Find the magnitude and direction of the net force

exerted on the loop by the magnetic field created by the wire.

Solution:

Note that dF1 and dF2 in the above figure have the same magnitude but opposite in direction,

because the magnetic field at these two infinitesimals are equal but the current I2 is in

opposite direction. As a result, the forces on line segments AB and DC will cancel out.

However, the forces on line segments AD and BC will not cancel out because the fields at

these two lines, B1 and B2, have different magnitude.

c 2

II

90sin c 2

I I BLI F

a)(c 2

IB and

c 2

IB

210

o10211 B

102

101

B1 B2

A B

C D

dF1

dF2

FB1 FB2

a)(c c 2

aII

a)(c c 2

c]-a)c[(II

a)(c 2

II -

c 2

II F F

a)(c 2

II

90sin a)(c 2

II BLI F

210

210

2102102 B1 B

210

o10222 B

5. Two long, parallel wires are attracted to each other by a force per unit length of f. One

wire carries a current of I1 to the right and is located along the line y = y0. The second wire

lies along the x axis. Determine the value of y for the line in the plane of the two wires along

which the total magnetic field is zero.

Solution:

2100

20

2100

20

10

2100

10

20

110

00

10

0

010

01

021

2

0

1

20

0

10

202

0

101

10

02

0

2102

0

210

0

1022

o22

0

101

I-fy2

fy2y

y)I-fy2(fy2

y)I

I-fy2 (

I

fy2

y)I-I

fy2 (y

I

fy2

(1)) (from )yy(I

fy2 y I

)yy(I y I

y

I

yy

I

y2

I

)yy(2

I 0 y at field B

page) of(out y2

I I toduey at field B

page) (into )yy(2

I I toduey at field B

(1)--- I

fy2I

y2

II

L

Ion acting Force f

Ly2

II

y2

ILI LBI 90sin LBIIon acting Force

y2

I I todue 0yat field B

y

y0 I1

I2

B=0

6. Three long wires (wire 1, wire 2, and wire 3) hang vertically. The distance between wire

1 and wire 2 is d. On the left, wire 1 carries an upward current of I1. To the right, wire 2

carries a downward current of I2. Wire 3 is located such that when it carries a certain current,

each wire experiences no net force. Find the following (a) the position of wire 3 (b) the

magnitude and direction of the current in wire 3.

Solution:

dI- I

Ix

)I- (I x dI

I x d)Ix(

I d

x I

d

dx (2)&(1)

(2)--- I d

xI

0 x

I -

d

I

0 x2

II -

d 2

II 0Ion Force

x2

II -

d 2

IIIon Force

(1)--- I d

dxI

0d

I

d)(x

I

0d 2

II

d)(x2

II 0Ion Force

d 2

II

d)(x2

II Ion Force

21

2

212

12

12

13

31

3202101

3202101

23

23

2103101

2103101

X is the position of I3 relative to I2. It can be positive or negative, depends on which one of

I1 and I2 is larger. If x is negative, I3 is at the left of I2. If x is positive, I3 is at the right of I2.

I1 I2 I3

d x

21

123

121

2313

21

2

I- I

III

I d

1d)

I- I

I(II

d

xI

(2) into Substitute

dI- I

Ix

I3 can be positive or negative, depends on which one of I1 and I2 is larger. If I3 is negative, it

is downward. If I3 is positive, it is upward.

7. The figure below is a cross-sectional view of a coaxial cable. The center conductor is

surrounded by a rubber layer, an outer conductor, and another rubber layer. In a particular

application, the current in the inner conductor is I1 out of the page and the current in the outer

conductor is I2 into the page. Assuming the distance d = 1.00 mm, answer the following.

(a) Determine the magnitude and direction of the magnetic field at point a. (b) Determine

the magnitude and direction of the magnetic field at point b.

Solution:

(a)

figure) (see :direction d 2

I B

I d 2B I sdB

:Law sAmpere'

10

100

B

(b)

21

21

210

2100

II if

II if figure) (see :direction

d 6

)I(I B

)I(I d)3(2B I sdB

:Law sAmpere'

B

8. The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner

radius of a and an outer radius of b. The toroid has N turns of large-diameter wire, each of

which carries a current of I.

(a) Find the magnitude of the magnetic field inside the toroid along the inner radius. (b) Find

the magnitude of the magnetic field inside the toroid along the outer radius.

(a) Apply Ampere’s law to the inner circle (red circle in the

figure at the right):

a2

NI B

NI a2B I sdB

0

0in0

(b) Apply Ampere’s law to the inner circle (red circle in the figure at the right):

b2

NI B

NI b 2B I sdB

0

0in0

9. Four long, parallel conductors carry equal currents of I. The figure below is an end view of

the conductors. The current direction is into the page at points A and B and out of the page at C

and D.

(a) Calculate the magnitude of the magnetic field at point P, located at the center of the square

of edge length .

(b) Determine the direction of the magnetic field at point P, located at the center of the

square of edge length .

Solution

(a) We have used right hand rule to determine the

direction of the magnetic field at the center of the square

due to each of the wires, as indicated in the figure.

All magnetic fields have the same magnitude, since each

wire carry the same magnitude of current and at equal

distance from the center of the square.

2

I

2

1

2

I- 45cosB- B

2

I

2

1

2

I 45sinB- B

2

I

2

1

2

I- 45cosB- B

2

I

2

1

2

I 45sinB B

2

I

2

1

2

I- 45cosB- B

2

I

2

1

2

I 45sinB B

2

I

2

1

2

I- 45cosB- B

2

I

2

1

2

I 45sinB- B

2

I

2

2

2

I BBBB

2

2 r 2 2r But

r 2

I BBBB

00DDy

00DDx

00CCy

00CCx

00BBy

00BBx

00AAy

00AAx

00DCBA

22

0DCBA

oo

oo

oo

oo

BC BD

BB BA x

y r

I2 |B|

I2)

2

I(4 BBBBB

0 BBBBB

0

00DyCyByAyy

DxCxBxAxx

(b) Direction of B is in the –y direction, i.e. downward.

B'

Br r'

1B' r'

Br

1 I

I

B' r' 2

Br 2 (1)/(2)

(2) -- I B' r' 2

-(1)- I Br 2

r' be distance Let the

0

0

0

0

d)d)(2R-(2R

2dI

d)d)(2R-(2R

d)-(2R-d)(2RI

)d2R

1

d-2R

1(

I

\d)(2R

I

d)-(2R

IB - B B

d)(2R

IB

)2

d(R2

IB I )B

2

d(R2

d)-(2R

IB

)2

d-(R2

IB I )B

2

d-(R2

right. at the figure in thenotation With

0

0

0

00XY

0X

0X0X

0Y

0Y0Y

10. The magnetic field r away from a long, straight wire carrying current I is B.

(a) At what distance is it B’? (b) At one instant, the two conductors in a long household

extension cord carry equal currents I in opposite directions. The two wires are distance d

apart. Find the magnetic field R away from the middle of the straight cord, in the plane of

the two wires. (c) At what distance is it one‐tenth as large? (d) The center wire in a coaxial

cable carries current I in one direction, and the sheath around it carries current I in the

opposite direction. What magnetic field does the cable create at points outside the cables?

(a)

(b)

dX Y

R

B=?

BY

BX

2

9d-40RR'

9d-40R2R'

9d-40R4R'

-d4R'10d-40R

10

1

-d4R'

d-4R

10

1

d)'(2R'-d)(2R

d)d)(2R-(2R

d)d)(2R-(2R

2dI

10

1

d)'(2R'-d)(2R

2dI

R' be distance Let the

22

22

222

2222

22

22

00

0 B

\0 Br 2

I)-(I Br 2 0

(c)

(d)

11. A certain superconducting magnet in the form of a solenoid of length L can generate a

magnetic field of B in its core when its coils carry a current of I. Find the number of turns in

the solenoid.

I

BL nL turnsofnumber Total

I

B n In B

0

0

0

0 loop square on the acting torqueTotal

0

0sin rF

Fr sideeach on Torqueo

12. A single-turn square loop of wire, L on each edge, carries a clockwise current of I’. The

loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the

solenoid. The solenoid has n turns per unit length and carries a clockwise current of I.

(a) Find the force on each side of the loop. (b) Find the magnitude of the torque acting on the

loop.

(a) By right hand rule, a clockwise current in the

solenoid will produce a magnetic field pointing into

the page, and as a result, the force acting on each

side of the square loop will be pointing away from

the center as shown in the figure. To calculate the

magnitude of this force:

LnII'LBI'90sin LBI'F

nI B

0o

0

(b)

I I’

F

F

F

F

13. It is desired to construct a solenoid that will have a resistance of R (at 20.0°C) and

produce a magnetic field of B at its center when it carries a current of I. The solenoid is to be

constructed from copper wire having a diameter of d. (a) If the radius of the solenoid is to be

R, determine the number of turns of wire needed. (b) If the radius of the solenoid is to be r,

determine the required length of the solenoid.

B8

Id

I

B8

d

n

N L

L

NnBut

I

B n nI B b)(

8

d

4

d

r2

1

r2N turnsofNumber

4

d

d

4R

)2

d(

R A

R

handbook) physicsin found be(can becopper ofy resistivit Let the a)(

02

0

2

0

0

22

2

22

r

Rr

R

r

RR

R

22B

2

X )kZjYiX()i ( BA

i A surface shaded theof vector Area

0 Z-ZY-YX-X

)kZjYiX()k (- )kZjYiX()k ( )kZjYiX()j(-

)kZjYiX()j ( )kZjYiX()i (- )kZjYiX()i (flux Total

k ,-k ,j ,-j ,i - ,i are facessix theof vector Area

222222

222

222

222222

14. A cube of edge length cm is positioned as shown in the figure below. A uniform

magnetic field given by = (X + Y + Z ) T exists throughout the region.

(a) Calculate the magnetic flux through the shaded face. (b) What is the total flux through the

six faces?

(a)

(b)

15. A solenoid of radius r and length has N turns and carries current I.

(a) Calculate the flux through the surface of a disk-shaped area of radius R that is positioned

perpendicular to and centered on the axis of the solenoid as in the figure (a) above. (b) Figure (b)

above shows an enlarged end view of the same solenoid. Calculate the flux through the tan

area, which is an annulus with an inner radius of a and outer radius of b .

IN

r rB

0B solenoid, theOutside

IN

nI B solenoid, theInside a)(

20

2

00

IN

)a-b( )a-b(B

solenoid theinside is area tan wholeThe

0B solenoid, theOutside

IN

nI B solenoid, theInside b)(

220

22

00

1. Three long, parallel conductors each carry a current of ...

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